\(\int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx\) [81]

Optimal result

Integrand size = 25, antiderivative size = 512 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \] Output:

5/8*(3*a^2-2*b^2)*e^(7/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^( 
1/4)/e^(1/2))/b^(7/2)/(-a^2+b^2)^(3/4)/d+5/8*(3*a^2-2*b^2)*e^(7/2)*arctanh 
(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(7/2)/(-a^2+b^2) 
^(3/4)/d-15/4*a*e^4*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+ 
c)^(1/2)/b^4/d/(e*sin(d*x+c))^(1/2)-5/8*a*(3*a^2-2*b^2)*e^4*EllipticPi(cos 
(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/ 
b^4/(a^2-b*(b-(-a^2+b^2)^(1/2)))/d/(e*sin(d*x+c))^(1/2)-5/8*a*(3*a^2-2*b^2 
)*e^4*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2 
))*sin(d*x+c)^(1/2)/b^4/(a^2-b*(b+(-a^2+b^2)^(1/2)))/d/(e*sin(d*x+c))^(1/2 
)+5/4*e^3*(3*a+2*b*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/b^3/d/(a+b*cos(d*x+c)) 
+1/2*e*(e*sin(d*x+c))^(5/2)/b/d/(a+b*cos(d*x+c))^2
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 18.28 (sec) , antiderivative size = 1954, normalized size of antiderivative = 3.82 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

(((-a^2 + b^2)/(2*b^3*(a + b*Cos[c + d*x])^2) + (9*a)/(4*b^3*(a + b*Cos[c 
+ d*x])))*Csc[c + d*x]^3*(e*Sin[c + d*x])^(7/2))/d - ((e*Sin[c + d*x])^(7/ 
2)*((14*a*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d*x]^2])*((a*(-2*ArcTan[1 
 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + 
(Sqrt[2]*Sqrt[b]*Sqrt[Sin[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[Sin[c + d*x]] + b*Sin[c + d*x 
]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + 
d*x]] + b*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(a^ 
2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/ 
(-a^2 + b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((-5*(a^2 - b^2 
)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + 
 b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + 
 d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d* 
x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 + 
 Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (12*b*C 
os[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(2*Arc 
Tan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTa 
n[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[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[Sin[c + d*x]] + I*b* 
Sin[c + d*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/...
 

Rubi [A] (warning: unable to verify)

Time = 2.16 (sec) , antiderivative size = 483, normalized size of antiderivative = 0.94, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 3172, 25, 3042, 3342, 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.

\(\displaystyle \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3172

\(\displaystyle \frac {5 e^2 \int -\frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2}dx}{4 b}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2}dx}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 b}\)

\(\Big \downarrow \) 3342

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (-\frac {e^2 \int -\frac {2 b+3 a \cos (c+d x)}{2 (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \int \frac {2 b+3 a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \int \frac {2 b-3 a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3346

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3181

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\)

Input:

Int[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

(e*(e*Sin[c + d*x])^(5/2))/(2*b*d*(a + b*Cos[c + d*x])^2) - (5*e^2*(-((e*( 
3*a + 2*b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(b^2*d*(a + b*Cos[c + d*x])) 
) + (e^2*((6*a*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*d*S 
qrt[e*Sin[c + d*x]]) - ((3*a^2 - 2*b^2)*((-2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqr 
t[e]*Sin[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]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]* 
(-a^2 + b^2)^(3/4)*e^(3/2))))/d + (a*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2 
]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(Sqrt[-a^2 + b^2]*(b - Sqrt 
[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (a*EllipticPi[(2*b)/(b + Sqrt[-a^2 
 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(Sqrt[-a^2 + b^2]*(b 
+ Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]])))/b))/(2*b^2)))/(4*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 221

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]
 

rule 266

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]
 

rule 756

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

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]
 

rule 3121

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]
 

rule 3172

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]
 

rule 3181

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]
 

rule 3284

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]
 

rule 3286

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]
 

rule 3342

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*C 
os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p 
 + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( 
p - 1)/(b^2*(m + 1)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin 
[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x 
], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
 

rule 3346

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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2589\) vs. \(2(445)=890\).

Time = 98.71 (sec) , antiderivative size = 2590, normalized size of antiderivative = 5.06

method	result	size
default	\(\text {Expression too large to display}\)	\(2590\)

Input:

int((e*sin(d*x+c))^(7/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
 

Output:

(2*e^3*b*(1/b^4*(e*sin(d*x+c))^(1/2)-e^2/b^4*(-1/8*(e*sin(d*x+c))^(1/2)*e^ 
2*(-11*cos(d*x+c)^2*a^2*b^2+2*b^4*cos(d*x+c)^2+7*a^4+2*a^2*b^2)/(-b^2*cos( 
d*x+c)^2*e^2+a^2*e^2)^2+5/64*(3*a^2-2*b^2)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2* 
e^2-b^2*e^2)*2^(1/2)*(ln((e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d* 
x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)-(e^2*(a^2-b^2 
)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*ar 
ctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^ 
(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1))))-(cos(d*x+c)^2*e 
*sin(d*x+c))^(1/2)*e^4*a*(-2/b^4*(5*a^2-3*b^2)*(-1/2/b/(-a^2+b^2)^(1/2)*(1 
-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e 
*sin(d*x+c))^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2), 
1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/b/(-a^2+b^2)^(1/2)*(1-sin(d*x+c) 
)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c) 
)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+ 
b^2)^(1/2)/b),1/2*2^(1/2)))+1/b^4*(11*a^4-14*a^2*b^2+3*b^4)*(1/2*b^2/e/a^2 
/(a^2-b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*b^2+a^2)+1/4/a 
^2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/ 
(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/ 
2))-5/8/(a^2-b^2)/b/(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c)) 
^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(1-(-a^2+b^2)...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*sin(d*x+c))**(7/2)/(a+b*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((e*sin(d*x + c))^(7/2)/(b*cos(d*x + c) + a)^3, x)
 

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*sin(d*x + c))^(7/2)/(b*cos(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((e*sin(c + d*x))^(7/2)/(a + b*cos(c + d*x))^3,x)
 

Output:

int((e*sin(c + d*x))^(7/2)/(a + b*cos(c + d*x))^3, x)
 

Reduce [F]

\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) e^{3} \] Input:

int((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x)
 

Output:

sqrt(e)*int((sqrt(sin(c + d*x))*sin(c + d*x)**3)/(cos(c + d*x)**3*b**3 + 3 
*cos(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*a**2*b + a**3),x)*e**3