\(\int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx\) [698]

Optimal result

Integrand size = 25, antiderivative size = 208 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=-\frac {b^2 (2 b c-3 a d) x}{d^3}+\frac {2 (b c-a d)^2 \left (2 b c^2+a c d-3 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \left (c^2-d^2\right )^{3/2} f}+\frac {b \left (2 a b c d-a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x)}{d^2 \left (c^2-d^2\right ) f}+\frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d \left (c^2-d^2\right ) f (c+d \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.74 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {-b^2 (2 b c-3 a d) (e+f x)+\frac {2 (b c-a d)^2 \left (2 b c^2+a c d-3 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}-b^3 d \cos (e+f x)+\frac {d (-b c+a d)^3 \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{d^3 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

-((((b^2*(2*b*c - 3*a*d)*(c^2 - d^2)*x)/d - (2*(b*c - a*d)^2*(2*b*c^2 + a* 
c*d - 3*b*d^2)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/( 
d*Sqrt[c^2 - d^2]*f))/d - (b*(2*a*b*c*d - a^2*d^2 - b^2*(2*c^2 - d^2))*Cos 
[e + f*x])/(d*f))/(d*(c^2 - d^2))) + ((b*c - a*d)^2*Cos[e + f*x]*(a + b*Si 
n[e + f*x]))/(d*(c^2 - d^2)*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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

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 = 3.61 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\frac {d^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} d \,c^{2}-b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} d \,c^{2}-b^{3} c^{3}\right )}{c^{2}-d^{2}}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{3} c \,d^{3}-3 a^{2} b \,d^{4}-3 a \,b^{2} c^{3} d +6 a \,b^{2} c \,d^{3}+2 b^{3} c^{4}-3 b^{3} c^{2} d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{3}}+\frac {2 b^{2} \left (-\frac {b d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (3 a d -2 b c \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{d^{3}}}{f}\)	\(302\)
default	\(\frac {\frac {\frac {2 \left (\frac {d^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} d \,c^{2}-b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} d \,c^{2}-b^{3} c^{3}\right )}{c^{2}-d^{2}}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{3} c \,d^{3}-3 a^{2} b \,d^{4}-3 a \,b^{2} c^{3} d +6 a \,b^{2} c \,d^{3}+2 b^{3} c^{4}-3 b^{3} c^{2} d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{3}}+\frac {2 b^{2} \left (-\frac {b d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (3 a d -2 b c \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{d^{3}}}{f}\)	\(302\)
risch	\(\text {Expression too large to display}\)	\(1185\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (203) = 406\).

