\(\int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx\) [706]

Optimal result

Integrand size = 25, antiderivative size = 156 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=-\frac {d \left (6 a b c d-2 a^2 d^2-b^2 \left (6 c^2+d^2\right )\right ) x}{2 b^3}+\frac {2 (b c-a d)^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} f}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{2 b^2 f}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f} \] Output:

-1/2*d*(6*a*b*c*d-2*a^2*d^2-b^2*(6*c^2+d^2))*x/b^3+2*(-a*d+b*c)^3*arctan(( 
b+a*tan(1/2*f*x+1/2*e))/(a^2-b^2)^(1/2))/b^3/(a^2-b^2)^(1/2)/f-1/2*d^2*(-2 
*a*d+5*b*c)*cos(f*x+e)/b^2/f-1/2*d^2*cos(f*x+e)*(c+d*sin(f*x+e))/b/f
 

Mathematica [A] (verified)

Time = 2.70 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=\frac {2 d \left (-6 a b c d+2 a^2 d^2+b^2 \left (6 c^2+d^2\right )\right ) (e+f x)+\frac {8 (b c-a d)^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 b d^2 (3 b c-a d) \cos (e+f x)-b^2 d^3 \sin (2 (e+f x))}{4 b^3 f} \] Input:

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

Output:

(2*d*(-6*a*b*c*d + 2*a^2*d^2 + b^2*(6*c^2 + d^2))*(e + f*x) + (8*(b*c - a* 
d)^3*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 4 
*b*d^2*(3*b*c - a*d)*Cos[e + f*x] - b^2*d^3*Sin[2*(e + f*x)])/(4*b^3*f)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3272, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3272

\(\displaystyle \frac {\int \frac {2 b c^3+a d^3+d^2 (5 b c-2 a d) \sin ^2(e+f x)-d \left (a c d-b \left (6 c^2+d^2\right )\right ) \sin (e+f x)}{a+b \sin (e+f x)}dx}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 b c^3+a d^3+d^2 (5 b c-2 a d) \sin (e+f x)^2-d \left (a c d-b \left (6 c^2+d^2\right )\right ) \sin (e+f x)}{a+b \sin (e+f x)}dx}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\int \frac {b \left (2 b c^3+a d^3\right )-d \left (-\left (\left (6 c^2+d^2\right ) b^2\right )+6 a c d b-2 a^2 d^2\right ) \sin (e+f x)}{a+b \sin (e+f x)}dx}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {b \left (2 b c^3+a d^3\right )-d \left (-\left (\left (6 c^2+d^2\right ) b^2\right )+6 a c d b-2 a^2 d^2\right ) \sin (e+f x)}{a+b \sin (e+f x)}dx}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {\frac {2 (b c-a d)^3 \int \frac {1}{a+b \sin (e+f x)}dx}{b}-\frac {d x \left (-2 a^2 d^2+6 a b c d-\left (b^2 \left (6 c^2+d^2\right )\right )\right )}{b}}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 (b c-a d)^3 \int \frac {1}{a+b \sin (e+f x)}dx}{b}-\frac {d x \left (-2 a^2 d^2+6 a b c d-\left (b^2 \left (6 c^2+d^2\right )\right )\right )}{b}}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {\frac {4 (b c-a d)^3 \int \frac {1}{a \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 b \tan \left (\frac {1}{2} (e+f x)\right )+a}d\tan \left (\frac {1}{2} (e+f x)\right )}{b f}-\frac {d x \left (-2 a^2 d^2+6 a b c d-\left (b^2 \left (6 c^2+d^2\right )\right )\right )}{b}}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {4 (b c-a d)^3 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (e+f x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b f \sqrt {a^2-b^2}}-\frac {d x \left (-2 a^2 d^2+6 a b c d-\left (b^2 \left (6 c^2+d^2\right )\right )\right )}{b}}{b}-\frac {d^2 (5 b c-2 a d) \cos (e+f x)}{b f}}{2 b}-\frac {d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f}\)

Input:

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

Output:

((-((d*(6*a*b*c*d - 2*a^2*d^2 - b^2*(6*c^2 + d^2))*x)/b) + (4*(b*c - a*d)^ 
3*ArcTan[(2*b + 2*a*Tan[(e + f*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - 
b^2]*f))/b - (d^2*(5*b*c - 2*a*d)*Cos[e + f*x])/(b*f))/(2*b) - (d^2*Cos[e 
+ f*x]*(c + d*Sin[e + f*x]))/(2*b*f)
 

