Integrand size = 40, antiderivative size = 272 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (4 a b B-6 a^2 C-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 b^4 d}+\frac {2 a^2 \left (2 a^2 b B-3 b^3 B-3 a^3 C+4 a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac {\left (2 a^2 b B-b^3 B-3 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a b B-3 a^2 C+b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac {a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
-1/2*(4*B*a*b-6*C*a^2-C*b^2)*arctanh(sin(d*x+c))/b^4/d+2*a^2*(2*B*a^2*b-3* B*b^3-3*C*a^3+4*C*a*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2 ))/(a-b)^(3/2)/b^4/(a+b)^(3/2)/d+(2*B*a^2*b-B*b^3-3*C*a^3+2*C*a*b^2)*tan(d *x+c)/b^3/(a^2-b^2)/d-1/2*(2*B*a*b-3*C*a^2+C*b^2)*sec(d*x+c)*tan(d*x+c)/b^ 2/(a^2-b^2)/d+a*(B*b-C*a)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d *x+c))
Time = 6.97 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 a^2 \left (-2 a^2 b B+3 b^3 B+3 a^3 C-4 a b^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} \left (-a^2+b^2\right ) d}+\frac {\left (4 a b B-6 a^2 C-b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac {\left (-4 a b B+6 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac {C}{4 b^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {C}{4 b^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b B \sin \left (\frac {1}{2} (c+d x)\right )-2 a C \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b B \sin \left (\frac {1}{2} (c+d x)\right )-2 a C \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-a^3 b B \sin (c+d x)+a^4 C \sin (c+d x)}{b^3 (-a+b) (a+b) d (b+a \cos (c+d x))} \]
(-2*a^2*(-2*a^2*b*B + 3*b^3*B + 3*a^3*C - 4*a*b^2*C)*ArcTanh[((-a + b)*Tan [(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*(-a^2 + b^2)*d) + (( 4*a*b*B - 6*a^2*C - b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*b^ 4*d) + ((-4*a*b*B + 6*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]])/(2*b^4*d) + C/(4*b^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - C/( 4*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (b*B*Sin[(c + d*x)/2] - 2*a*C*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ( b*B*Sin[(c + d*x)/2] - 2*a*C*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (-(a^3*b*B*Sin[c + d*x]) + a^4*C*Sin[c + d*x])/(b^3*( -a + b)*(a + b)*d*(b + a*Cos[c + d*x]))
Time = 1.90 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.425, Rules used = {3042, 4560, 3042, 4517, 3042, 4580, 25, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4517 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (-3 C a^2+2 b B a+b^2 C\right ) \sec ^2(c+d x)\right )-b (b B-a C) \sec (c+d x)+2 a (b B-a C)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (3 C a^2-2 b B a-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (b B-a C)\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec (c+d x) \left (-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \sec ^2(c+d x)-b \left (-C a^2+2 b B a-b^2 C\right ) \sec (c+d x)+a \left (-3 C a^2+2 b B a+b^2 C\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \sec ^2(c+d x)-b \left (-C a^2+2 b B a-b^2 C\right ) \sec (c+d x)+a \left (-3 C a^2+2 b B a+b^2 C\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-2 \left (-3 C a^3+2 b B a^2+2 b^2 C a-b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-C a^2+2 b B a-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (-3 C a^2+2 b B a+b^2 C\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (-3 C a^2+2 b B a+b^2 C\right )+\left (a^2-b^2\right ) \left (-6 C a^2+4 b B a-b^2 C\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (-3 C a^2+2 b B a+b^2 C\right )+\left (a^2-b^2\right ) \left (-6 C a^2+4 b B a-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \int \sec (c+d x)dx}{b}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {-\frac {\left (-3 a^2 C+2 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 b d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 C+4 a b B-b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-3 a^3 C+2 a^2 b B+4 a b^2 C-3 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 \left (-3 a^3 C+2 a^2 b B+2 a b^2 C-b^3 B\right ) \tan (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}\) |
