\(\int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx\) [1373]

Optimal result

Integrand size = 33, antiderivative size = 621 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^3 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}+\frac {a^3 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}-\frac {2 a^3 g \sqrt {g \cos (e+f x)}}{b^4 f}+\frac {2 a (g \cos (e+f x))^{5/2}}{5 b^2 f g}-\frac {2 a^4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^5 f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^3 f \sqrt {g \cos (e+f x)}}+\frac {4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{21 b f \sqrt {g \cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^3 f}+\frac {4 g \sqrt {g \cos (e+f x)} \sin (e+f x)}{21 b f}-\frac {2 (g \cos (e+f x))^{5/2} \sin (e+f x)}{7 b f g} \] Output:

a^3*(-a^2+b^2)^(1/4)*g^(3/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2 
)^(1/4)/g^(1/2))/b^(9/2)/f+a^3*(-a^2+b^2)^(1/4)*g^(3/2)*arctanh(b^(1/2)*(g 
*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(9/2)/f-2*a^3*g*(g*cos(f*x+ 
e))^(1/2)/b^4/f+2/5*a*(g*cos(f*x+e))^(5/2)/b^2/f/g-2*a^4*g^2*cos(f*x+e)^(1 
/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^5/f/(g*cos(f*x+e))^(1/2)+2/3* 
a^2*g^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^3/f/(g*c 
os(f*x+e))^(1/2)+4/21*g^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2 
^(1/2))/b/f/(g*cos(f*x+e))^(1/2)+a^4*(a^2-b^2)*g^2*cos(f*x+e)^(1/2)*Ellipt 
icPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b^5/(a^2-b*(b-(- 
a^2+b^2)^(1/2)))/f/(g*cos(f*x+e))^(1/2)+a^4*(a^2-b^2)*g^2*cos(f*x+e)^(1/2) 
*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/b^5/(a^2- 
b*(b+(-a^2+b^2)^(1/2)))/f/(g*cos(f*x+e))^(1/2)+2/3*a^2*g*(g*cos(f*x+e))^(1 
/2)*sin(f*x+e)/b^3/f+4/21*g*(g*cos(f*x+e))^(1/2)*sin(f*x+e)/b/f-2/7*(g*cos 
(f*x+e))^(5/2)*sin(f*x+e)/b/f/g
 

Mathematica [C] (warning: unable to verify)

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

Time = 27.31 (sec) , antiderivative size = 1991, normalized size of antiderivative = 3.21 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

Integrate[((g*Cos[e + f*x])^(3/2)*Sin[e + f*x]^3)/(a + b*Sin[e + f*x]),x]
 

Output:

-1/420*((g*Cos[e + f*x])^(3/2)*((-2*(70*a^3 - 19*a*b^2)*(a + b*Sqrt[1 - Co 
s[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2 
, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + 
 f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Co 
s[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + 
f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/ 
2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x 
]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 
 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + 
 ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*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[e + f*x]] + I*b*Cos[e 
 + f*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[ 
Cos[e + f*x]] + I*b*Cos[e + f*x]]))/(-a^2 + b^2)^(3/4))*Sin[e + f*x])/(Sqr 
t[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((210*a^3 - 21*a*b^2)*(a + b 
*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*A 
rcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/ 
2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ((1 + I)*S 
qrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4 
)) + (4*Sqrt[Cos[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f* 
x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(5/2))/(5*(a^2 - ...
 

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 621, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 2009}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^3 (g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3377

\(\displaystyle \int \left (-\frac {a^3 (g \cos (e+f x))^{3/2}}{b^3 (a+b \sin (e+f x))}+\frac {a^2 (g \cos (e+f x))^{3/2}}{b^3}-\frac {a \sin (e+f x) (g \cos (e+f x))^{3/2}}{b^2}+\frac {\sin ^2(e+f x) (g \cos (e+f x))^{3/2}}{b}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^5 f \sqrt {g \cos (e+f x)}}-\frac {2 a^3 g \sqrt {g \cos (e+f x)}}{b^4 f}+\frac {2 a^2 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^3 f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^3 f}+\frac {a^4 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^4 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 g^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac {a^3 g^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac {2 a (g \cos (e+f x))^{5/2}}{5 b^2 f g}+\frac {4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{21 b f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) (g \cos (e+f x))^{5/2}}{7 b f g}+\frac {4 g \sin (e+f x) \sqrt {g \cos (e+f x)}}{21 b f}\)

