\(\int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx\) [565]

Optimal result

Integrand size = 27, antiderivative size = 194 \[ \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {3 (c-d) \left (c^2+6 c d+25 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d)^2 (3 c+13 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(c-9 d) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}} \] Output:

-3/32*(c-d)*(c^2+6*c*d+25*d^2)*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^(5/2)/f-1/16*(c-d)^2*(3*c+13*d)*cos(f*x+e)/a 
/f/(a+a*sin(f*x+e))^(3/2)+1/4*(c-9*d)*d^2*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e) 
)^(1/2)-1/4*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^(5/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.13 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.06 \[ \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-11 c^3 \cos \left (\frac {1}{2} (e+f x)\right )+9 c^2 d \cos \left (\frac {1}{2} (e+f x)\right )+15 c d^2 \cos \left (\frac {1}{2} (e+f x)\right )-45 d^3 \cos \left (\frac {1}{2} (e+f x)\right )-3 c^3 \cos \left (\frac {3}{2} (e+f x)\right )-15 c^2 d \cos \left (\frac {3}{2} (e+f x)\right )+39 c d^2 \cos \left (\frac {3}{2} (e+f x)\right )-69 d^3 \cos \left (\frac {3}{2} (e+f x)\right )+16 d^3 \cos \left (\frac {5}{2} (e+f x)\right )+11 c^3 \sin \left (\frac {1}{2} (e+f x)\right )-9 c^2 d \sin \left (\frac {1}{2} (e+f x)\right )-15 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+45 d^3 \sin \left (\frac {1}{2} (e+f x)\right )+(6+6 i) (-1)^{3/4} \left (c^3+5 c^2 d+19 c d^2-25 d^3\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-3 c^3 \sin \left (\frac {3}{2} (e+f x)\right )-15 c^2 d \sin \left (\frac {3}{2} (e+f x)\right )+39 c d^2 \sin \left (\frac {3}{2} (e+f x)\right )-69 d^3 \sin \left (\frac {3}{2} (e+f x)\right )-16 d^3 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{32 f (a (1+\sin (e+f x)))^{5/2}} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-11*c^3*Cos[(e + f*x)/2] + 9*c^2*d 
*Cos[(e + f*x)/2] + 15*c*d^2*Cos[(e + f*x)/2] - 45*d^3*Cos[(e + f*x)/2] - 
3*c^3*Cos[(3*(e + f*x))/2] - 15*c^2*d*Cos[(3*(e + f*x))/2] + 39*c*d^2*Cos[ 
(3*(e + f*x))/2] - 69*d^3*Cos[(3*(e + f*x))/2] + 16*d^3*Cos[(5*(e + f*x))/ 
2] + 11*c^3*Sin[(e + f*x)/2] - 9*c^2*d*Sin[(e + f*x)/2] - 15*c*d^2*Sin[(e 
+ f*x)/2] + 45*d^3*Sin[(e + f*x)/2] + (6 + 6*I)*(-1)^(3/4)*(c^3 + 5*c^2*d 
+ 19*c*d^2 - 25*d^3)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4] 
)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 3*c^3*Sin[(3*(e + f*x))/2] - 
15*c^2*d*Sin[(3*(e + f*x))/2] + 39*c*d^2*Sin[(3*(e + f*x))/2] - 69*d^3*Sin 
[(3*(e + f*x))/2] - 16*d^3*Sin[(5*(e + f*x))/2]))/(32*f*(a*(1 + Sin[e + f* 
x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3498, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int -\frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{2 (\sin (e+f x) a+a)^{3/2}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\int \frac {-a (c-9 d) d^2 \sin ^2(e+f x)+\left (a d \left (3 c^2+9 d c-4 d^2\right )-a c (c-9 d) d\right ) \sin (e+f x)+a c \left (3 c^2+9 d c-4 d^2\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-a (c-9 d) d^2 \sin (e+f x)^2+\left (a d \left (3 c^2+9 d c-4 d^2\right )-a c (c-9 d) d\right ) \sin (e+f x)+a c \left (3 c^2+9 d c-4 d^2\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3498

\(\displaystyle \frac {-\frac {\int -\frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {a (3 c+13 d) (c-d)^2 \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {3 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {3 \sqrt {2} a^{3/2} (c-d) \left (c^2+6 c d+25 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]) 
^(5/2)) + (-1/2*(a*(c - d)^2*(3*c + 13*d)*Cos[e + f*x])/(f*(a + a*Sin[e + 
f*x])^(3/2)) + ((-3*Sqrt[2]*a^(3/2)*(c - d)*(c^2 + 6*c*d + 25*d^2)*ArcTanh 
[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f + (8*a^2*(c 
 - 9*d)*d^2*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]))/(4*a^2))/(8*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 3244

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*Sin[e 
+ f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* 
(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* 
Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) 
 + b*c*(m + n))*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] && LtQ[m, -1] 
&& GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3447

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

rule 3498

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* 
B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 
/(a^2*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b 
*B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
 B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(687\) vs. \(2(171)=342\).

