\(\int \frac {(a+b \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [783]

Optimal result

Integrand size = 29, antiderivative size = 497 \[ \int \frac {(a+b \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 (b c-a d) \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {8 (a-b) \sqrt {a+b} (a c-b d) E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (c-d)^2 (c+d)^{3/2} (b c-a d) f}+\frac {2 (a-b) \sqrt {a+b} (a (3 c+d)-b (c+3 d)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (c-d)^2 (c+d)^{3/2} (b c-a d) f} \] Output:

-2/3*(-a*d+b*c)*cos(f*x+e)*(a+b*sin(f*x+e))^(1/2)/(c^2-d^2)/f/(c+d*sin(f*x 
+e))^(3/2)-8/3*(a-b)*(a+b)^(1/2)*(a*c-b*d)*EllipticE((c+d)^(1/2)*(a+b*sin( 
f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d)) 
^(1/2))*sec(f*x+e)*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2 
)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^(1/2)*(c+d*sin(f*x+e 
))/(c-d)^2/(c+d)^(3/2)/(-a*d+b*c)/f+2/3*(a-b)*(a+b)^(1/2)*(a*(3*c+d)-b*(c+ 
3*d))*EllipticF((c+d)^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f* 
x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d))^(1/2))*sec(f*x+e)*((-a*d+b*c)*(1-sin 
(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b)/( 
c+d*sin(f*x+e)))^(1/2)*(c+d*sin(f*x+e))/(c-d)^2/(c+d)^(3/2)/(-a*d+b*c)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2012\) vs. \(2(497)=994\).

Time = 6.51 (sec) , antiderivative size = 2012, normalized size of antiderivative = 4.05 \[ \int \frac {(a+b \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + b*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((-2*(b*c*Cos[e + f*x] 
- a*d*Cos[e + f*x]))/(3*(c^2 - d^2)*(c + d*Sin[e + f*x])^2) - (8*(-(a*c*d* 
Cos[e + f*x]) + b*d^2*Cos[e + f*x]))/(3*(c^2 - d^2)^2*(c + d*Sin[e + f*x]) 
)))/f + ((-4*(-(b*c) + a*d)*(3*a^2*c^2 + b^2*c^2 - 4*a*b*c*d + a^2*d^2 - b 
^2*d^2)*Sqrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcS 
in[Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) 
 + a*d)]/Sqrt[2]], (2*(-(b*c) + a*d))/((a + b)*(-c + d))]*Sec[e + f*x]*Sin 
[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + d)*Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Si 
n[e + f*x]))/(-(b*c) + a*d)]*Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c 
+ d*Sin[e + f*x]))/(-(b*c) + a*d)])/((a + b)*(c + d)*Sqrt[a + b*Sin[e + f* 
x]]*Sqrt[c + d*Sin[e + f*x]]) - 4*(-(b*c) + a*d)*(4*a*b*c^2 + 4*a^2*c*d - 
4*b^2*c*d - 4*a*b*d^2)*((Sqrt[((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d 
)]*EllipticF[ArcSin[Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e 
 + f*x]))/(-(b*c) + a*d)]/Sqrt[2]], (2*(-(b*c) + a*d))/((a + b)*(-c + d))] 
*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + d)*Csc[(-e + Pi/2 - f* 
x)/2]^2*(a + b*Sin[e + f*x]))/(-(b*c) + a*d)]*Sqrt[((-a - b)*Csc[(-e + Pi/ 
2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + a*d)])/((a + b)*(c + d)*Sqrt 
[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - (Sqrt[((c + d)*Cot[(-e + 
Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticPi[(-(b*c) + a*d)/((a + b)*d), ArcSin[ 
Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c)...
 