Input:

Int[((g*Cos[e + f*x])^(3/2)*Sin[e + f*x]^3)/(a + b*Sin[e + f*x]),x]
 

Output:

(a^3*(-a^2 + b^2)^(1/4)*g^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a 
^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*f) + (a^3*(-a^2 + b^2)^(1/4)*g^(3/2)*A 
rcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9 
/2)*f) - (2*a^3*g*Sqrt[g*Cos[e + f*x]])/(b^4*f) + (2*a*(g*Cos[e + f*x])^(5 
/2))/(5*b^2*f*g) - (2*a^4*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] 
)/(b^5*f*Sqrt[g*Cos[e + f*x]]) + (2*a^2*g^2*Sqrt[Cos[e + f*x]]*EllipticF[( 
e + f*x)/2, 2])/(3*b^3*f*Sqrt[g*Cos[e + f*x]]) + (4*g^2*Sqrt[Cos[e + f*x]] 
*EllipticF[(e + f*x)/2, 2])/(21*b*f*Sqrt[g*Cos[e + f*x]]) + (a^4*(a^2 - b^ 
2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f* 
x)/2, 2])/(b^5*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) + 
(a^4*(a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + 
b^2]), (e + f*x)/2, 2])/(b^5*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos 
[e + f*x]]) + (2*a^2*g*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^3*f) + (4*g 
*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(21*b*f) - (2*(g*Cos[e + f*x])^(5/2)*S 
in[e + f*x])/(7*b*f*g)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3377

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1121\) vs. \(2(550)=1100\).

Time = 7.21 (sec) , antiderivative size = 1122, normalized size of antiderivative = 1.81

method	result	size
default	\(\text {Expression too large to display}\)	\(1122\)

Input:

int((g*cos(f*x+e))^(3/2)*sin(f*x+e)^3/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

(-16*g^2*a*(-1/40/b^4/g*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(4*sin(1/2*f*x 
+1/2*e)^4*b^2-4*sin(1/2*f*x+1/2*e)^2*b^2-5*a^2+b^2)+1/4*a^2*(a^2-b^2)/b^4* 
(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(-ln((-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*co 
s(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2 
)/b^2)^(1/2)+g)/((g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1 
/2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+g))-2*arcta 
n(2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)+ 
2*arctan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^( 
1/2)+1))/(16*a^2-16*b^2)/g)+32*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+ 
1/2*e)^2)^(1/2)*g^2*b*(-1/120*(sin(1/2*f*x+1/2*e)^2-1)/b^4/(-2*g*sin(1/2*f 
*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)*(24*cos(1/2*f*x+1/2*e)*sin(1/2*f 
*x+1/2*e)^6*b^2-44*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^4*b^2+16*cos(1/2* 
f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2*b^2+15*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2 
))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*a^2-5*El 
lipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin 
(1/2*f*x+1/2*e)^2)^(1/2)*b^2-15*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(-1+ 
2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*a^2+9*EllipticE 
(cos(1/2*f*x+1/2*e),2^(1/2))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f* 
x+1/2*e)^2)^(1/2)*b^2)+1/64*a^2*(sin(1/2*f*x+1/2*e)^2*a^2-sin(1/2*f*x+1/2* 
e)^2*b^2-a^2+b^2)/b^6*sum(_alpha/(2*_alpha^2-1)*(2^(1/2)/(g*(2*_alpha^2...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((g*cos(f*x+e))**(3/2)*sin(f*x+e)**3/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((g*cos(f*x + e))^(3/2)*sin(f*x + e)^3/(b*sin(f*x + e) + a), x)
 

Giac [F]

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

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

Output:

integrate((g*cos(f*x + e))^(3/2)*sin(f*x + e)^3/(b*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

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

int((sin(e + f*x)^3*(g*cos(e + f*x))^(3/2))/(a + b*sin(e + f*x)),x)
 

Output:

int((sin(e + f*x)^3*(g*cos(e + f*x))^(3/2))/(a + b*sin(e + f*x)), x)
 

Reduce [F]

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

int((g*cos(f*x+e))^(3/2)*sin(f*x+e)^3/(a+b*sin(f*x+e)),x)
 

Output:

sqrt(g)*int((sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x)**3)/(sin(e + f*x 
)*b + a),x)*g