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 3272

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
 + n))   Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d 
*(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 
 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m 
] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] 
&& NeQ[c, 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 = 3.17 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.47

method	result	size
derivativedivides	\(\frac {\frac {2 \left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} d \,c^{2}+b^{3} c^{3}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}+\frac {2 d \left (\frac {\frac {d^{2} b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (a b \,d^{2}-3 b^{2} c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {d^{2} b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+a b \,d^{2}-3 b^{2} c d}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a^{2} d^{2}-6 a b c d +6 b^{2} c^{2}+d^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{b^{3}}}{f}\)	\(229\)
default	\(\frac {\frac {2 \left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} d \,c^{2}+b^{3} c^{3}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}+\frac {2 d \left (\frac {\frac {d^{2} b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (a b \,d^{2}-3 b^{2} c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {d^{2} b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+a b \,d^{2}-3 b^{2} c d}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a^{2} d^{2}-6 a b c d +6 b^{2} c^{2}+d^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{b^{3}}}{f}\)	\(229\)
risch	\(\frac {d^{3} x \,a^{2}}{b^{3}}-\frac {3 d^{2} x a c}{b^{2}}+\frac {3 d x \,c^{2}}{b}+\frac {d^{3} x}{2 b}+\frac {d^{3} {\mathrm e}^{i \left (f x +e \right )} a}{2 b^{2} f}-\frac {3 d^{2} {\mathrm e}^{i \left (f x +e \right )} c}{2 b f}+\frac {d^{3} {\mathrm e}^{-i \left (f x +e \right )} a}{2 b^{2} f}-\frac {3 d^{2} {\mathrm e}^{-i \left (f x +e \right )} c}{2 b f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{3} d^{3}}{\sqrt {-a^{2}+b^{2}}\, f \,b^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2} c \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,b^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a d \,c^{2}}{\sqrt {-a^{2}+b^{2}}\, f b}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) c^{3}}{\sqrt {-a^{2}+b^{2}}\, f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{3} d^{3}}{\sqrt {-a^{2}+b^{2}}\, f \,b^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2} c \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a d \,c^{2}}{\sqrt {-a^{2}+b^{2}}\, f b}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) c^{3}}{\sqrt {-a^{2}+b^{2}}\, f}-\frac {d^{3} \sin \left (2 f x +2 e \right )}{4 b f}\)	\(733\)

Input:

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

Output:

1/f*(2*(-a^3*d^3+3*a^2*b*c*d^2-3*a*b^2*c^2*d+b^3*c^3)/b^3/(a^2-b^2)^(1/2)* 
arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))+2*d/b^3*((1/2*d^2 
*b^2*tan(1/2*f*x+1/2*e)^3+(a*b*d^2-3*b^2*c*d)*tan(1/2*f*x+1/2*e)^2-1/2*d^2 
*b^2*tan(1/2*f*x+1/2*e)+a*b*d^2-3*b^2*c*d)/(1+tan(1/2*f*x+1/2*e)^2)^2+1/2* 
(2*a^2*d^2-6*a*b*c*d+6*b^2*c^2+b^2*d^2)*arctan(tan(1/2*f*x+1/2*e))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 566, normalized size of antiderivative = 3.63 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=\left [-\frac {{\left (a^{2} b^{2} - b^{4}\right )} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (6 \, {\left (a^{2} b^{2} - b^{4}\right )} c^{2} d - 6 \, {\left (a^{3} b - a b^{3}\right )} c d^{2} + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d^{3}\right )} f x - {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (3 \, {\left (a^{2} b^{2} - b^{4}\right )} c d^{2} - {\left (a^{3} b - a b^{3}\right )} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} f}, -\frac {{\left (a^{2} b^{2} - b^{4}\right )} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (6 \, {\left (a^{2} b^{2} - b^{4}\right )} c^{2} d - 6 \, {\left (a^{3} b - a b^{3}\right )} c d^{2} + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d^{3}\right )} f x + 2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + 2 \, {\left (3 \, {\left (a^{2} b^{2} - b^{4}\right )} c d^{2} - {\left (a^{3} b - a b^{3}\right )} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} f}\right ] \] Input:

