\(\int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx\) [457]

Optimal result

Integrand size = 25, antiderivative size = 161 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=-\frac {a^3 (2 c-3 d) x}{d^3}+\frac {2 a^3 (c-d)^2 (2 c+3 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 (c+d) \sqrt {c^2-d^2} f}-\frac {2 a^3 c \cos (e+f x)}{d^2 (c+d) f}+\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))} \] Output:

-a^3*(2*c-3*d)*x/d^3+2*a^3*(c-d)^2*(2*c+3*d)*arctan((d+c*tan(1/2*f*x+1/2*e 
))/(c^2-d^2)^(1/2))/d^3/(c+d)/(c^2-d^2)^(1/2)/f-2*a^3*c*cos(f*x+e)/d^2/(c+ 
d)/f+(c-d)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/(c+d)/f/(c+d*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 5.70 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {a^3 (1+\sin (e+f x))^3 \left ((-2 c+3 d) (e+f x)+\frac {2 (c-d)^2 (2 c+3 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}-d \cos (e+f x)-\frac {(c-d)^2 d \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}\right )}{d^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6} \] Input:

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

Output:

(a^3*(1 + Sin[e + f*x])^3*((-2*c + 3*d)*(e + f*x) + (2*(c - d)^2*(2*c + 3* 
d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*Sqrt[c^2 - d 
^2]) - d*Cos[e + f*x] - ((c - d)^2*d*Cos[e + f*x])/((c + d)*(c + d*Sin[e + 
 f*x]))))/(d^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3241, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {(c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{d f (c+d) (c+d \sin (e+f x))}-\frac {a \left (\frac {\frac {a^2 x (2 c-3 d) (c+d)}{d}-\frac {2 a^2 (c-d)^2 (2 c+3 d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}}{d}+\frac {2 a^2 c \cos (e+f x)}{d f}\right )}{d (c+d)}\)

Input:

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

Output:

-((a*(((a^2*(2*c - 3*d)*(c + d)*x)/d - (2*a^2*(c - d)^2*(2*c + 3*d)*ArcTan 
[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/(d*Sqrt[c^2 - d^2]*f)) 
/d + (2*a^2*c*Cos[e + f*x])/(d*f)))/(d*(c + d))) + ((c - d)*Cos[e + f*x]*( 
a^3 + a^3*Sin[e + f*x]))/(d*(c + d)*f*(c + d*Sin[e + f*x]))
 

Defintions of rubi rules used

rule 217

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3214

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

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 3502

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

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\frac {-\frac {d^{2} \left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (c^{2}-2 c d +d^{2}\right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 c^{3}-c^{2} d -4 c \,d^{2}+3 d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{3}}-\frac {\frac {d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (2 c -3 d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{3}}\right )}{f}\)	\(209\)
default	\(\frac {2 a^{3} \left (\frac {\frac {-\frac {d^{2} \left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (c^{2}-2 c d +d^{2}\right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 c^{3}-c^{2} d -4 c \,d^{2}+3 d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{3}}-\frac {\frac {d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (2 c -3 d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{3}}\right )}{f}\)	\(209\)
risch	\(-\frac {2 a^{3} x c}{d^{3}}+\frac {3 a^{3} x}{d^{2}}-\frac {a^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 d^{2} f}-\frac {a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 d^{2} f}+\frac {2 i a^{3} \left (c^{2}-2 c d +d^{2}\right ) \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d^{3} \left (c +d \right ) f \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )}+\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c^{2}}{\left (c +d \right )^{2} f \,d^{3}}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right )^{2} f \,d^{2}}-\frac {3 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right )^{2} f d}-\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c^{2}}{\left (c +d \right )^{2} f \,d^{3}}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right )^{2} f \,d^{2}}+\frac {3 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right )^{2} f d}\)	\(510\)

Input:

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

Output:

