Integrand size = 25, antiderivative size = 459 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{9/2} d}-\frac {7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{9/2} d}+\frac {7 \left (5 a^2-3 b^2\right ) e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)}}-\frac {7 a^2 \left (a^2-b^2\right ) e^5 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {7 a^2 \left (a^2-b^2\right ) e^5 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))} \]
7/2*a*(-a^2+b^2)^(3/4)*e^(9/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b ^2)^(1/4)/e^(1/2))/b^(9/2)/d-7/2*a*(-a^2+b^2)^(3/4)*e^(9/2)*arctanh(b^(1/2 )*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d+7/15*e^3*(e*cos (d*x+c))^(3/2)*(5*a-3*b*sin(d*x+c))/b^3/d-e*(e*cos(d*x+c))^(7/2)/b/d/(a+b* sin(d*x+c))-7/2*a^2*(a^2-b^2)*e^5*(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))*co s(d*x+c)^(1/2)/b^5/d/(b-(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)-7/2*a^2*(a^ 2-b^2)*e^5*(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^5/d/(b +(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)+7/5*(5*a^2-3*b^2)*e^4*(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))* (e*cos(d*x+c))^(1/2)/b^4/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 20.59 (sec) , antiderivative size = 835, normalized size of antiderivative = 1.82 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {7 (e \cos (c+d x))^{9/2} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \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 )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (c+d x)}{\sqrt {1-\cos ^2(c+d x)} (a+b \sin (c+d x))}-\frac {\left (5 a^2-3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \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 )\right ) \sin ^2(c+d x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}\right )}{10 b^3 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {(e \cos (c+d x))^{9/2} \sec ^4(c+d x) \left (\frac {4 a \cos (c+d x)}{3 b^3}+\frac {a^2 \cos (c+d x)-b^2 \cos (c+d x)}{b^3 (a+b \sin (c+d x))}-\frac {\sin (2 (c+d x))}{5 b^2}\right )}{d} \]
(7*(e*Cos[c + d*x])^(9/2)*((-4*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((a*Ap pellF1[3/4, 1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2) ]*Cos[c + d*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(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]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[c + d*x])/(Sqr t[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) - ((5*a^2 - 3*b^2)*(a + b*Sqrt [1 - Cos[c + d*x]^2])*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[c + d*x]^ 2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2) + 3*Sqrt[2]*a*(a^ 2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b ^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2) ^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos [c + d*x]] + b*Cos[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))*Sin[c + d*x]^2)/(12*b^ (3/2)*(-a^2 + b^2)*(1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(10*b^3*d* Cos[c + d*x]^(9/2)) + ((e*Cos[c + d*x])^(9/2)*Sec[c + d*x]^4*((4*a*Cos[c + d*x])/(3*b^3) + (a^2*Cos[c + d*x] - b^2*Cos[c + d*x])/(b^3*(a + b*Sin[c + d*x])) - Sin[2*(c + d*x)]/(5*b^2)))/d
Time = 2.08 (sec) , antiderivative size = 446, normalized size of antiderivative = 0.97, 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, 3344, 27, 3042, 3346, 3042, 3121, 3042, 3119, 3180, 266, 827, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle -\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2} \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2} \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle -\frac {7 e^2 \left (\frac {2 e^2 \int -\frac {\sqrt {e \cos (c+d x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (c+d x)\right )}{2 (a+b \sin (c+d x))}dx}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {e \cos (c+d x)}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3180 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {b e \int \frac {\sqrt {e \cos (c+d x)}}{b^2 \cos ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2}d(e \cos (c+d x))}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b e \int \frac {e^2 \cos ^2(c+d x)}{b^2 e^4 \cos ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \cos (c+d x)}}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b e \left (\frac {\int \frac {1}{b e^2 \cos ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \cos (c+d x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}\right )}{b}\right )}{5 b^2}-\frac {2 e (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^2 d}\right )}{2 b}-\frac {e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\) |
-((e*(e*Cos[c + d*x])^(7/2))/(b*d*(a + b*Sin[c + d*x]))) - (7*e^2*(-1/5*(e ^2*((2*(5*a^2 - 3*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b* d*Sqrt[Cos[c + d*x]]) - (5*a*(a^2 - b^2)*((2*b*e*(ArcTan[(Sqrt[b]*Sqrt[e]* Cos[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e]) - ArcTanh[(Sqrt[b]*Sqrt[e]*Cos[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a ^2 + b^2)^(1/4)*Sqrt[e])))/d + (a*e*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e* Cos[c + d*x]]) + (a*e*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) ))/b))/b^2 - (2*e*(e*Cos[c + d*x])^(3/2)*(5*a - 3*b*Sin[c + d*x]))/(15*b^2 *d)))/(2*b)
3.6.85.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ )]), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[a*(g/(2*b)) Int[1/(S qrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Simp[a*(g/(2*b)) In t[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Simp[b*(g/f) Su bst[Int[Sqrt[x]/(g^2*(a^2 - b^2) + b^2*x^2), x], x, g*Cos[e + f*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[g*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 , 0] && 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 = 17.40 (sec) , antiderivative size = 2107, normalized size of antiderivative = 4.59
(8*e^5*a*b*(1/6/b^4*(-2*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*sin(1/2*d*x+1/ 2*c)^2+(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2))/e-1/8/b^6*(a^2-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)))+1/64/b^6*(16*(e^2*(a^2-b^2 )/b^2)^(1/4)*(cos(1/2*d*x+1/2*c)^2-1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2) *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)*e*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))))/(e^2*(a^2-b^2)/b^2)^(1/4)*(a-b)*(a+b)/e/(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*(e*(2*cos(1/2*d*x+1 /2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*e^5*(1/5/b^4/(-2*sin(1/2*d*x+1/2...
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]