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

Output:

[-1/2*((a^2*b^2 - b^4)*d^3*cos(f*x + e)*sin(f*x + e) - (6*(a^2*b^2 - b^4)* 
c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (2*a^4 - a^2*b^2 - b^4)*d^3)*f*x - (b^3* 
c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-a^2 + b^2)*log(-((2*a 
^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 - 2*(a*cos(f*x + 
 e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 
 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*(3*(a^2*b^2 - b^4)*c*d^2 - (a^3*b - 
a*b^3)*d^3)*cos(f*x + e))/((a^2*b^3 - b^5)*f), -1/2*((a^2*b^2 - b^4)*d^3*c 
os(f*x + e)*sin(f*x + e) - (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c* 
d^2 + (2*a^4 - a^2*b^2 - b^4)*d^3)*f*x + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^ 
2*b*c*d^2 - a^3*d^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^ 
2 - b^2)*cos(f*x + e))) + 2*(3*(a^2*b^2 - b^4)*c*d^2 - (a^3*b - a*b^3)*d^3 
)*cos(f*x + e))/((a^2*b^3 - b^5)*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((c+d*sin(f*x+e))^3/(a+b*sin(f*x+e)),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*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.56 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=\frac {\frac {{\left (6 \, b^{2} c^{2} d - 6 \, a b c d^{2} + 2 \, a^{2} d^{3} + b^{2} d^{3}\right )} {\left (f x + e\right )}}{b^{3}} + \frac {4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {2 \, {\left (b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, a d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, b c d^{2} + 2 \, a d^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, f} \] Input:

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

Output:

1/2*((6*b^2*c^2*d - 6*a*b*c*d^2 + 2*a^2*d^3 + b^2*d^3)*(f*x + e)/b^3 + 4*( 
b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(pi*floor(1/2*(f*x + e) 
/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/ 
(sqrt(a^2 - b^2)*b^3) + 2*(b*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*b*c*d^2*tan(1/ 
2*f*x + 1/2*e)^2 + 2*a*d^3*tan(1/2*f*x + 1/2*e)^2 - b*d^3*tan(1/2*f*x + 1/ 
2*e) - 6*b*c*d^2 + 2*a*d^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*b^2))/f
 

Mupad [B] (verification not implemented)

