\(\int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx\) [1273]

Optimal result

Integrand size = 37, antiderivative size = 159 \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\frac {2 \left (a^2+b^2\right ) \sqrt {d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac {4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}+\frac {\left (2 a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (4 e-\pi )+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{3 f g^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Output:

2/3*(a^2+b^2)*(d*sin(f*x+e))^(1/2)/d/f/g/(g*cos(f*x+e))^(3/2)+4/3*a*b*(d*s 
in(f*x+e))^(3/2)/d^2/f/g/(g*cos(f*x+e))^(3/2)+1/3*(2*a^2-b^2)*InverseJacob 
iAM(e-1/4*Pi+f*x,2^(1/2))*sin(2*f*x+2*e)^(1/2)/f/g^2/(g*cos(f*x+e))^(1/2)/ 
(d*sin(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\frac {2 \left (15 a^2 \cos ^2(e+f x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{4},\frac {5}{4},\sin ^2(e+f x)\right )+b \sin (e+f x) \left (10 a+3 b \cos ^2(e+f x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {9}{4},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right ) \tan (e+f x)}{15 f g^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Input:

Integrate[(a + b*Sin[e + f*x])^2/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f* 
x]]),x]
 

Output:

(2*(15*a^2*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[1/4, 7/4, 5/4, Sin[e + 
 f*x]^2] + b*Sin[e + f*x]*(10*a + 3*b*(Cos[e + f*x]^2)^(3/4)*Hypergeometri 
c2F1[5/4, 7/4, 9/4, Sin[e + f*x]^2]*Sin[e + f*x]))*Tan[e + f*x])/(15*f*g^2 
*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.85 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.23, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3390, 3042, 3043, 3681, 362, 266, 768, 858, 807, 230}

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.

\(\displaystyle \int \frac {(a+b \sin (e+f x))^2}{\sqrt {d \sin (e+f x)} (g \cos (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sin (e+f x))^2}{\sqrt {d \sin (e+f x)} (g \cos (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3390

\(\displaystyle \int \frac {a^2+b^2 \sin ^2(e+f x)}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}}dx+\frac {2 a b \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{5/2}}dx}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^2+b^2 \sin (e+f x)^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}}dx+\frac {2 a b \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{5/2}}dx}{d}\)

\(\Big \downarrow \) 3043

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

\(\Big \downarrow \) 3681

\(\displaystyle \frac {\cos ^2(e+f x)^{3/4} \int \frac {a^2+b^2 \sin ^2(e+f x)}{\sqrt {d \sin (e+f x)} \left (1-\sin ^2(e+f x)\right )^{7/4}}d\sin (e+f x)}{f g (g \cos (e+f x))^{3/2}}+\frac {4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}\)

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 230

\(\displaystyle \frac {\cos ^2(e+f x)^{3/4} \left (\frac {2 \left (a^2+b^2\right ) \sqrt {d \sin (e+f x)}}{3 d \left (1-\sin ^2(e+f x)\right )^{3/4}}-\frac {2 \left (2 a^2-b^2\right ) (d \sin (e+f x))^{3/2} \left (1-d^2 \csc ^4(e+f x)\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (d^2 \sin (e+f x)\right ),2\right )}{3 d^2 \left (1-\sin ^2(e+f x)\right )^{3/4}}\right )}{f g (g \cos (e+f x))^{3/2}}+\frac {4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}\)

Input:

Int[(a + b*Sin[e + f*x])^2/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f*x]]),x 
]
 

Output:

(4*a*b*(d*Sin[e + f*x])^(3/2))/(3*d^2*f*g*(g*Cos[e + f*x])^(3/2)) + ((Cos[ 
e + f*x]^2)^(3/4)*((2*(a^2 + b^2)*Sqrt[d*Sin[e + f*x]])/(3*d*(1 - Sin[e + 
f*x]^2)^(3/4)) - (2*(2*a^2 - b^2)*(1 - d^2*Csc[e + f*x]^4)^(3/4)*EllipticF 
[ArcSin[d^2*Sin[e + f*x]]/2, 2]*(d*Sin[e + f*x])^(3/2))/(3*d^2*(1 - Sin[e 
+ f*x]^2)^(3/4))))/(f*g*(g*Cos[e + f*x])^(3/2))
 

Defintions of rubi rules used

rule 230

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] 
))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

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 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 3042

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

rule 3043

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( 
m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ 
(a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & 
& NeQ[m, -1]
 

rule 3390

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) 
Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* 
x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, 
e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
 

rule 3681

Int[(cos[(e_.) + (f_.)*(x_)]*(c_.))^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = 
 FreeFactors[Sin[e + f*x], x]}, Simp[ff*c^(2*IntPart[(m - 1)/2] + 1)*((c*Co 
s[e + f*x])^(2*FracPart[(m - 1)/2])/(f*(Cos[e + f*x]^2)^FracPart[(m - 1)/2] 
))   Subst[Int[(d*ff*x)^n*(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x] 
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m]
 

Maple [A] (verified)