Time = 0.14 (sec) , antiderivative size = 1062, normalized size of antiderivative = 5.11 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*(2*(2*b^3*c^6 - 3*a*b^2*c^5*d - 4*b^3*c^4*d^2 + 6*a*b^2*c^3*d^3 + 2* 
b^3*c^2*d^4 - 3*a*b^2*c*d^5)*f*x + (2*b^3*c^5 - 3*a*b^2*c^4*d - 3*b^3*c^3* 
d^2 - 3*a^2*b*c*d^4 + (a^3 + 6*a*b^2)*c^2*d^3 + (2*b^3*c^4*d - 3*a*b^2*c^3 
*d^2 - 3*b^3*c^2*d^3 - 3*a^2*b*d^5 + (a^3 + 6*a*b^2)*c*d^4)*sin(f*x + e))* 
sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - 
c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d 
^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*(2*b^3*c^5 
*d - 3*a*b^2*c^4*d^2 + a^3*d^6 + 3*(a^2*b - b^3)*c^3*d^3 - (a^3 - 3*a*b^2) 
*c^2*d^4 - (3*a^2*b - b^3)*c*d^5)*cos(f*x + e) + 2*((2*b^3*c^5*d - 3*a*b^2 
*c^4*d^2 - 4*b^3*c^3*d^3 + 6*a*b^2*c^2*d^4 + 2*b^3*c*d^5 - 3*a*b^2*d^6)*f* 
x + (b^3*c^4*d^2 - 2*b^3*c^2*d^4 + b^3*d^6)*cos(f*x + e))*sin(f*x + e))/(( 
c^4*d^4 - 2*c^2*d^6 + d^8)*f*sin(f*x + e) + (c^5*d^3 - 2*c^3*d^5 + c*d^7)* 
f), -((2*b^3*c^6 - 3*a*b^2*c^5*d - 4*b^3*c^4*d^2 + 6*a*b^2*c^3*d^3 + 2*b^3 
*c^2*d^4 - 3*a*b^2*c*d^5)*f*x + (2*b^3*c^5 - 3*a*b^2*c^4*d - 3*b^3*c^3*d^2 
 - 3*a^2*b*c*d^4 + (a^3 + 6*a*b^2)*c^2*d^3 + (2*b^3*c^4*d - 3*a*b^2*c^3*d^ 
2 - 3*b^3*c^2*d^3 - 3*a^2*b*d^5 + (a^3 + 6*a*b^2)*c*d^4)*sin(f*x + e))*sqr 
t(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) 
+ (2*b^3*c^5*d - 3*a*b^2*c^4*d^2 + a^3*d^6 + 3*(a^2*b - b^3)*c^3*d^3 - (a^ 
3 - 3*a*b^2)*c^2*d^4 - (3*a^2*b - b^3)*c*d^5)*cos(f*x + e) + ((2*b^3*c^5*d 
 - 3*a*b^2*c^4*d^2 - 4*b^3*c^3*d^3 + 6*a*b^2*c^2*d^4 + 2*b^3*c*d^5 - 3*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+b*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. 564 vs. \(2 (203) = 406\).

Time = 0.16 (sec) , antiderivative size = 564, normalized size of antiderivative = 2.71 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(2*b^3*c^4 - 3*a*b^2*c^3*d - 3*b^3*c^2*d^2 + a^3*c*d^3 + 6*a*b^2*c*d^3 
- 3*a^2*b*d^4)*(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^2*d^3 - d^5)*sqrt(c^2 - d^2)) - 
2*(b^3*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) 
^3 + 3*a^2*b*c*d^3*tan(1/2*f*x + 1/2*e)^3 - a^3*d^4*tan(1/2*f*x + 1/2*e)^3 
 + 2*b^3*c^4*tan(1/2*f*x + 1/2*e)^2 - 3*a*b^2*c^3*d*tan(1/2*f*x + 1/2*e)^2 
 + 3*a^2*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 - b^3*c^2*d^2*tan(1/2*f*x + 1/2* 
e)^2 - a^3*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 3*b^3*c^3*d*tan(1/2*f*x + 1/2*e) 
 - 3*a*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) + 3*a^2*b*c*d^3*tan(1/2*f*x + 1/2* 
e) - 2*b^3*c*d^3*tan(1/2*f*x + 1/2*e) - a^3*d^4*tan(1/2*f*x + 1/2*e) + 2*b 
^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - b^3*c^2*d^2 - a^3*c*d^3)/((c^3* 
d^2 - c*d^4)*(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)) - (2*b^3*c - 3*a*b 
^2*d)*(f*x + e)/d^3)/f
 

Mupad [B] (verification not implemented)

