Integrand size = 25, antiderivative size = 519 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\frac {\left (a^2+2 b^2\right ) e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{3/2} \left (-a^2+b^2\right )^{7/4} d}+\frac {\left (a^2+2 b^2\right ) e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{3/2} \left (-a^2+b^2\right )^{7/4} d}-\frac {a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right ) d \sqrt {e \cos (c+d x)}}+\frac {a \left (a^2+2 b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{8 b^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}+\frac {a \left (a^2+2 b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{8 b^2 \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac {a e \sqrt {e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
1/8*(a^2+2*b^2)*e^(3/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/ 4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(7/4)/d+1/8*(a^2+2*b^2)*e^(3/2)*arctanh(b^( 1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(7/ 4)/d-1/4*a*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(s in(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b^2/(a^2-b^2)/d/(e*cos(d*x+c)) ^(1/2)+1/8*a*(a^2+2*b^2)*e^2*(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*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x +c)^(1/2)/b^2/(a^2-b^2)/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2 )+1/8*a*(a^2+2*b^2)*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El lipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^( 1/2)/b^2/(a^2-b^2)/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-1/2 *e*(e*cos(d*x+c))^(1/2)/b/d/(a+b*sin(d*x+c))^2+1/4*a*e*(e*cos(d*x+c))^(1/2 )/b/(a^2-b^2)/d/(a+b*sin(d*x+c))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 20.48 (sec) , antiderivative size = 1211, normalized size of antiderivative = 2.33 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\frac {(e \cos (c+d x))^{3/2} \sec (c+d x) \left (-\frac {1}{2 b (a+b \sin (c+d x))^2}-\frac {a}{4 b \left (-a^2+b^2\right ) (a+b \sin (c+d x))}\right )}{d}-\frac {(e \cos (c+d x))^{3/2} \left (\frac {4 b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)}}{\sqrt {1-\cos ^2(c+d x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )-\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )}{\left (-a^2+b^2\right )^{3/4}}\right ) \sin (c+d x)}{\sqrt {1-\cos ^2(c+d x)} (a+b \sin (c+d x))}-\frac {2 a \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^2(c+d x)}}{\left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}+\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{\left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}\right )}{8 (a-b) b (a+b) d \cos ^{\frac {3}{2}}(c+d x)} \]
((e*Cos[c + d*x])^(3/2)*Sec[c + d*x]*(-1/2*1/(b*(a + b*Sin[c + d*x])^2) - a/(4*b*(-a^2 + b^2)*(a + b*Sin[c + d*x]))))/d - ((e*Cos[c + d*x])^(3/2)*(( 4*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d* x]])/(Sqrt[1 - Cos[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, C os[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^ 2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) - ((1/8 - I/ 8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2) ^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^( 1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos [c + d*x]] + I*b*Cos[c + d*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(- a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]]))/(-a^2 + b^2)^(3/ 4))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) - (2*a*( a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5 /4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]]* Sqrt[1 - Cos[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos [c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, - 1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (a^2 ...
Time = 2.04 (sec) , antiderivative size = 468, normalized size of antiderivative = 0.90, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3172, 3042, 3343, 27, 3042, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 218, 221, 3042, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle -\frac {e^2 \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}dx}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}dx}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3343 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\int \frac {2 b-a \sin (c+d x)}{2 \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{a^2-b^2}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\int \frac {2 b-a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\int \frac {2 b-a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{b}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}-\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (\frac {b e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b^2 \cos ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \cos (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (\frac {2 b e \int \frac {1}{b^2 e^4 \cos ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \cos (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \cos ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle -\frac {e^2 \left (-\frac {\frac {\left (a^2+2 b^2\right ) \left (\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}\right )}{b}-\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {a \sqrt {e \cos (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{4 b}-\frac {e \sqrt {e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}\) |
-1/2*(e*Sqrt[e*Cos[c + d*x]])/(b*d*(a + b*Sin[c + d*x])^2) - (e^2*(-1/2*(( -2*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(b*d*Sqrt[e*Cos[c + d*x ]]) + ((a^2 + 2*b^2)*((2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e]*Cos[c + d*x])/( -a^2 + b^2)^(1/4)]/(Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2)) - ArcTanh[(Sqrt[b] *Sqrt[e]*Cos[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*e ^(3/2))))/d + (a*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2] ), (c + d*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[ c + d*x]]) - (a*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]) , (c + d*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]])))/b)/(a^2 - b^2) - (a*Sqrt[e*Cos[c + d*x]])/((a^2 - b^2)*d*e*(a + b*Sin[c + d*x]))))/(4*b)
3.6.99.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q) Int[1/( Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f) Subst[ Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S imp[a/(2*q) Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / ; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1)) Int[(g*Cos[e + f*x])^p *(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ [a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* (x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int [(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b Int[(g*Cos[e + f*x])^p/( a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 10.64 (sec) , antiderivative size = 2986, normalized size of antiderivative = 5.75
(-4*e^2*b*(1/b^2*(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*e*cos(1/2*d*x+1/ 2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1 /2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/ b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^ (1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2 )/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/2*d* x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))) /(16*a^2-16*b^2)/e-1/64*(5*a^2-b^2)/b^2*(3*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e +(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2 *(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/ 4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+ 2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^( 1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c) ^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4)))*(4*cos( 1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)*2^(1/2)*(e^2*(a^2-b^2 )/b^2)^(1/4)+(8*a^2-8*b^2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2))/e/(a-b)^2/( a+b)^2/(4*cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)+1/128*a ^2/(a^2-b^2)^2/b^2*(21*(e^2*(a^2-b^2)/b^2)^(1/4)*(4*sin(1/2*d*x+1/2*c)^4*b ^2-4*sin(1/2*d*x+1/2*c)^2*b^2+a^2)^2*2^(1/2)*ln((2*e*cos(1/2*d*x+1/2*c)^2- e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+...
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]