(a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])) + (-1/2*((2*a*b*B - 3*a^2*C + b^2*C)*Sec[c + d*x]*Tan[c + d*x])/( b*d) - ((((a^2 - b^2)*(4*a*b*B - 6*a^2*C - b^2*C)*ArcTanh[Sin[c + d*x]])/( b*d) - (4*a^2*(2*a^2*b*B - 3*b^3*B - 3*a^3*C + 4*a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2 *(2*a^2*b*B - b^3*B - 3*a^3*C + 2*a*b^2*C)*Tan[c + d*x])/(b*d))/(2*b))/(b* (a^2 - b^2))
3.9.1.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 0.76 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (4 B a b -6 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {2 a^{2} \left (\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 B \,a^{2} b -3 B \,b^{3}-3 a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-4 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}}{d}\) | \(330\) |
| default | \(\frac {\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (4 B a b -6 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {2 a^{2} \left (\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 B \,a^{2} b -3 B \,b^{3}-3 a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-4 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}}{d}\) | \(330\) |
| risch | \(\text {Expression too large to display}\) | \(1276\) |
1/d*(1/2*C/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(2*B*b-4*C*a-C*b)/b^3/(tan(1/2 *d*x+1/2*c)-1)+1/2*(4*B*a*b-6*C*a^2-C*b^2)/b^4*ln(tan(1/2*d*x+1/2*c)-1)-2* a^2/b^4*(a*b*(B*b-C*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2* a-tan(1/2*d*x+1/2*c)^2*b-a-b)-(2*B*a^2*b-3*B*b^3-3*C*a^3+4*C*a*b^2)/(a+b)/ (a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^( 1/2)))-1/2*C/b^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*B*b-4*C*a-C*b)/b^3/(tan(1 /2*d*x+1/2*c)+1)+1/2/b^4*(-4*B*a*b+6*C*a^2+C*b^2)*ln(tan(1/2*d*x+1/2*c)+1) )
Leaf count of result is larger than twice the leaf count of optimal. 643 vs. \(2 (260) = 520\).
Time = 11.64 (sec) , antiderivative size = 1343, normalized size of antiderivative = 4.94 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
[1/4*(2*((3*C*a^6 - 2*B*a^5*b - 4*C*a^4*b^2 + 3*B*a^3*b^3)*cos(d*x + c)^3 + (3*C*a^5*b - 2*B*a^4*b^2 - 4*C*a^3*b^3 + 3*B*a^2*b^4)*cos(d*x + c)^2)*sq rt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*s qrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d *x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + ((6*C*a^7 - 4*B*a^6*b - 11*C*a^5* b^2 + 8*B*a^4*b^3 + 4*C*a^3*b^4 - 4*B*a^2*b^5 + C*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 - 11*C*a^4*b^3 + 8*B*a^3*b^4 + 4*C*a^2*b^5 - 4*B* a*b^6 + C*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((6*C*a^7 - 4*B*a^6 *b - 11*C*a^5*b^2 + 8*B*a^4*b^3 + 4*C*a^3*b^4 - 4*B*a^2*b^5 + C*a*b^6)*cos (d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 - 11*C*a^4*b^3 + 8*B*a^3*b^4 + 4*C* a^2*b^5 - 4*B*a*b^6 + C*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(C *a^4*b^3 - 2*C*a^2*b^5 + C*b^7 - 2*(3*C*a^6*b - 2*B*a^5*b^2 - 5*C*a^4*b^3 + 3*B*a^3*b^4 + 2*C*a^2*b^5 - B*a*b^6)*cos(d*x + c)^2 - (3*C*a^5*b^2 - 2*B *a^4*b^3 - 6*C*a^3*b^4 + 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))* sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2), -1/4*(4*((3*C*a^6 - 2*B*a^5*b - 4*C*a ^4*b^2 + 3*B*a^3*b^3)*cos(d*x + c)^3 + (3*C*a^5*b - 2*B*a^4*b^2 - 4*C*a^3* b^3 + 3*B*a^2*b^4)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^ 2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - ((6*C*a^7 - 4*B*a^6* b - 11*C*a^5*b^2 + 8*B*a^4*b^3 + 4*C*a^3*b^4 - 4*B*a^2*b^5 + C*a*b^6)*c...