2/f*a^3*(1/d^3*((-d^2*(c^2-2*c*d+d^2)/(c+d)/c*tan(1/2*f*x+1/2*e)-d*(c^2-2* 
c*d+d^2)/(c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(2*c^3-c 
^2*d-4*c*d^2+3*d^3)/(c+d)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2* 
e)+2*d)/(c^2-d^2)^(1/2)))-1/d^3*(d/(1+tan(1/2*f*x+1/2*e)^2)+(2*c-3*d)*arct 
an(tan(1/2*f*x+1/2*e))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 645, normalized size of antiderivative = 4.01 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\left [-\frac {2 \, {\left (2 \, a^{3} c^{3} - a^{3} c^{2} d - 3 \, a^{3} c d^{2}\right )} f x + {\left (2 \, a^{3} c^{3} + a^{3} c^{2} d - 3 \, a^{3} c d^{2} + {\left (2 \, a^{3} c^{2} d + a^{3} c d^{2} - 3 \, a^{3} d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (2 \, a^{3} c^{2} d - a^{3} c d^{2} + a^{3} d^{3}\right )} \cos \left (f x + e\right ) + 2 \, {\left ({\left (2 \, a^{3} c^{2} d - a^{3} c d^{2} - 3 \, a^{3} d^{3}\right )} f x + {\left (a^{3} c d^{2} + a^{3} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c d^{4} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{2} d^{3} + c d^{4}\right )} f\right )}}, -\frac {{\left (2 \, a^{3} c^{3} - a^{3} c^{2} d - 3 \, a^{3} c d^{2}\right )} f x + {\left (2 \, a^{3} c^{3} + a^{3} c^{2} d - 3 \, a^{3} c d^{2} + {\left (2 \, a^{3} c^{2} d + a^{3} c d^{2} - 3 \, a^{3} d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) + {\left (2 \, a^{3} c^{2} d - a^{3} c d^{2} + a^{3} d^{3}\right )} \cos \left (f x + e\right ) + {\left ({\left (2 \, a^{3} c^{2} d - a^{3} c d^{2} - 3 \, a^{3} d^{3}\right )} f x + {\left (a^{3} c d^{2} + a^{3} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{{\left (c d^{4} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{2} d^{3} + c d^{4}\right )} f}\right ] \] Input:

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

Output:

[-1/2*(2*(2*a^3*c^3 - a^3*c^2*d - 3*a^3*c*d^2)*f*x + (2*a^3*c^3 + a^3*c^2* 
d - 3*a^3*c*d^2 + (2*a^3*c^2*d + a^3*c*d^2 - 3*a^3*d^3)*sin(f*x + e))*sqrt 
(-(c - d)/(c + d))*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) 
- c^2 - d^2 + 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f 
*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) 
- c^2 - d^2)) + 2*(2*a^3*c^2*d - a^3*c*d^2 + a^3*d^3)*cos(f*x + e) + 2*((2 
*a^3*c^2*d - a^3*c*d^2 - 3*a^3*d^3)*f*x + (a^3*c*d^2 + a^3*d^3)*cos(f*x + 
e))*sin(f*x + e))/((c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*f), -( 
(2*a^3*c^3 - a^3*c^2*d - 3*a^3*c*d^2)*f*x + (2*a^3*c^3 + a^3*c^2*d - 3*a^3 
*c*d^2 + (2*a^3*c^2*d + a^3*c*d^2 - 3*a^3*d^3)*sin(f*x + e))*sqrt((c - d)/ 
(c + d))*arctan(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f 
*x + e))) + (2*a^3*c^2*d - a^3*c*d^2 + a^3*d^3)*cos(f*x + e) + ((2*a^3*c^2 
*d - a^3*c*d^2 - 3*a^3*d^3)*f*x + (a^3*c*d^2 + a^3*d^3)*cos(f*x + e))*sin( 
f*x + e))/((c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (156) = 312\).