Time = 22.39 (sec) , antiderivative size = 5902, normalized size of antiderivative = 37.83 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(a*d^3 - 3*b*c*d^2))/b^2 + (d^3*tan(e/2 + (f*x)/2)^3)/b + (2*tan(e/2 + 
 (f*x)/2)^2*(a*d^3 - 3*b*c*d^2))/b^2 - (d^3*tan(e/2 + (f*x)/2))/b)/(f*(2*t 
an(e/2 + (f*x)/2)^2 + tan(e/2 + (f*x)/2)^4 + 1)) + (atan((((a^2*d^3*1i + ( 
b^2*d*(6*c^2 + d^2)*1i)/2 - a*b*c*d^2*3i)*((8*(a^2*b^6*d^6 + 4*a^4*b^4*d^6 
 + 4*a^6*b^2*d^6 - 12*a^3*b^5*c*d^5 - 24*a^5*b^3*c*d^5 + 12*a^2*b^6*c^2*d^ 
4 + 36*a^2*b^6*c^4*d^2 - 72*a^3*b^5*c^3*d^3 + 60*a^4*b^4*c^2*d^4))/b^5 + ( 
8*tan(e/2 + (f*x)/2)*(2*a*b^8*d^6 - 4*a*b^8*c^6 + 7*a^3*b^6*d^6 + 4*a^5*b^ 
4*d^6 - 8*a^7*b^2*d^6 + 24*a*b^8*c^2*d^4 + 72*a*b^8*c^4*d^2 - 24*a^2*b^7*c 
*d^5 + 24*a^2*b^7*c^5*d - 36*a^4*b^5*c*d^5 + 48*a^6*b^3*c*d^5 - 144*a^2*b^ 
7*c^3*d^3 + 108*a^3*b^6*c^2*d^4 - 96*a^3*b^6*c^4*d^2 + 152*a^4*b^5*c^3*d^3 
 - 120*a^5*b^4*c^2*d^4))/b^6 + ((a^2*d^3*1i + (b^2*d*(6*c^2 + d^2)*1i)/2 - 
 a*b*c*d^2*3i)*((8*tan(e/2 + (f*x)/2)*(8*a*b^9*c^3 - 8*a^4*b^6*d^3 - 24*a^ 
2*b^8*c^2*d + 24*a^3*b^7*c*d^2))/b^6 - (8*(2*a*b^8*d^3 - 4*a^2*b^7*c^3 + 2 
*a^3*b^6*d^3 - 12*a^2*b^7*c*d^2 + 12*a*b^8*c^2*d))/b^5 + ((32*a^2*b^3 + (8 
*tan(e/2 + (f*x)/2)*(12*a*b^10 - 8*a^3*b^8))/b^6)*(a^2*d^3*1i + (b^2*d*(6* 
c^2 + d^2)*1i)/2 - a*b*c*d^2*3i))/b^3))/b^3)*1i)/b^3 + ((a^2*d^3*1i + (b^2 
*d*(6*c^2 + d^2)*1i)/2 - a*b*c*d^2*3i)*((8*(a^2*b^6*d^6 + 4*a^4*b^4*d^6 + 
4*a^6*b^2*d^6 - 12*a^3*b^5*c*d^5 - 24*a^5*b^3*c*d^5 + 12*a^2*b^6*c^2*d^4 + 
 36*a^2*b^6*c^4*d^2 - 72*a^3*b^5*c^3*d^3 + 60*a^4*b^4*c^2*d^4))/b^5 + (8*t 
an(e/2 + (f*x)/2)*(2*a*b^8*d^6 - 4*a*b^8*c^6 + 7*a^3*b^6*d^6 + 4*a^5*b^...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.99 \[ \int \frac {(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} d^{3}+12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b c \,d^{2}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a \,b^{2} c^{2} d +4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b^{3} c^{3}-\cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2} b^{2} d^{3}+\cos \left (f x +e \right ) \sin \left (f x +e \right ) b^{4} d^{3}+2 \cos \left (f x +e \right ) a^{3} b \,d^{3}-6 \cos \left (f x +e \right ) a^{2} b^{2} c \,d^{2}-2 \cos \left (f x +e \right ) a \,b^{3} d^{3}+6 \cos \left (f x +e \right ) b^{4} c \,d^{2}+2 a^{4} d^{3} e +2 a^{4} d^{3} f x -6 a^{3} b c \,d^{2} e -6 a^{3} b c \,d^{2} f x +6 a^{2} b^{2} c^{2} d e +6 a^{2} b^{2} c^{2} d f x -a^{2} b^{2} d^{3} e -a^{2} b^{2} d^{3} f x +6 a \,b^{3} c \,d^{2} e +6 a \,b^{3} c \,d^{2} f x -6 b^{4} c^{2} d e -6 b^{4} c^{2} d f x -b^{4} d^{3} e -b^{4} d^{3} f x}{2 b^{3} f \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a 
**3*d**3 + 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - 
b**2))*a**2*b*c*d**2 - 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/ 
sqrt(a**2 - b**2))*a*b**2*c**2*d + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x) 
/2)*a + b)/sqrt(a**2 - b**2))*b**3*c**3 - cos(e + f*x)*sin(e + f*x)*a**2*b 
**2*d**3 + cos(e + f*x)*sin(e + f*x)*b**4*d**3 + 2*cos(e + f*x)*a**3*b*d** 
3 - 6*cos(e + f*x)*a**2*b**2*c*d**2 - 2*cos(e + f*x)*a*b**3*d**3 + 6*cos(e 
 + f*x)*b**4*c*d**2 + 2*a**4*d**3*e + 2*a**4*d**3*f*x - 6*a**3*b*c*d**2*e 
- 6*a**3*b*c*d**2*f*x + 6*a**2*b**2*c**2*d*e + 6*a**2*b**2*c**2*d*f*x - a* 
*2*b**2*d**3*e - a**2*b**2*d**3*f*x + 6*a*b**3*c*d**2*e + 6*a*b**3*c*d**2* 
f*x - 6*b**4*c**2*d*e - 6*b**4*c**2*d*f*x - b**4*d**3*e - b**4*d**3*f*x)/( 
2*b**3*f*(a**2 - b**2))