Integrand size = 21, antiderivative size = 236 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {3}{8} a \left (a^2-12 b^2\right ) x+\frac {3 b \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac {a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d} \] Output:
3/8*a*(a^2-12*b^2)*x+3/2*b*(2*a^2-b^2)*arctanh(sin(d*x+c))/d-1/2*b*(17*a^2 -b^2)*sin(d*x+c)/d-1/8*a*(21*a^2-2*b^2)*cos(d*x+c)*sin(d*x+c)/d-1/4*(6*a^2 -b^2)*(b+a*cos(d*x+c))^2*sin(d*x+c)/b/d-1/4*(4*a^2-b^2)*(b+a*cos(d*x+c))^3 *sin(d*x+c)/b^2/d+a*(b+a*cos(d*x+c))^4*tan(d*x+c)/b^2/d+1/2*(b+a*cos(d*x+c ))^4*sec(d*x+c)*tan(d*x+c)/b/d
Leaf count is larger than twice the leaf count of optimal. \(696\) vs. \(2(236)=472\).
Time = 6.53 (sec) , antiderivative size = 696, normalized size of antiderivative = 2.95 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {3 a \left (a^2-12 b^2\right ) (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{8 d (b+a \cos (c+d x))^3}+\frac {3 \left (-2 a^2 b+b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}-\frac {3 \left (-2 a^2 b+b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b \left (-15 a^2+4 b^2\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d (b+a \cos (c+d x))^3}-\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (2 (c+d x))}{4 d (b+a \cos (c+d x))^3}+\frac {a^2 b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (3 (c+d x))}{4 d (b+a \cos (c+d x))^3}+\frac {a^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (4 (c+d x))}{32 d (b+a \cos (c+d x))^3} \] Input:
Integrate[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^4,x]
Output:
(3*a*(a^2 - 12*b^2)*(c + d*x)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(8*d* (b + a*Cos[c + d*x])^3) + (3*(-2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c + d* x])^3) - (3*(-2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c + d*x])^3) + (b^3*Cos [c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d* x])^3*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2])) - (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*C os[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3 *(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (b*(-15*a^2 + 4*b^2)*Cos[c + d*x ]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d*(b + a*Cos[c + d*x])^3) - (a *(a^2 - 3*b^2)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[2*(c + d*x)])/(4* d*(b + a*Cos[c + d*x])^3) + (a^2*b*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*S in[3*(c + d*x)])/(4*d*(b + a*Cos[c + d*x])^3) + (a^3*Cos[c + d*x]^3*(a + b *Sec[c + d*x])^3*Sin[4*(c + d*x)])/(32*d*(b + a*Cos[c + d*x])^3)
Time = 1.82 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.10, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4360, 25, 25, 3042, 3372, 25, 3042, 3528, 27, 3042, 3528, 3042, 3512, 3042, 3502, 27, 3042, 3214, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^4(c+d x) (a+b \sec (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \sin (c+d x) \tan ^3(c+d x) \left (-(-a \cos (c+d x)-b)^3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -(b+a \cos (c+d x))^3 \sin (c+d x) \tan ^3(c+d x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \sin (c+d x) \tan ^3(c+d x) (a \cos (c+d x)+b)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 3372 |
\(\displaystyle -\frac {\int -(b+a \cos (c+d x))^3 \left (-2 \left (4 a^2-b^2\right ) \cos ^2(c+d x)-3 a b \cos (c+d x)+3 \left (2 a^2-b^2\right )\right ) \sec (c+d x)dx}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int (b+a \cos (c+d x))^3 \left (-2 \left (4 a^2-b^2\right ) \cos ^2(c+d x)-3 a b \cos (c+d x)+3 \left (2 a^2-b^2\right )\right ) \sec (c+d x)dx}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (-2 \left (4 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 a b \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (2 a^2-b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {\frac {1}{4} \int 6 (b+a \cos (c+d x))^2 \left (-3 a \cos (c+d x) b^2-\left (6 a^2-b^2\right ) \cos ^2(c+d x) b+2 \left (2 a^2-b^2\right ) b\right ) \sec (c+d x)dx-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3}{2} \int (b+a \cos (c+d x))^2 \left (-3 a \cos (c+d x) b^2-\left (6 a^2-b^2\right ) \cos ^2(c+d x) b+2 \left (2 a^2-b^2\right ) b\right ) \sec (c+d x)dx-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-3 a \sin \left (c+d x+\frac {\pi }{2}\right ) b^2-\left (6 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b+2 \left (2 a^2-b^2\right ) b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \int (b+a \cos (c+d x)) \left (-13 a \cos (c+d x) b^3-\left (21 a^2-2 b^2\right ) \cos ^2(c+d x) b^2+6 \left (2 a^2-b^2\right ) b^2\right ) \sec (c+d x)dx-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-13 a \sin \left (c+d x+\frac {\pi }{2}\right ) b^3-\left (21 a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2+6 \left (2 a^2-b^2\right ) b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (-4 \left (17 a^2-b^2\right ) \cos ^2(c+d x) b^3+12 \left (2 a^2-b^2\right ) b^3+3 a \left (a^2-12 b^2\right ) \cos (c+d x) b^2\right ) \sec (c+d x)dx-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {-4 \left (17 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^3+12 \left (2 a^2-b^2\right ) b^3+3 a \left (a^2-12 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (4 \left (2 a^2-b^2\right ) b^3+a \left (a^2-12 b^2\right ) \cos (c+d x) b^2\right ) \sec (c+d x)dx-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (4 \left (2 a^2-b^2\right ) b^3+a \left (a^2-12 b^2\right ) \cos (c+d x) b^2\right ) \sec (c+d x)dx-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {4 \left (2 a^2-b^2\right ) b^3+a \left (a^2-12 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (4 b^3 \left (2 a^2-b^2\right ) \int \sec (c+d x)dx+a b^2 x \left (a^2-12 b^2\right )\right )-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (4 b^3 \left (2 a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a b^2 x \left (a^2-12 b^2\right )\right )-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {4 b^3 \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+a b^2 x \left (a^2-12 b^2\right )\right )-\frac {4 b^3 \left (17 a^2-b^2\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\right )-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{2 d}}{2 b^2}+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d}\) |
Input:
Int[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^4,x]
Output:
(-1/2*((4*a^2 - b^2)*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/d + (3*(-1/3*(b* (6*a^2 - b^2)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*(a*b^2*(21*a^ 2 - 2*b^2)*Cos[c + d*x]*Sin[c + d*x])/d + (3*(a*b^2*(a^2 - 12*b^2)*x + (4* b^3*(2*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/d) - (4*b^3*(17*a^2 - b^2)*Sin[c + d*x])/d)/2)/3))/2)/(2*b^2) + (a*(b + a*Cos[c + d*x])^4*Tan[c + d*x])/(b^ 2*d) + ((b + a*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*b*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(a + b* Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] + (-Si mp[b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x] )^(n + 2)/(a^2*d^2*f*(n + 1)*(n + 2))), x] - Simp[1/(a^2*d^2*(n + 1)*(n + 2 )) Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n]) && !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.94 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(194\) |
default | \(\frac {a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(194\) |
parts | \(\frac {a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}+\frac {3 a^{2} b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(202\) |
parallelrisch | \(\frac {-192 b \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+192 b \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 a d x \left (a^{2}-12 b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-15 a^{3}+240 a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (-104 a^{2} b +32 b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (-6 a^{3}+24 a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+8 \sin \left (5 d x +5 c \right ) a^{2} b +a^{3} \sin \left (6 d x +6 c \right )+\left (-112 a^{2} b +96 b^{3}\right ) \sin \left (d x +c \right )+24 a d x \left (a^{2}-12 b^{2}\right )}{64 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(240\) |
risch | \(\frac {3 a^{3} x}{8}-\frac {9 a \,b^{2} x}{2}-\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b}{8 d}+\frac {i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {15 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b}{8 d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} a^{2} b}{8 d}-\frac {i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{3}}{2 d}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a \,b^{2}}{8 d}-\frac {i {\mathrm e}^{3 i \left (d x +c \right )} a^{2} b}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{3}}{2 d}-\frac {i b^{2} \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}-6 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}\) | \(365\) |
norman | \(\frac {\left (\frac {3}{8} a^{3}-\frac {9}{2} a \,b^{2}\right ) x +\left (-\frac {3}{2} a^{3}+18 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a^{3}+\frac {9}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a^{3}+\frac {9}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a^{3}-9 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a^{3}-9 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a^{3}-\frac {9}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {3 \left (a^{3}-8 a^{2} b -12 a \,b^{2}+4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {3 \left (a^{3}+8 a^{2} b -12 a \,b^{2}-4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 a^{3}-56 a^{2} b -60 a \,b^{2}+28 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (5 a^{3}+56 a^{2} b -60 a \,b^{2}-28 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {\left (15 a^{3}-40 a^{2} b +12 a \,b^{2}-12 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {\left (15 a^{3}+40 a^{2} b +12 a \,b^{2}+12 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {3 b \left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {3 b \left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(484\) |
Input:
