3.478 \(\int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=298 \[ -\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))}-\frac {2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^4 (c+d) \sqrt {c^2-d^2}}-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))} \]

[Out]

-1/15*d*(2*c^3-12*c^2*d+43*c*d^2+72*d^3)*cos(f*x+e)/a^3/(c-d)^4/(c+d)/f/(c+d*sin(f*x+e))-1/5*cos(f*x+e)/(c-d)/
f/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))-1/15*(2*c-9*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)
)-1/15*(2*c^2-12*c*d+45*d^2)*cos(f*x+e)/(c-d)^3/f/(a^3+a^3*sin(f*x+e))/(c+d*sin(f*x+e))-2*d^3*(4*c+3*d)*arctan
((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a^3/(c-d)^4/(c+d)/f/(c^2-d^2)^(1/2)

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Rubi [A]  time = 0.73, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2766, 2978, 2754, 12, 2660, 618, 204} \[ -\frac {2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^4 (c+d) \sqrt {c^2-d^2}}-\frac {d \left (-12 c^2 d+2 c^3+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2),x]

[Out]

(-2*d^3*(4*c + 3*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^4*(c + d)*Sqrt[c^2 - d^2]*f
) - (d*(2*c^3 - 12*c^2*d + 43*c*d^2 + 72*d^3)*Cos[e + f*x])/(15*a^3*(c - d)^4*(c + d)*f*(c + d*Sin[e + f*x]))
- Cos[e + f*x]/(5*(c - d)*f*(a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])) - ((2*c - 9*d)*Cos[e + f*x])/(15*a*(c
- d)^2*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])) - ((2*c^2 - 12*c*d + 45*d^2)*Cos[e + f*x])/(15*(c - d)^
3*f*(a^3 + a^3*Sin[e + f*x])*(c + d*Sin[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2754

