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

Optimal. Leaf size=143 \[ -\frac {2 a^3 (c-d)^3 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d^3 f \sqrt {c^2-d^2}}+\frac {a^3 x \left (2 c^2-6 c d+7 d^2\right )}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f} \]

[Out]

1/2*a^3*(2*c^2-6*c*d+7*d^2)*x/d^3+1/2*a^3*(2*c-5*d)*cos(f*x+e)/d^2/f-1/2*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/f-2
*a^3*(c-d)^3*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/d^3/f/(c^2-d^2)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2763, 2968, 3023, 2735, 2660, 618, 204} \[ -\frac {2 a^3 (c-d)^3 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d^3 f \sqrt {c^2-d^2}}+\frac {a^3 x \left (2 c^2-6 c d+7 d^2\right )}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(2*c^2 - 6*c*d + 7*d^2)*x)/(2*d^3) - (2*a^3*(c - d)^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(
d^3*Sqrt[c^2 - d^2]*f) + (a^3*(2*c - 5*d)*Cos[e + f*x])/(2*d^2*f) - (Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(2
*d*f)

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(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 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 2968

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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[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]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx &=-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}+\frac {\int \frac {(a+a \sin (e+f x)) \left (a^2 (c+2 d)-a^2 (2 c-5 d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{2 d}\\ &=-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}+\frac {\int \frac {a^3 (c+2 d)+\left (-a^3 (2 c-5 d)+a^3 (c+2 d)\right ) \sin (e+f x)-a^3 (2 c-5 d) \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx}{2 d}\\ &=\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}+\frac {\int \frac {a^3 d (c+2 d)+a^3 \left (2 c^2-6 c d+7 d^2\right ) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{2 d^2}\\ &=\frac {a^3 \left (2 c^2-6 c d+7 d^2\right ) x}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}-\frac {\left (a^3 (c-d)^3\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{d^3}\\ &=\frac {a^3 \left (2 c^2-6 c d+7 d^2\right ) x}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}-\frac {\left (2 a^3 (c-d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{d^3 f}\\ &=\frac {a^3 \left (2 c^2-6 c d+7 d^2\right ) x}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}+\frac {\left (4 a^3 (c-d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d^3 f}\\ &=\frac {a^3 \left (2 c^2-6 c d+7 d^2\right ) x}{2 d^3}-\frac {2 a^3 (c-d)^3 \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \sqrt {c^2-d^2} f}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d f}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 162, normalized size = 1.13 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\sqrt {c^2-d^2} \left (2 \left (2 c^2-6 c d+7 d^2\right ) (e+f x)+4 d (c-3 d) \cos (e+f x)+d^2 (-\sin (2 (e+f x)))\right )-8 (c-d)^3 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )\right )}{4 d^3 f \sqrt {c^2-d^2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[e + f*x])^3*(-8*(c - d)^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]] + Sqrt[c^2 - d^2]*(2*
(2*c^2 - 6*c*d + 7*d^2)*(e + f*x) + 4*(c - 3*d)*d*Cos[e + f*x] - d^2*Sin[2*(e + f*x)])))/(4*d^3*Sqrt[c^2 - d^2
]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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fricas [A]  time = 0.51, size = 404, normalized size = 2.83 \[ \left [-\frac {a^{3} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} f x - {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}, -\frac {a^{3} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} f x - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^3*d^2*cos(f*x + e)*sin(f*x + e) - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*f*x - (a^3*c^2 - 2*a^3*c*d + a^
3*d^2)*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*(a^3*c*d - 3*a^3*d^2)*cos(f*x + e))/(d^3*f), -1/2*(a^3*d^2*cos(f*x + e)*sin(f*x
+ e) - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*f*x - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*sqrt((c - d)/(c + d))*arct
an(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) - 2*(a^3*c*d - 3*a^3*d^2)*cos(f*x + e))
/(d^3*f)]

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giac [A]  time = 0.26, size = 239, normalized size = 1.67 \[ \frac {\frac {{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} {\left (f x + e\right )}}{d^{3}} - \frac {4 \, {\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{3}} + \frac {2 \, {\left (a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{3} c - 6 \, a^{3} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*(f*x + e)/d^3 - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - 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)))/(sqrt(c^2 - d^2)
*d^3) + 2*(a^3*d*tan(1/2*f*x + 1/2*e)^3 + 2*a^3*c*tan(1/2*f*x + 1/2*e)^2 - 6*a^3*d*tan(1/2*f*x + 1/2*e)^2 - a^
3*d*tan(1/2*f*x + 1/2*e) + 2*a^3*c - 6*a^3*d)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*d^2))/f