Time = 25.47 (sec) , antiderivative size = 8953, normalized size of antiderivative = 43.04 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(a^3*d^3 - 2*b^3*c^3 + b^3*c*d^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(d^ 
2*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2)^2*(a^3*d^3 - 2*b^3*c^3 + b^3*c*d^2 
+ 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(d^2*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2 
)*(a^3*d^3 - 3*b^3*c^3 + 2*b^3*c*d^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(c* 
d*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2)^3*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2* 
d - 3*a^2*b*c*d^2))/(c*d*(c^2 - d^2)))/(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*b^2*atan(((b^2*(3*a*d - 2*b*c)*((32*(4*b^6*c^4*d^6 - 8*b^6*c^6*d^4 
 + 4*b^6*c^8*d^2 - 12*a*b^5*c^3*d^7 + 24*a*b^5*c^5*d^5 - 12*a*b^5*c^7*d^3 
+ 9*a^2*b^4*c^2*d^8 - 18*a^2*b^4*c^4*d^6 + 9*a^2*b^4*c^6*d^4))/(d^9 - 2*c^ 
2*d^7 + c^4*d^5) - (32*tan(e/2 + (f*x)/2)*(a^6*c^3*d^8 - 8*b^6*c^3*d^8 + 2 
9*b^6*c^5*d^6 - 28*b^6*c^7*d^4 + 8*b^6*c^9*d^2 + 24*a*b^5*c^2*d^9 - 96*a*b 
^5*c^4*d^7 + 90*a*b^5*c^6*d^5 - 24*a*b^5*c^8*d^3 - 18*a^2*b^4*c*d^10 + 9*a 
^4*b^2*c*d^10 - 6*a^5*b*c^2*d^9 + 99*a^2*b^4*c^3*d^8 - 84*a^2*b^4*c^5*d^6 
+ 18*a^2*b^4*c^7*d^4 - 36*a^3*b^3*c^2*d^9 + 12*a^3*b^3*c^4*d^7 + 4*a^3*b^3 
*c^6*d^5 + 12*a^4*b^2*c^3*d^8 - 6*a^4*b^2*c^5*d^6))/(d^10 - 2*c^2*d^8 + c^ 
4*d^6) + (b^2*(3*a*d - 2*b*c)*((32*tan(e/2 + (f*x)/2)*(2*a^3*c^2*d^11 - 2* 
a^3*c^4*d^9 - 6*b^3*c^3*d^10 + 10*b^3*c^5*d^8 - 4*b^3*c^7*d^6 + 12*a*b^2*c 
^2*d^11 - 18*a*b^2*c^4*d^9 + 6*a*b^2*c^6*d^7 + 6*a^2*b*c^3*d^10 - 6*a^2*b* 
c*d^12))/(d^10 - 2*c^2*d^8 + c^4*d^6) - (32*(a^3*c^5*d^7 - a^3*c^3*d^9 ...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1138, normalized size of antiderivative = 5.47 \[ \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin( 
e + f*x)*a**3*c*d**4 - 6*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/s 
qrt(c**2 - d**2))*sin(e + f*x)*a**2*b*d**5 - 6*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*a*b**2*c**3*d**2 + 12 
*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e 
+ f*x)*a*b**2*c*d**4 + 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/s 
qrt(c**2 - d**2))*sin(e + f*x)*b**3*c**4*d - 6*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b**3*c**2*d**3 + 2*sq 
rt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a**3*c**2 
*d**3 - 6*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2 
))*a**2*b*c*d**4 - 6*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt( 
c**2 - d**2))*a*b**2*c**4*d + 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)* 
c + d)/sqrt(c**2 - d**2))*a*b**2*c**2*d**3 + 4*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b**3*c**5 - 6*sqrt(c**2 - d**2)*at 
an((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b**3*c**3*d**2 - cos(e + f* 
x)*sin(e + f*x)*b**3*c**4*d**2 + 2*cos(e + f*x)*sin(e + f*x)*b**3*c**2*d** 
4 - cos(e + f*x)*sin(e + f*x)*b**3*d**6 + cos(e + f*x)*a**3*c**2*d**4 - co 
s(e + f*x)*a**3*d**6 - 3*cos(e + f*x)*a**2*b*c**3*d**3 + 3*cos(e + f*x)*a* 
*2*b*c*d**5 + 3*cos(e + f*x)*a*b**2*c**4*d**2 - 3*cos(e + f*x)*a*b**2*c**2 
*d**4 - 2*cos(e + f*x)*b**3*c**5*d + 3*cos(e + f*x)*b**3*c**3*d**3 - co...