Integrand size = 43, antiderivative size = 286 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {(3 a A-6 b B-8 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (6 a b B+a^2 (3 A-2 C)+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {a (3 A b+2 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d} \]
-1/3*b*(3*A-2*C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d-1/3*(3*A*a-6*B*b-8*C* a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1 /2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/ (a+b))^(1/2)+1/3*(6*B*a*b+a^2*(3*A-2*C)+2*b^2*(3*A+C))*(cos(1/2*d*x+1/2*c) ^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b) )^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+a*(3*A*b+ 2*B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2* d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+ b*cos(d*x+c))^(1/2)+A*(a+b*cos(d*x+c))^(3/2)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 12.29 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.52 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\frac {8 \left (3 A b^2+6 a b B+3 a^2 C+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (12 a^2 B+6 b^2 B+a b (15 A+8 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i (-3 a A+6 b B+8 a C) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} (3 a A+2 b C \cos (c+d x)) \tan (c+d x)}{12 d} \]
Integrate[(a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^2,x]
((8*(3*A*b^2 + 6*a*b*B + 3*a^2*C + b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b )]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(1 2*a^2*B + 6*b^2*B + a*b*(15*A + 8*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*E llipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + ((2*I )*(-3*a*A + 6*b*B + 8*a*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)]) + 4* Sqrt[a + b*Cos[c + d*x]]*(3*a*A + 2*b*C*Cos[c + d*x])*Tan[c + d*x])/(12*d)
Time = 2.56 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.488, Rules used = {3042, 3526, 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.
\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} \left (-b (3 A-2 C) \cos ^2(c+d x)+2 (b B+a C) \cos (c+d x)+3 A b+2 a B\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \sqrt {a+b \cos (c+d x)} \left (-b (3 A-2 C) \cos ^2(c+d x)+2 (b B+a C) \cos (c+d x)+3 A b+2 a B\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (-b (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 (b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b+2 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {\left (-b (3 a A-6 b B-8 a C) \cos ^2(c+d x)+2 \left (3 C a^2+6 b B a+3 A b^2+b^2 C\right ) \cos (c+d x)+3 a (3 A b+2 a B)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {\left (-b (3 a A-6 b B-8 a C) \cos ^2(c+d x)+2 \left (3 C a^2+6 b B a+3 A b^2+b^2 C\right ) \cos (c+d x)+3 a (3 A b+2 a B)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {-b (3 a A-6 b B-8 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (3 C a^2+6 b B a+3 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a (3 A b+2 a B)}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-\frac {\int -\frac {\left (3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\left ((3 a A-8 a C-6 b B) \int \sqrt {a+b \cos (c+d x)}dx\right )\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {\left (3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-(3 a A-8 a C-6 b B) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-(3 a A-8 a C-6 b B) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {(3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {(3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 a b (3 A b+2 a B)+b \left ((3 A-2 C) a^2+6 b B a+2 b^2 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+3 a b (2 a B+3 A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a b (2 a B+3 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+3 a b (2 a B+3 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+3 a b (2 a B+3 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a b (2 a B+3 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a b (2 a B+3 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a b (2 a B+3 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {2 b \left (a^2 (3 A-2 C)+6 a b B+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {6 a b (2 a B+3 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 (3 a A-8 a C-6 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
(((-2*(3*a*A - 6*b*B - 8*a*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*(6*a*b* B + a^2*(3*A - 2*C) + 2*b^2*(3*A + 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]]) + (6*a *b*(3*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*(3*A - 2* C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/2 + (A*(a + b*Cos[c + d*x ])^(3/2)*Tan[c + d*x])/d
3.11.25.3.1 Defintions of rubi rules used
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. \(1440\) vs. \(2(351)=702\).
Time = 6.98 (sec) , antiderivative size = 1441, normalized size of antiderivative = 5.04
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1441\) |
default | \(\text {Expression too large to display}\) | \(1635\) |
int((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, method=_RETURNVERBOSE)
-A*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*a*b*cos( 1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+(-2*a^2-2*a*b)*sin(1/2*d*x+1/2*c)^2*co s(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)*(si n(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^2+2*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^2-EllipticE(cos (1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+b*EllipticE(cos(1/2*d*x+1/2*c),(-2 *b/(a-b))^(1/2))*a-3*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a *b)*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^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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-( 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)*b*Ellipt icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-3*a*b*(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)*EllipticPi(cos(1 /2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/(-1+2*cos(1/2*d*x+1/2*c)^2)/(-2*b*sin (1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2 *b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d-2*B*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*s in(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x...
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="fricas")
integral((C*b*cos(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2 + A*a + (B*a + A *b)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^2, x)
Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)*sec(d*x + c)^2, x)
\[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)*sec(d*x + c)^2, x)
Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]