\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.34 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.41 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, C a^{5} - 2 \, B a^{4} b - 4 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {{\left (6 \, C a^{2} - 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {{\left (6 \, C a^{2} - 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {2 \, {\left (4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, d} \]
-1/2*(4*(3*C*a^5 - 2*B*a^4*b - 4*C*a^3*b^2 + 3*B*a^2*b^3)*(pi*floor(1/2*(d *x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*ta n(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - 4*(C*a^4*tan(1/2*d*x + 1/2*c) - B*a^3*b*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (6 *C*a^2 - 4*B*a*b + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*C*a^ 2 - 4*B*a*b + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*C*a*tan (1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 1/2 *c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan( 1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3))/d
Time = 28.30 (sec) , antiderivative size = 6677, normalized size of antiderivative = 24.55 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)^5*(6*C*a^4 - 2*B*b^4 + C*b^4 + 2*B*a^2*b^2 - 5*C*a^ 2*b^2 + 2*B*a*b^3 - 4*B*a^3*b + 3*C*a*b^3 - 3*C*a^3*b))/((a*b^3 - b^4)*(a + b)) + (2*tan(c/2 + (d*x)/2)^3*(C*b^4 - 6*C*a^4 + 3*C*a^2*b^2 - 2*B*a*b^3 + 4*B*a^3*b))/(b*(a*b^2 - b^3)*(a + b)) + (tan(c/2 + (d*x)/2)*(2*B*b^4 + 6*C*a^4 + C*b^4 - 2*B*a^2*b^2 - 5*C*a^2*b^2 + 2*B*a*b^3 - 4*B*a^3*b - 3*C* a*b^3 + 3*C*a^3*b))/(b^3*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)^ 2*(3*a + b) - tan(c/2 + (d*x)/2)^6*(a - b) + tan(c/2 + (d*x)/2)^4*(3*a - b ))) - (atan(((((8*tan(c/2 + (d*x)/2)*(72*C^2*a^10 + C^2*b^10 - 2*C^2*a*b^9 - 72*C^2*a^9*b + 16*B^2*a^2*b^8 - 32*B^2*a^3*b^7 + 20*B^2*a^4*b^6 + 64*B^ 2*a^5*b^5 - 64*B^2*a^6*b^4 - 32*B^2*a^7*b^3 + 32*B^2*a^8*b^2 + 11*C^2*a^2* b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b^6 - 26*C^2*a^5*b^5 + 17*C^2*a^6*b^4 + 120*C^2*a^7*b^3 - 120*C^2*a^8*b^2 - 8*B*C*a*b^9 - 96*B*C*a^9*b + 16*B*C*a^ 2*b^8 - 40*B*C*a^3*b^7 + 64*B*C*a^4*b^6 - 40*B*C*a^5*b^5 - 176*B*C*a^6*b^4 + 176*B*C*a^7*b^3 + 96*B*C*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (((8*(2*C*b^15 + 12*B*a^2*b^13 + 12*B*a^3*b^12 - 20*B*a^4*b^11 - 4*B*a^5*b ^10 + 8*B*a^6*b^9 + 6*C*a^2*b^13 - 16*C*a^3*b^12 - 14*C*a^4*b^11 + 28*C*a^ 5*b^10 + 6*C*a^6*b^9 - 12*C*a^7*b^8 - 8*B*a*b^14))/(a*b^11 + b^12 - a^2*b^ 10 - a^3*b^9) - (4*tan(c/2 + (d*x)/2)*(6*C*a^2 + C*b^2 - 4*B*a*b)*(8*a*b^1 3 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/(b^4* (a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(6*C*a^2 + C*b^2 - 4*B*a*b))/(2*b^4...