Time = 0.15 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.36 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, a^{3} c^{3} - a^{3} c^{2} d - 4 \, a^{3} c d^{2} + 3 \, a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{3} + d^{4}\right )} \sqrt {c^{2} - d^{2}}} - \frac {2 \, {\left (a^{3} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{3} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a^{3} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{3} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{3} c^{3} - a^{3} c^{2} d + a^{3} c d^{2}\right )}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )} {\left (c^{2} d^{2} + c d^{3}\right )}} - \frac {{\left (2 \, a^{3} c - 3 \, a^{3} d\right )} {\left (f x + e\right )}}{d^{3}}}{f} \] Input:

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

Output:

(2*(2*a^3*c^3 - a^3*c^2*d - 4*a^3*c*d^2 + 3*a^3*d^3)*(pi*floor(1/2*(f*x + 
e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)) 
)/((c*d^3 + d^4)*sqrt(c^2 - d^2)) - 2*(a^3*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 
2*a^3*c*d^2*tan(1/2*f*x + 1/2*e)^3 + a^3*d^3*tan(1/2*f*x + 1/2*e)^3 + 2*a^ 
3*c^3*tan(1/2*f*x + 1/2*e)^2 - a^3*c^2*d*tan(1/2*f*x + 1/2*e)^2 + a^3*c*d^ 
2*tan(1/2*f*x + 1/2*e)^2 + 3*a^3*c^2*d*tan(1/2*f*x + 1/2*e) + a^3*d^3*tan( 
1/2*f*x + 1/2*e) + 2*a^3*c^3 - a^3*c^2*d + a^3*c*d^2)/((c*tan(1/2*f*x + 1/ 
2*e)^4 + 2*d*tan(1/2*f*x + 1/2*e)^3 + 2*c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan 
(1/2*f*x + 1/2*e) + c)*(c^2*d^2 + c*d^3)) - (2*a^3*c - 3*a^3*d)*(f*x + e)/ 
d^3)/f
 

Mupad [B] (verification not implemented)