int((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(-1/4*(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)+3/8*d*x+3/8*c)+3*a ^2*b*(-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+3*a*b^2*(sin (d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2* c)+b^3*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2* ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.83 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {3 \, {\left (a^{3} - 12 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{2} b \cos \left (d x + c\right )^{4} + 24 \, a b^{2} \cos \left (d x + c\right ) - {\left (5 \, a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 4 \, b^{3} - 8 \, {\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="fricas")
Output:
1/8*(3*(a^3 - 12*a*b^2)*d*x*cos(d*x + c)^2 + 6*(2*a^2*b - b^3)*cos(d*x + c )^2*log(sin(d*x + c) + 1) - 6*(2*a^2*b - b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + (2*a^3*cos(d*x + c)^5 + 8*a^2*b*cos(d*x + c)^4 + 24*a*b^2*cos( d*x + c) - (5*a^3 - 12*a*b^2)*cos(d*x + c)^3 + 4*b^3 - 8*(4*a^2*b - b^3)*c os(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*sec(d*x+c))**3*sin(d*x+c)**4,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.78 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} b - 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} - 8 \, b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )}}{32 \, d} \] Input:
integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="maxima")
Output:
1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a^3 - 16*(2* sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin (d*x + c))*a^2*b - 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2 *tan(d*x + c))*a*b^2 - 8*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log( sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin(d*x + c)))/d
Time = 0.24 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.83 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="giac")
Output:
1/8*(3*(a^3 - 12*a*b^2)*(d*x + c) + 12*(2*a^2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(2*a^2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 8*(6*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*t an(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(3*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 8*b^3*tan(1/2*d*x + 1/2*c)^7 + 11*a^3*t an(1/2*d*x + 1/2*c)^5 - 104*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*a*b^2*tan(1/ 2*d*x + 1/2*c)^5 + 24*b^3*tan(1/2*d*x + 1/2*c)^5 - 11*a^3*tan(1/2*d*x + 1/ 2*c)^3 - 104*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^ 3 + 24*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^3*tan(1/2*d*x + 1/2*c) - 24*a^2*b* tan(1/2*d*x + 1/2*c) + 12*a*b^2*tan(1/2*d*x + 1/2*c) + 8*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 11.62 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}-\frac {3\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {5\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}-\frac {4\,a^2\,b\,\sin \left (c+d\,x\right )}{d}-\frac {9\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {a^2\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{d} \] Input:
int(sin(c + d*x)^4*(a + b/cos(c + d*x))^3,x)
Output:
(b^3*sin(c + d*x))/d + (3*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))) /(4*d) - (3*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a^3*cos (c + d*x)^3*sin(c + d*x))/(4*d) + (b^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) - (5*a^3*cos(c + d*x)*sin(c + d*x))/(8*d) - (4*a^2*b*sin(c + d*x))/d - (9* a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*a^2*b*atanh(sin( c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*a*b^2*cos(c + d*x)*sin(c + d*x) )/(2*d) + (3*a*b^2*sin(c + d*x))/(d*cos(c + d*x)) + (a^2*b*cos(c + d*x)^2* sin(c + d*x))/d
Time = 0.19 (sec) , antiderivative size = 545, normalized size of antiderivative = 2.31 \[ \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x)
Output:
( - 2*cos(c + d*x)**2*sin(c + d*x)**5*a**3 - cos(c + d*x)**2*sin(c + d*x)* *3*a**3 + 3*cos(c + d*x)**2*sin(c + d*x)*a**3 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + 12*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2*b**3 + 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* a**2*b - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 + 24*cos(c + d*x)* log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b - 12*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 24*cos(c + d*x)*log(tan((c + d*x )/2) + 1)*a**2*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**3 - 8*cos( c + d*x)*sin(c + d*x)**5*a**2*b - 16*cos(c + d*x)*sin(c + d*x)**3*a**2*b + 8*cos(c + d*x)*sin(c + d*x)**3*b**3 + 3*cos(c + d*x)*sin(c + d*x)**2*a**3 *d*x - 36*cos(c + d*x)*sin(c + d*x)**2*a*b**2*c - 36*cos(c + d*x)*sin(c + d*x)**2*a*b**2*d*x + 24*cos(c + d*x)*sin(c + d*x)*a**2*b - 12*cos(c + d*x) *sin(c + d*x)*b**3 - 3*cos(c + d*x)*a**3*d*x + 36*cos(c + d*x)*a*b**2*c + 36*cos(c + d*x)*a*b**2*d*x - 12*sin(c + d*x)**5*a*b**2 + 48*sin(c + d*x)** 3*a*b**2 - 36*sin(c + d*x)*a*b**2)/(8*cos(c + d*x)*d*(sin(c + d*x)**2 - 1) )