Time = 4.67 (sec) , antiderivative size = 688, normalized size of antiderivative = 3.55

method	result	size
default	\(-\frac {\left (\left (-64 a^{\frac {3}{2}} d^{3} \sqrt {a -\sin \left (f x +e \right ) a}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d -57 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}+75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right ) \cos \left (f x +e \right )^{2}+2 \sin \left (f x +e \right ) \left (64 a^{\frac {3}{2}} d^{3} \sqrt {a -\sin \left (f x +e \right ) a}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +57 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right )+20 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c^{3}+36 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c^{2} d -132 a^{\frac {3}{2}} d^{2} c \sqrt {a -\sin \left (f x +e \right ) a}+204 a^{\frac {3}{2}} d^{3} \sqrt {a -\sin \left (f x +e \right ) a}-6 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c^{3}-30 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c^{2} d +78 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c \,d^{2}-42 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} d^{3} \sqrt {a}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}+30 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +114 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(688\)
parts	\(\text {Expression too large to display}\)	\(840\)

Input:

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

Output:

-1/32*((-64*a^(3/2)*d^3*(a-sin(f*x+e)*a)^(1/2)-3*2^(1/2)*arctanh(1/2*(a-si 
n(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3-15*2^(1/2)*arctanh(1/2*(a-sin(f 
*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d-57*2^(1/2)*arctanh(1/2*(a-sin(f* 
x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d^2+75*2^(1/2)*arctanh(1/2*(a-sin(f*x 
+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3)*cos(f*x+e)^2+2*sin(f*x+e)*(64*a^(3/ 
2)*d^3*(a-sin(f*x+e)*a)^(1/2)+3*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2) 
*2^(1/2)/a^(1/2))*a^2*c^3+15*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^ 
(1/2)/a^(1/2))*a^2*c^2*d+57*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^( 
1/2)/a^(1/2))*a^2*c*d^2-75*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1 
/2)/a^(1/2))*a^2*d^3)+20*(a-sin(f*x+e)*a)^(1/2)*a^(3/2)*c^3+36*(a-sin(f*x+ 
e)*a)^(1/2)*a^(3/2)*c^2*d-132*a^(3/2)*d^2*c*(a-sin(f*x+e)*a)^(1/2)+204*a^( 
3/2)*d^3*(a-sin(f*x+e)*a)^(1/2)-6*(a-sin(f*x+e)*a)^(3/2)*a^(1/2)*c^3-30*(a 
-sin(f*x+e)*a)^(3/2)*a^(1/2)*c^2*d+78*(a-sin(f*x+e)*a)^(3/2)*a^(1/2)*c*d^2 
-42*(a-sin(f*x+e)*a)^(3/2)*d^3*a^(1/2)+6*2^(1/2)*arctanh(1/2*(a-sin(f*x+e) 
*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+30*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a) 
^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d+114*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a) 
^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d^2-150*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a) 
^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2)/(1+sin 
(f*x+e))/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. 608 vs. \(2 (171) = 342\).

Time = 0.10 (sec) , antiderivative size = 608, normalized size of antiderivative = 3.13 \[ \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/64*(3*sqrt(2)*((c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e)^3 - 4*c 
^3 - 20*c^2*d - 76*c*d^2 + 100*d^3 + 3*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3) 
*cos(f*x + e)^2 - 2*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e) - (4* 
c^3 + 20*c^2*d + 76*c*d^2 - 100*d^3 - (c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)* 
cos(f*x + e)^2 + 2*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*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)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*si 
n(f*x + e) - cos(f*x + e) - 2)) + 4*(32*d^3*cos(f*x + e)^3 - 4*c^3 + 12*c^ 
2*d - 12*c*d^2 + 4*d^3 - (3*c^3 + 15*c^2*d - 39*c*d^2 + 53*d^3)*cos(f*x + 
e)^2 - (7*c^3 + 3*c^2*d - 27*c*d^2 + 81*d^3)*cos(f*x + e) - (32*d^3*cos(f* 
x + e)^2 - 4*c^3 + 12*c^2*d - 12*c*d^2 + 4*d^3 + (3*c^3 + 15*c^2*d - 39*c* 
d^2 + 85*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f 
*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f 
+ (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c^{3}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) d^{3}+3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c \,d^{2}+3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c^{2} d \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 
+ 3*sin(e + f*x) + 1),x)*c**3 + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)** 
3)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*d**3 + 3* 
int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 + 3*sin(e + 
f*x)**2 + 3*sin(e + f*x) + 1),x)*c*d**2 + 3*int((sqrt(sin(e + f*x) + 1)*si 
n(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)* 
c**2*d))/a**3