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maple [B]  time = 0.26, size = 480, normalized size = 3.36 \[ -\frac {2 a^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) c^{3}}{f \,d^{3} \sqrt {c^{2}-d^{2}}}+\frac {6 a^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) c^{2}}{f \,d^{2} \sqrt {c^{2}-d^{2}}}-\frac {6 a^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) c}{f d \sqrt {c^{2}-d^{2}}}+\frac {2 a^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{f \sqrt {c^{2}-d^{2}}}+\frac {a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f d \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {2 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c}{f \,d^{2} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {6 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f d \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f d \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {2 a^{3} c}{f \,d^{2} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {6 a^{3}}{f d \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {2 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2}}{f \,d^{3}}-\frac {6 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c}{f \,d^{2}}+\frac {7 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*a^3/d^3/(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))*c^3+6/f*a^3/d^2/(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))*c^2-6/f*a^3/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))*c+2/f*a^3/(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/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3+2/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^
2*tan(1/2*f*x+1/2*e)^2*c-6/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2-1/f*a^3/d/(1+tan(1/2*f*x+1/
2*e)^2)^2*tan(1/2*f*x+1/2*e)+2/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^2*c-6/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^2+2/f
*a^3/d^3*arctan(tan(1/2*f*x+1/2*e))*c^2-6/f*a^3/d^2*arctan(tan(1/2*f*x+1/2*e))*c+7/f*a^3/d*arctan(tan(1/2*f*x+
1/2*e))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.22, size = 3382, normalized size = 23.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(a^3*c - 3*a^3*d))/d^2 + (a^3*tan(e/2 + (f*x)/2)^3)/d - (a^3*tan(e/2 + (f*x)/2))/d + (2*tan(e/2 + (f*x)/2)
^2*(a^3*c - 3*a^3*d))/d^2)/(f*(2*tan(e/2 + (f*x)/2)^2 + tan(e/2 + (f*x)/2)^4 + 1)) + (2*a^3*atan(((a^3*((8*(49
*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*
(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*
c^7*d^2))/d^6 + (a^3*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d
^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*
d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (
7*d^2)/2))/d^3 + (a^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))
/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*
d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (
8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 +
(8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)
/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2))/d^3)/((16*(2*a^9*c^7 - 14*a^9*c*d^6 - 8*a^9*c^6*d + 47*a^9*c^2*d^5 - 5
5*a^9*c^3*d^4 + 21*a^9*c^4*d^3 + 7*a^9*c^5*d^2))/d^5 + (16*tan(e/2 + (f*x)/2)*(8*a^9*c^8 - 98*a^9*c*d^7 - 72*a
^9*c^7*d + 462*a^9*c^2*d^6 - 926*a^9*c^3*d^5 + 1034*a^9*c^4*d^4 - 704*a^9*c^5*d^3 + 296*a^9*c^6*d^2))/d^6 - (a
^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 +
(f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^
3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c
^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)
/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 -
3*c*d + (7*d^2)/2)*1i)/d^3 + (a^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*
a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 -
116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*
d^6))/d^5 - (8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*
(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*
c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3))*(c^2 - 3*c*d + (7*d^2)/2))/(d^3*f) + (a^3*atan((
(a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6
*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 11
6*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*tan(e/2 + (f*x)/2)*
(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a
^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)
^(1/2))/(d^3*(c + d))))/(d^3*(c + d)))*1i)/(d^3*(c + d)) + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6
- 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^
8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^
6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (8*tan(e/2 + (f
*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 + (8*tan(e/2 +
(f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d)))*1i)/(d^3*(c
+ d)))/((16*(2*a^9*c^7 - 14*a^9*c*d^6 - 8*a^9*c^6*d + 47*a^9*c^2*d^5 - 55*a^9*c^3*d^4 + 21*a^9*c^4*d^3 + 7*a^9
*c^5*d^2))/d^5 + (16*tan(e/2 + (f*x)/2)*(8*a^9*c^8 - 98*a^9*c*d^7 - 72*a^9*c^7*d + 462*a^9*c^2*d^6 - 926*a^9*c
^3*d^5 + 1034*a^9*c^4*d^4 - 704*a^9*c^5*d^3 + 296*a^9*c^6*d^2))/d^6 - (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*
a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(
94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c
^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c
^3*d^7 - 8*a^3*c^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8
*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d))))/
(d^3*(c + d)) + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6
*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*
a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(14
*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a
^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c
+ d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d))))/(d^3*(c + d))))*(-(c + d)*(c - d)^5)^(1/2)*2i)/(d^3*f*
(c + d))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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