\(\int \frac {\sqrt {c+d \sin (e+f x)}}{(a+b \sin (e+f x))^{5/2}} \, dx\) [804]

Optimal result

Integrand size = 29, antiderivative size = 489 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(a+b \sin (e+f x))^{5/2}} \, dx=\frac {2 (c-d) \sqrt {c+d} \left (4 a b c-3 a^2 d-b^2 d\right ) E\left (\arcsin \left (\frac {\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}{\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}\right )|\frac {(a-b) (c+d)}{(a+b) (c-d)}\right ) \sec (e+f x) \sqrt {-\frac {(b c-a d) (1-\sin (e+f x))}{(c+d) (a+b \sin (e+f x))}} \sqrt {\frac {(b c-a d) (1+\sin (e+f x))}{(c-d) (a+b \sin (e+f x))}} (a+b \sin (e+f x))}{3 (a-b)^2 (a+b)^{3/2} (b c-a d)^2 f}+\frac {2 b \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^{3/2}}+\frac {2 (3 a+b) (c-d) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{3 (a-b)^2 \sqrt {a+b} \sqrt {c+d} (b c-a d) f} \] Output:

2/3*(c-d)*(c+d)^(1/2)*(-3*a^2*d+4*a*b*c-b^2*d)*EllipticE((a+b)^(1/2)*(c+d* 
sin(f*x+e))^(1/2)/(c+d)^(1/2)/(a+b*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))/(c+d)/(a+b*sin(f*x+e))) 
^(1/2)*((-a*d+b*c)*(1+sin(f*x+e))/(c-d)/(a+b*sin(f*x+e)))^(1/2)*(a+b*sin(f 
*x+e))/(a-b)^2/(a+b)^(3/2)/(-a*d+b*c)^2/f+2/3*b*cos(f*x+e)*(c+d*sin(f*x+e) 
)^(1/2)/(a^2-b^2)/f/(a+b*sin(f*x+e))^(3/2)+2/3*(3*a+b)*(c-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))/(a-b)^2/(a+b)^(1/2)/(c+d)^(1/2)/(-a*d+b*c)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2067\) vs. \(2(489)=978\).

Time = 6.56 (sec) , antiderivative size = 2067, normalized size of antiderivative = 4.23 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(a+b \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((2*b*Cos[e + f*x])/(3* 
(a^2 - b^2)*(a + b*Sin[e + f*x])^2) + (2*(-4*a*b^2*c*Cos[e + f*x] + 3*a^2* 
b*d*Cos[e + f*x] + b^3*d*Cos[e + f*x]))/(3*(a^2 - b^2)^2*(-(b*c) + a*d)*(a 
 + b*Sin[e + f*x]))))/f + ((-4*(-(b*c) + a*d)*(-3*a^2*b*c^2 - b^3*c^2 + 3* 
a^3*c*d + a*b^2*c*d - a^2*b*d^2 + b^3*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)*C 
sc[(-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]]) - 4*(-(b* 
c) + a*d)*(-4*a*b^2*c^2 - a^2*b*c*d + b^3*c*d + 3*a^3*d^2 + a*b^2*d^2)*((S 
qrt[((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 ...
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 501, 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, 3275, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3275

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3297

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

\(\Big \downarrow \) 3475

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

Input:

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

Output:

(2*b*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*(a^2 - b^2)*f*(a + b*Sin[e 
+ f*x])^(3/2)) + ((2*(c - d)*Sqrt[c + d]*(4*a*b*c - 3*a^2*d - b^2*d)*Ellip 
ticE[ArcSin[(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])/(Sqrt[c + d]*Sqrt[a + b 
*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]))/((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]))]*(a + b*Sin[ 
e + f*x]))/((a - b)*Sqrt[a + b]*(b*c - a*d)^2*f) + (2*Sqrt[a + b]*(3*a + b 
)*(c - 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*Sin[e + f*x]) 
))]*(c + d*Sin[e + f*x]))/((a - b)*Sqrt[c + d]*(b*c - a*d)*f))/(3*(a^2 - b 
^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 3275

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] 
)^(m + 1)*((c + d*Sin[e + f*x])^n/(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 - 1)*Simp[a*c*(m + 1) + b*d*n + (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] 
 - b*d*(m + n + 2)*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[0, n, 1] && 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. \(96892\) vs. \(2(449)=898\).

Time = 10.99 (sec) , antiderivative size = 96893, normalized size of antiderivative = 198.15

method	result	size
default	\(\text {Expression too large to display}\)	\(96893\)

Input:

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

Output:

result too large to display
 

Fricas [F]

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

integrate((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((c+d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((c+d*sin(f*x+e))^(1/2)/(a+b*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)**3*b 
**3 + 3*sin(e + f*x)**2*a*b**2 + 3*sin(e + f*x)*a**2*b + a**3),x)