Time = 6.00 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {\left (2 \cos \left (f x +e \right )+2\right ) \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, a^{2} \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-1-\cos \left (f x +e \right )\right ) \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, b^{2} \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+4 a b \sin \left (f x +e \right ) \tan \left (f x +e \right )+2 a^{2} \tan \left (f x +e \right )+2 b^{2} \tan \left (f x +e \right )}{3 g^{2} f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {d \sin \left (f x +e \right )}}\)	\(254\)
parts	\(\frac {2 a^{2} \left (\left (1+\cos \left (f x +e \right )\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+\tan \left (f x +e \right )\right )}{3 f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {d \sin \left (f x +e \right )}\, g^{2}}+\frac {b^{2} \left (\left (-1-\cos \left (f x +e \right )\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \tan \left (f x +e \right )\right )}{3 f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {d \sin \left (f x +e \right )}\, g^{2}}+\frac {4 a b \sin \left (f x +e \right ) \tan \left (f x +e \right )}{3 f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {d \sin \left (f x +e \right )}\, g^{2}}\)	\(300\)

Input:

int((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x,method= 
_RETURNVERBOSE)
 

Output:

1/3/g^2/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)*((2*cos(f*x+e)+2)*(-2* 
csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(csc(f*x+e 
)-cot(f*x+e)+1)^(1/2)*a^2*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^ 
(1/2))+(-1-cos(f*x+e))*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+c 
ot(f*x+e))^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*b^2*EllipticF((csc(f*x+e) 
-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))+4*a*b*sin(f*x+e)*tan(f*x+e)+2*a^2*tan(f* 
x+e)+2*b^2*tan(f*x+e))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=-\frac {{\left (2 \, a^{2} - b^{2}\right )} \sqrt {i \, d g} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + {\left (2 \, a^{2} - b^{2}\right )} \sqrt {-i \, d g} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - 2 \, {\left (2 \, a b \sin \left (f x + e\right ) + a^{2} + b^{2}\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {d \sin \left (f x + e\right )}}{3 \, d f g^{3} \cos \left (f x + e\right )^{2}} \] Input:

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, 
algorithm="fricas")
 

Output:

-1/3*((2*a^2 - b^2)*sqrt(I*d*g)*cos(f*x + e)^2*elliptic_f(arcsin(cos(f*x + 
 e) + I*sin(f*x + e)), -1) + (2*a^2 - b^2)*sqrt(-I*d*g)*cos(f*x + e)^2*ell 
iptic_f(arcsin(cos(f*x + e) - I*sin(f*x + e)), -1) - 2*(2*a*b*sin(f*x + e) 
 + a^2 + b^2)*sqrt(g*cos(f*x + e))*sqrt(d*sin(f*x + e)))/(d*f*g^3*cos(f*x 
+ e)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((a+b*sin(f*x+e))**2/(g*cos(f*x+e))**(5/2)/(d*sin(f*x+e))**(1/2), 
x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right )}} \,d x } \] Input:

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, 
algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e) + a)^2/((g*cos(f*x + e))^(5/2)*sqrt(d*sin(f*x + 
e))), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, 
algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {d\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int((a + b*sin(e + f*x))^2/((g*cos(e + f*x))^(5/2)*(d*sin(e + f*x))^(1/2)) 
,x)
 

Output:

int((a + b*sin(e + f*x))^2/((g*cos(e + f*x))^(5/2)*(d*sin(e + f*x))^(1/2)) 
, x)
 

Reduce [F]

\[ \int \frac {(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx=\frac {\sqrt {g}\, \sqrt {d}\, \left (6 \cos \left (f x +e \right )^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right )}}{\cos \left (f x +e \right )^{3}}d x \right ) a b f -\cos \left (f x +e \right )^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right )}}{\cos \left (f x +e \right ) \sin \left (f x +e \right )}d x \right ) b^{2} f +3 \cos \left (f x +e \right )^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right )}}{\cos \left (f x +e \right )^{3} \sin \left (f x +e \right )}d x \right ) a^{2} f +2 \sqrt {\sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right )}\, b^{2}\right )}{3 \cos \left (f x +e \right )^{2} d f \,g^{3}} \] Input:

int((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x)
 

Output:

(sqrt(g)*sqrt(d)*(6*cos(e + f*x)**2*int((sqrt(sin(e + f*x))*sqrt(cos(e + f 
*x)))/cos(e + f*x)**3,x)*a*b*f - cos(e + f*x)**2*int((sqrt(sin(e + f*x))*s 
qrt(cos(e + f*x)))/(cos(e + f*x)*sin(e + f*x)),x)*b**2*f + 3*cos(e + f*x)* 
*2*int((sqrt(sin(e + f*x))*sqrt(cos(e + f*x)))/(cos(e + f*x)**3*sin(e + f* 
x)),x)*a**2*f + 2*sqrt(sin(e + f*x))*sqrt(cos(e + f*x))*b**2))/(3*cos(e + 
f*x)**2*d*f*g**3)