\(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{3/2}} \, dx\) [501]

Optimal result

Integrand size = 27, antiderivative size = 189 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{3/2}} \, dx=\frac {2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^2 c E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {4 a^2 (c-d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{d^2 f \sqrt {c+d \sin (e+f x)}} \] Output:

2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(1/2)-4*a^2*c*EllipticE( 
cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/ 
d^2/(c+d)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4*a^2*(c-d)*InverseJacobiAM(1/2 
*e-1/4*Pi+1/2*f*x,2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/ 
d^2/f/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 3.05 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.93 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{3/2}} \, dx=-\frac {2 a^2 (1+\sin (e+f x))^2 \left (2 c (c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-(c-d) \left (d \cos (e+f x)+2 (c+d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )\right )}{d^2 (c+d) f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c+d \sin (e+f x)}} \] Input:

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

Output:

(-2*a^2*(1 + Sin[e + f*x])^2*(2*c*(c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, 
 (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - (c - d)*(d*Cos[e + f* 
x] + 2*(c + d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d 
*Sin[e + f*x])/(c + d)])))/(d^2*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x 
)/2])^4*Sqrt[c + d*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3241, 25, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2 a \left (\frac {2 a c \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {a \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}\right )}{d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2 a \left (\frac {2 a c \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {a \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}\right )}{d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\frac {2 a c \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {a \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}\right )}{d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {c+d \sin (e+f x)}}\)

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(2*a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]) + (2*a 
*((2*a*c*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f 
*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*a*(c^2 - d^2)*Elliptic 
F[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/( 
d*f*Sqrt[c + d*Sin[e + f*x]])))/(d*(c + d))
 

Defintions of rubi rules used

rule 25

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

rule 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*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]
 

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
 && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(462\) vs. \(2(184)=368\).

Time = 2.67 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.45

method	result	size
default	\(-\frac {2 \left (2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {d \left (-1+\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c^{3}-2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {d \left (-1+\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c \,d^{2}-2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {d \left (-1+\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c^{2} d +2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {d \left (-1+\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) d^{3}+c \,d^{2} \sin \left (f x +e \right )^{2}-d^{3} \sin \left (f x +e \right )^{2}-c \,d^{2}+d^{3}\right ) a^{2}}{d^{3} \left (c +d \right ) \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(463\)
parts	\(\text {Expression too large to display}\)	\(1648\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.92 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

2/3*(3*(a^2*c*d^2 - a^2*d^3)*sqrt(d*sin(f*x + e) + c)*cos(f*x + e) - 2*(2* 
a^2*c^3 - 3*a^2*c*d^2 + (2*a^2*c^2*d - 3*a^2*d^3)*sin(f*x + e))*sqrt(1/2*I 
*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d 
^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) - 2*(2*a^2 
*c^3 - 3*a^2*c*d^2 + (2*a^2*c^2*d - 3*a^2*d^3)*sin(f*x + e))*sqrt(-1/2*I*d 
)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^ 
2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) - 6*(I*a^2* 
c*d^2*sin(f*x + e) + I*a^2*c^2*d)*sqrt(1/2*I*d)*weierstrassZeta(-4/3*(4*c^ 
2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3* 
(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e 
) - 3*I*d*sin(f*x + e) - 2*I*c)/d)) - 6*(-I*a^2*c*d^2*sin(f*x + e) - I*a^2 
*c^2*d)*sqrt(-1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8 
*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/ 
27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) 
+ 2*I*c)/d)))/((c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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