Integrand size = 43, antiderivative size = 357 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (20 a^2 b B+10 b^3 B+a^3 (15 A-16 C)+4 a b^2 (15 A+4 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 (5 A b+2 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d} \] Output:
1/15*(70*B*a*b-a^2*(15*A-46*C)+6*b^2*(5*A+3*C))*(a+b*cos(d*x+c))^(1/2)*Ell ipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/d/((a+b*cos(d*x+c))/(a+ b))^(1/2)+1/15*(20*B*a^2*b+10*B*b^3+a^3*(15*A-16*C)+4*a*b^2*(15*A+4*C))*(( a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b ))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+a^2*(5*A*b+2*B*a)*((a+b*cos(d*x+c))/(a+ b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b* cos(d*x+c))^(1/2)-1/15*b*(15*A*a-10*B*b-16*C*a)*(a+b*cos(d*x+c))^(1/2)*sin (d*x+c)/d-1/5*b*(5*A-2*C)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+A*(a+b*cos(d *x+c))^(5/2)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 13.33 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.41 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\frac {8 \left (45 a^2 b B+5 b^3 B+15 a^3 C+a b^2 (45 A+17 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (60 a^3 B+70 a b^2 B+6 b^3 (5 A+3 C)+a^2 b (135 A+46 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (70 a b B+6 b^2 (5 A+3 C)+a^2 (-15 A+46 C)\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+60 \sqrt {a+b \cos (c+d x)} \left (\frac {2}{15} b (5 b B+11 a C) \sin (c+d x)+\frac {1}{5} b^2 C \sin (2 (c+d x))+a^2 A \tan (c+d x)\right )}{60 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^2,x]
Output:
((8*(45*a^2*b*B + 5*b^3*B + 15*a^3*C + a*b^2*(45*A + 17*C))*Sqrt[(a + b*Co s[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos [c + d*x]] + (2*(60*a^3*B + 70*a*b^2*B + 6*b^3*(5*A + 3*C) + a^2*b*(135*A + 46*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2* b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + ((2*I)*(70*a*b*B + 6*b^2*(5*A + 3* C) + a^2*(-15*A + 46*C))*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-(( b*(1 + Cos[c + d*x]))/(a - b))]*Csc[c + d*x]*(-2*a*(a - b)*EllipticE[I*Arc Sinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*( -2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqr t[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*Sqrt[-(a + b)^(-1)]) + 60 *Sqrt[a + b*Cos[c + d*x]]*((2*b*(5*b*B + 11*a*C)*Sin[c + d*x])/15 + (b^2*C *Sin[2*(c + d*x)])/5 + a^2*A*Tan[c + d*x]))/(60*d)
Time = 3.31 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.05, number of steps used = 24, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.558, Rules used = {3042, 3526, 27, 3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}
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.
\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \int \frac {1}{2} (a+b \cos (c+d x))^{3/2} \left (-b (5 A-2 C) \cos ^2(c+d x)+2 (b B+a C) \cos (c+d x)+5 A b+2 a B\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int (a+b \cos (c+d x))^{3/2} \left (-b (5 A-2 C) \cos ^2(c+d x)+2 (b B+a C) \cos (c+d x)+5 A b+2 a B\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-b (5 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 (b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 A b+2 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} \left (-b (15 a A-10 b B-16 a C) \cos ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \cos (c+d x)+5 a (5 A b+2 a B)\right ) \sec (c+d x)dx-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \sqrt {a+b \cos (c+d x)} \left (-b (15 a A-10 b B-16 a C) \cos ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \cos (c+d x)+5 a (5 A b+2 a B)\right ) \sec (c+d x)dx-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (-b (15 a A-10 b B-16 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a (5 A b+2 a B)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {\left (15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \cos ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {\left (15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \cos ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx+\frac {\int \frac {\left (15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\int \frac {15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\int \frac {15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\int \frac {15 b (5 A b+2 a B) a^2+b \left ((15 A-16 C) a^3+20 b B a^2+4 b^2 (15 A+4 C) a+10 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 a^2 b (2 a B+5 A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\frac {15 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\frac {15 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\frac {30 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}\right )-\frac {2 b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d}\) |
Input:
Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^2,x]
Output:
((-2*b*(5*A - 2*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (((2*( 70*a*b*B - a^2*(15*A - 46*C) + 6*b^2*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]] *EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((2*b*(20*a^2*b*B + 10*b^3*B + a^3*(15*A - 16*C) + 4*a*b^2*(15*A + 4*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (30*a^2*b*(5*A*b + 2*a*B)*Sqrt[(a + b *Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt [a + b*Cos[c + d*x]]))/b)/3 - (2*b*(15*a*A - 10*b*B - 16*a*C)*Sqrt[a + b*C os[c + d*x]]*Sin[c + d*x])/(3*d))/5)/2 + (A*(a + b*Cos[c + d*x])^(5/2)*Tan [c + d*x])/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2151\) vs. \(2(345)=690\).
Time = 61.95 (sec) , antiderivative size = 2152, normalized size of antiderivative = 6.03
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2152\) |
default | \(\text {Expression too large to display}\) | \(2274\) |
Input:
int((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, method=_RETURNVERBOSE)
Output:
-A*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*a^2*b*co s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+(-2*a^3-2*a^2*b)*sin(1/2*d*x+1/2*c)^ 2*cos(1/2*d*x+1/2*c)-2*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2) *(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^( 1/2))*a^3+4*b^2*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-Ellipti cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+b*EllipticE(cos(1/2*d*x+1/2* c),(-2*b/(a-b))^(1/2))*a^2+2*b^2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b)) ^(1/2))*a-2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-5*b*Ellip ticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2)*sin(1/2*d*x+1/2*c)^2+( sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^ (1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+4*b^2*(sin(1/2* d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*El lipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1 /2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2* d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b )*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),( -2*b/(a-b))^(1/2))*a^2+2*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin( 1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a- b))^(1/2))*a-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c) ^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*...
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="fricas")
Output:
integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c ))*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^2, x)
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x +c)**2,x)
Output:
Timed out
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/ 2)*sec(d*x + c)^2, x)
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/ 2)*sec(d*x + c)^2, x)
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int(((a + b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^2,x)
Output:
int(((a + b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^2, x)
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=3 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) a^{2} b +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) b^{2} c +2 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) a b c +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) b^{3}+\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a^{2} c +3 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a \,b^{2}+\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}d x \right ) a^{3} \] Input:
int((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)
Output:
3*int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**2,x)*a**2*b + in t(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**4*sec(c + d*x)**2,x)*b**2*c + 2*i nt(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x)**2,x)*a*b*c + int (sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x)**2,x)*b**3 + int(sq rt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**2,x)*a**2*c + 3*int(s qrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**2,x)*a*b**2 + int(sq rt(cos(c + d*x)*b + a)*sec(c + d*x)**2,x)*a**3