\(\int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx\) [559]

Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {(c-d) (c+7 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(c-5 d) d \cos (e+f x)}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a+a \sin (e+f x))^{3/2}} \] Output:

-1/4*(c-d)*(c+7*d)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e)) 
^(1/2))*2^(1/2)/a^(3/2)/f+1/2*(c-5*d)*d*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(1 
/2)-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))/f/(a+a*sin(f*x+e))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.05 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.73 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-(c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} \left (c^2+6 c d-7 d^2\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-4 d^2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+4 d^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )}{2 f (a (1+\sin (e+f x)))^{3/2}} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)^2*Sin[(e + f*x)/2] - (c 
- d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (1 + I)*(-1)^(3/4)*(c^2 + 6 
*c*d - 7*d^2)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 4*d^2*Cos[(e + f*x)/2]*(Cos[(e + f*x 
)/2] + Sin[(e + f*x)/2])^2 + 4*d^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Si 
n[(e + f*x)/2])^2))/(2*f*(a*(1 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3239, 27, 3042, 3230, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3239

\(\displaystyle -\frac {\int -\frac {a \left (c^2+5 d c-2 d^2\right )-a (c-5 d) d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {a (c-d) (c+7 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (c-d) (c+7 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {2 a (c-d) (c+7 d) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {\sqrt {2} \sqrt {a} (c-d) (c+7 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(f*(a + a*Sin[e + f*x])^( 
3/2)) + (-((Sqrt[2]*Sqrt[a]*(c - d)*(c + 7*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x 
])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f) + (2*a*(c - 5*d)*d*Cos[e + f*x] 
)/(f*Sqrt[a + a*Sin[e + f*x]]))/(4*a^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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

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

rule 3230

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

rule 3239

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(315\) vs. \(2(119)=238\).

Time = 2.01 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.29

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,c^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c d -7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,d^{2}+8 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, d^{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,c^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c d -7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,d^{2}+2 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, c^{2}-4 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, c d +10 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, d^{2}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{4 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(316\)
parts	\(-\frac {c^{2} \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (f x +e \right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+2 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{4 a^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}+\frac {d^{2} \left (7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (f x +e \right )+7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -8 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, \sin \left (f x +e \right )-10 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{4 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}-\frac {c d \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (f x +e \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -2 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{2 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(402\)

Input:

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

Output:

-1/4/a^(5/2)*(sin(f*x+e)*(2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/ 
2)/a^(1/2))*a*c^2+6*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^( 
1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))* 
a*d^2+8*(a-sin(f*x+e)*a)^(1/2)*a^(1/2)*d^2)+2^(1/2)*arctanh(1/2*(a-sin(f*x 
+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a*c^2+6*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a) 
^(1/2)*2^(1/2)/a^(1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2) 
*2^(1/2)/a^(1/2))*a*d^2+2*(a-sin(f*x+e)*a)^(1/2)*a^(1/2)*c^2-4*(a-sin(f*x+ 
e)*a)^(1/2)*a^(1/2)*c*d+10*(a-sin(f*x+e)*a)^(1/2)*a^(1/2)*d^2)*(-a*(-1+sin 
(f*x+e)))^(1/2)/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (119) = 238\).

Time = 0.09 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.75 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c^{2} - 12 \, c d + 14 \, d^{2} - {\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (2 \, c^{2} + 12 \, c d - 14 \, d^{2} + {\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (4 \, d^{2} \cos \left (f x + e\right )^{2} + c^{2} - 2 \, c d + d^{2} + {\left (c^{2} - 2 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left (4 \, d^{2} \cos \left (f x + e\right ) - c^{2} + 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

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

Output:

-1/8*(sqrt(2)*((c^2 + 6*c*d - 7*d^2)*cos(f*x + e)^2 - 2*c^2 - 12*c*d + 14* 
d^2 - (c^2 + 6*c*d - 7*d^2)*cos(f*x + e) - (2*c^2 + 12*c*d - 14*d^2 + (c^2 
 + 6*c*d - 7*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e) 
^2 + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + 
e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(c 
os(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*( 
4*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 5*d^2)*cos(f*x + 
 e) + (4*d^2*cos(f*x + e) - c^2 + 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f* 
x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f 
*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
 

Sympy [F]

\[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) c^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) d^{2}+2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) c d \right )}{a^{2}} \] Input:

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

Output:

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