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 509, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3278, 27, 3042, 3477, 3042, 3297, 3475}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3278

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3297

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

\(\Big \downarrow \) 3475

\(\displaystyle \frac {\frac {2 (a-b) \sqrt {a+b} (a (3 c+d)-b (c+3 d)) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}-\frac {8 (a-b) \sqrt {a+b} (a c-b d) \sec (e+f x) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}}{3 \left (c^2-d^2\right )}-\frac {2 (b c-a d) \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}\)

Input:

Int[(a + b*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

(-2*(b*c - a*d)*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]])/(3*(c^2 - d^2)*f*(c 
 + d*Sin[e + f*x])^(3/2)) + ((-8*(a - b)*Sqrt[a + b]*(a*c - b*d)*EllipticE 
[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sin 
[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b*c 
 - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - 
 a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + 
f*x]))/((c - d)*Sqrt[c + d]*(b*c - a*d)*f) + (2*(a - b)*Sqrt[a + b]*(a*(3* 
c + d) - b*(c + 3*d))*EllipticF[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x 
]])/(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c 
 + d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d 
*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*S 
in[e + f*x])))]*(c + d*Sin[e + f*x]))/((c - d)*Sqrt[c + d]*(b*c - a*d)*f)) 
/(3*(c^2 - d^2))
 

Defintions of rubi rules used

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 3042

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

rule 3278

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*(a + b*Si 
n[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 - b^2))), 
 x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + 
d*Sin[e + f*x])^(n - 2)*Simp[c*(a*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) 
+ (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d*(b*c - a 
*d)*(m + n + 1)*Sin[e + f*x]^2, 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] && LtQ[m, -1 
] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]
 

rule 3297

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ 
.) + (f_.)*(x_)]]), x_Symbol] :> Simp[2*((c + d*Sin[e + f*x])/(f*(b*c - a*d 
)*Rt[(c + d)/(a + b), 2]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 - Sin[e + f*x] 
)/((a + b)*(c + d*Sin[e + f*x])))]*Sqrt[(-(b*c - a*d))*((1 + Sin[e + f*x])/ 
((a - b)*(c + d*Sin[e + f*x])))]*EllipticF[ArcSin[Rt[(c + d)/(a + b), 2]*(S 
qrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]])], (a + b)*((c - d)/((a - 
b)*(c + d)))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && N 
eQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/(a + b)]
 

rule 3475

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_) + (b_.)*sin[(e_.) + (f_.) 
*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[-2*A*(c - d)*((a + b*Sin[e + f*x])/(f*(b*c - a*d)^2*Rt[(a + b)/(c + d), 2 
]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 + Sin[e + f*x])/((c - d)*(a + b*Sin[e 
 + f*x])))]*Sqrt[(-(b*c - a*d))*((1 - Sin[e + f*x])/((c + d)*(a + b*Sin[e + 
 f*x])))]*EllipticE[ArcSin[Rt[(a + b)/(c + d), 2]*(Sqrt[c + d*Sin[e + f*x]] 
/Sqrt[a + b*Sin[e + f*x]])], (a - b)*((c + d)/((a + b)*(c - d)))], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] 
&& NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(a + b)/(c + d)]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(92144\) vs. \(2(457)=914\).

Time = 10.73 (sec) , antiderivative size = 92145, normalized size of antiderivative = 185.40

method	result	size
default	\(\text {Expression too large to display}\)	\(92145\)

Input:

int((a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [F]

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

integrate((a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

integral(-(b*sin(f*x + e) + a)^(3/2)*sqrt(d*sin(f*x + e) + c)/(3*c*d^2*cos 
(f*x + e)^2 - c^3 - 3*c*d^2 + (d^3*cos(f*x + e)^2 - 3*c^2*d - d^3)*sin(f*x 
 + e)), x)
 

Sympy [F]

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

integrate((a+b*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**(5/2),x)
 

Output:

Integral((a + b*sin(e + f*x))**(3/2)/(c + d*sin(e + f*x))**(5/2), x)
 

Maxima [F]

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

integrate((a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

integrate((b*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^(5/2), x)
 

Giac [F]

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

integrate((a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

integrate((b*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + b*sin(e + f*x))^(3/2)/(c + d*sin(e + f*x))^(5/2),x)
 

Output:

int((a + b*sin(e + f*x))^(3/2)/(c + d*sin(e + f*x))^(5/2), x)
 

Reduce [F]

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

int((a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)
 

Output:

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