Time = 20.76 (sec) , antiderivative size = 5079, normalized size of antiderivative = 31.55 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*(2*a^3*c^2 + a^3*d^2 - a^3*c*d))/(d^2*(c + d)) + (2*tan(e/2 + (f*x)/ 
2)^2*(2*a^3*c^2 + a^3*d^2 - a^3*c*d))/(d^2*(c + d)) + (2*tan(e/2 + (f*x)/2 
)*(3*a^3*c^2 + a^3*d^2))/(c*d*(c + d)) + (2*tan(e/2 + (f*x)/2)^3*(a^3*c^2 
+ a^3*d^2 - 2*a^3*c*d))/(c*d*(c + d)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + 2* 
c*tan(e/2 + (f*x)/2)^2 + c*tan(e/2 + (f*x)/2)^4 + 2*d*tan(e/2 + (f*x)/2)^3 
)) - (2*a^3*atan(((a^3*(2*c - 3*d)*((32*(9*a^6*c^2*d^6 + 6*a^6*c^3*d^5 - 1 
1*a^6*c^4*d^4 - 4*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/(2*c*d^6 + d^7 + c^2*d^5) 
+ (32*tan(e/2 + (f*x)/2)*(9*a^6*c*d^8 + 36*a^6*c^2*d^7 - 41*a^6*c^3*d^6 - 
34*a^6*c^4*d^5 + 34*a^6*c^5*d^4 + 8*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/(2*c*d^7 
 + d^8 + c^2*d^6) + (a^3*(2*c - 3*d)*((32*tan(e/2 + (f*x)/2)*(6*a^3*c*d^10 
 - 2*a^3*c^2*d^9 - 10*a^3*c^3*d^8 + 2*a^3*c^4*d^7 + 4*a^3*c^5*d^6))/(2*c*d 
^7 + d^8 + c^2*d^6) - (32*(3*a^3*c*d^9 + a^3*c^2*d^8 - 3*a^3*c^3*d^7 - a^3 
*c^4*d^6))/(2*c*d^6 + d^7 + c^2*d^5) + (a^3*((32*(c^2*d^10 + 2*c^3*d^9 + c 
^4*d^8))/(2*c*d^6 + d^7 + c^2*d^5) + (32*tan(e/2 + (f*x)/2)*(3*c*d^12 + 6* 
c^2*d^11 + c^3*d^10 - 4*c^4*d^9 - 2*c^5*d^8))/(2*c*d^7 + d^8 + c^2*d^6))*( 
2*c - 3*d)*1i)/d^3)*1i)/d^3))/d^3 + (a^3*(2*c - 3*d)*((32*(9*a^6*c^2*d^6 + 
 6*a^6*c^3*d^5 - 11*a^6*c^4*d^4 - 4*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/(2*c*d^6 
 + d^7 + c^2*d^5) + (32*tan(e/2 + (f*x)/2)*(9*a^6*c*d^8 + 36*a^6*c^2*d^7 - 
 41*a^6*c^3*d^6 - 34*a^6*c^4*d^5 + 34*a^6*c^5*d^4 + 8*a^6*c^6*d^3 - 8*a^6* 
c^7*d^2))/(2*c*d^7 + d^8 + c^2*d^6) + (a^3*(2*c - 3*d)*((32*(3*a^3*c*d^...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 574, normalized size of antiderivative = 3.57 \[ \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {a^{3} \left (-\cos \left (f x +e \right ) \sin \left (f x +e \right ) d^{4}-2 \cos \left (f x +e \right ) c^{3} d -\cos \left (f x +e \right ) c^{2} d^{2}-2 c^{4} f x -\sin \left (f x +e \right ) d^{4}-\sin \left (f x +e \right ) c^{2} d^{2} f x -\cos \left (f x +e \right ) d^{4}-c^{3} d -2 c^{2} d^{2}-c \,d^{3}-\cos \left (f x +e \right ) \sin \left (f x +e \right ) c^{2} d^{2}-2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) c \,d^{3}-c^{3} d f x +4 c^{2} d^{2} f x +3 c \,d^{3} f x -\sin \left (f x +e \right ) c^{2} d^{2}-2 \sin \left (f x +e \right ) c \,d^{3}-6 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) d^{3}+2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) c^{2} d -6 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) c \,d^{2}+3 \sin \left (f x +e \right ) d^{4} f x +4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) c^{3}+4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) c^{2} d +2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) c \,d^{2}-2 \sin \left (f x +e \right ) c^{3} d f x +4 \sin \left (f x +e \right ) c \,d^{3} f x \right )}{d^{3} f \left (\sin \left (f x +e \right ) c^{2} d +2 \sin \left (f x +e \right ) c \,d^{2}+\sin \left (f x +e \right ) d^{3}+c^{3}+2 c^{2} d +c \,d^{2}\right )} \] Input:

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

Output:

(a**3*(4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2) 
)*sin(e + f*x)*c**2*d + 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/ 
sqrt(c**2 - d**2))*sin(e + f*x)*c*d**2 - 6*sqrt(c**2 - d**2)*atan((tan((e 
+ f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*d**3 + 4*sqrt(c**2 - d**2 
)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*c**3 + 2*sqrt(c**2 - d* 
*2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*c**2*d - 6*sqrt(c**2 
- d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*c*d**2 - cos(e + 
f*x)*sin(e + f*x)*c**2*d**2 - 2*cos(e + f*x)*sin(e + f*x)*c*d**3 - cos(e + 
 f*x)*sin(e + f*x)*d**4 - 2*cos(e + f*x)*c**3*d - cos(e + f*x)*c**2*d**2 - 
 cos(e + f*x)*d**4 - 2*sin(e + f*x)*c**3*d*f*x - sin(e + f*x)*c**2*d**2*f* 
x - sin(e + f*x)*c**2*d**2 + 4*sin(e + f*x)*c*d**3*f*x - 2*sin(e + f*x)*c* 
d**3 + 3*sin(e + f*x)*d**4*f*x - sin(e + f*x)*d**4 - 2*c**4*f*x - c**3*d*f 
*x - c**3*d + 4*c**2*d**2*f*x - 2*c**2*d**2 + 3*c*d**3*f*x - c*d**3))/(d** 
3*f*(sin(e + f*x)*c**2*d + 2*sin(e + f*x)*c*d**2 + sin(e + f*x)*d**3 + c** 
3 + 2*c**2*d + c*d**2))