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

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

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

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {\int \frac {-2 a (c-3 d)-3 a d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx}{5 a^2 (c-d)}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac {\int \frac {a^2 \left (2 c^2-8 c d+27 d^2\right )+2 a^2 (2 c-9 d) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac {\int \frac {-2 a^3 (c-36 d) d^2-a^3 d \left (2 c^2-12 c d+45 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}+\frac {\int -\frac {15 a^3 d^3 (4 c+3 d)}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^4 (c+d)}\\ &=-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac {\left (d^3 (4 c+3 d)\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^4 (c+d)}\\ &=-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac {\left (2 d^3 (4 c+3 d)\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 )}{a^3 (c-d)^4 (c+d) f}\\ &=-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}+\frac {\left (4 d^3 (4 c+3 d)\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 )}{a^3 (c-d)^4 (c+d) f}\\ &=-\frac {2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^4 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 2.55, size = 361, normalized size = 1.21 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 \left (2 c^2-14 c d+57 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4-\frac {30 d^3 (4 c+3 d) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}-\frac {15 d^4 \cos (e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d) (c+d \sin (e+f x))}+6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-2 (c-6 d) (c-d) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3+4 (c-6 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-3 (c-d)^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{15 a^3 f (c-d)^4 (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(c - d)^2*Sin[(e + f*x)/2] - 3*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2]) + 4*(c - 6*d)*(c - d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*(c - 6*d)*(c - d
)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 2*(2*c^2 - 14*c*d + 57*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^4 - (30*d^3*(4*c + 3*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5)/((c + d)*Sqrt[c^2 - d^2]) - (15*d^4*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
/((c + d)*(c + d*Sin[e + f*x]))))/(15*a^3*(c - d)^4*f*(1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.60, size = 3235, normalized size = 10.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/30*(6*c^6 - 12*c^5*d - 6*c^4*d^2 + 24*c^3*d^3 - 6*c^2*d^4 - 12*c*d^5 + 6*d^6 - 2*(2*c^5*d - 12*c^4*d^2 + 4
1*c^3*d^3 + 84*c^2*d^4 - 43*c*d^5 - 72*d^6)*cos(f*x + e)^4 - 2*(2*c^6 - 6*c^5*d + 5*c^4*d^2 + 147*c^3*d^3 + 16
4*c^2*d^4 - 141*c*d^5 - 171*d^6)*cos(f*x + e)^3 + 2*(4*c^6 - 19*c^5*d + 22*c^4*d^2 + 128*c^3*d^3 + 64*c^2*d^4
- 109*c*d^5 - 90*d^6)*cos(f*x + e)^2 + 15*(16*c^2*d^3 + 28*c*d^4 + 12*d^5 + (4*c*d^4 + 3*d^5)*cos(f*x + e)^4 -
(4*c^2*d^3 + 11*c*d^4 + 6*d^5)*cos(f*x + e)^3 - (12*c^2*d^3 + 29*c*d^4 + 15*d^5)*cos(f*x + e)^2 + 2*(4*c^2*d^
3 + 7*c*d^4 + 3*d^5)*cos(f*x + e) + (16*c^2*d^3 + 28*c*d^4 + 12*d^5 - (4*c*d^4 + 3*d^5)*cos(f*x + e)^3 - (4*c^
2*d^3 + 15*c*d^4 + 9*d^5)*cos(f*x + e)^2 + 2*(4*c^2*d^3 + 7*c*d^4 + 3*d^5)*cos(f*x + e))*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)) + 6*(3*c^6 - 11*c^
5*d + 12*c^4*d^2 + 82*c^3*d^3 + 47*c^2*d^4 - 71*c*d^5 - 62*d^6)*cos(f*x + e) - 2*(3*c^6 - 6*c^5*d - 3*c^4*d^2
+ 12*c^3*d^3 - 3*c^2*d^4 - 6*c*d^5 + 3*d^6 + (2*c^5*d - 12*c^4*d^2 + 41*c^3*d^3 + 84*c^2*d^4 - 43*c*d^5 - 72*d
^6)*cos(f*x + e)^3 - (2*c^6 - 8*c^5*d + 17*c^4*d^2 + 106*c^3*d^3 + 80*c^2*d^4 - 98*c*d^5 - 99*d^6)*cos(f*x + e
)^2 - 3*(2*c^6 - 9*c^5*d + 13*c^4*d^2 + 78*c^3*d^3 + 48*c^2*d^4 - 69*c*d^5 - 63*d^6)*cos(f*x + e))*sin(f*x + e
))/((a^3*c^7*d - 3*a^3*c^6*d^2 + a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 5*a^3*c^3*d^5 - a^3*c^2*d^6 + 3*a^3*c*d^7 - a^3
*d^8)*f*cos(f*x + e)^4 - (a^3*c^8 - a^3*c^7*d - 5*a^3*c^6*d^2 + 7*a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 11*a^3*c^3*d^5
+ a^3*c^2*d^6 + 5*a^3*c*d^7 - 2*a^3*d^8)*f*cos(f*x + e)^3 - (3*a^3*c^8 - 4*a^3*c^7*d - 12*a^3*c^6*d^2 + 20*a^
3*c^5*d^3 + 10*a^3*c^4*d^4 - 28*a^3*c^3*d^5 + 4*a^3*c^2*d^6 + 12*a^3*c*d^7 - 5*a^3*d^8)*f*cos(f*x + e)^2 + 2*(
a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)
*f*cos(f*x + e) + 4*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2
*a^3*c*d^7 - a^3*d^8)*f - ((a^3*c^7*d - 3*a^3*c^6*d^2 + a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 5*a^3*c^3*d^5 - a^3*c^2*
d^6 + 3*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e)^3 + (a^3*c^8 - 8*a^3*c^6*d^2 + 8*a^3*c^5*d^3 + 10*a^3*c^4*d^4 - 16
*a^3*c^3*d^5 + 8*a^3*c*d^7 - 3*a^3*d^8)*f*cos(f*x + e)^2 - 2*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^
5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e) - 4*(a^3*c^8 - 2*a^3*c^7*d - 2*a
^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f)*sin(f*x + e)), -1/15*(3
*c^6 - 6*c^5*d - 3*c^4*d^2 + 12*c^3*d^3 - 3*c^2*d^4 - 6*c*d^5 + 3*d^6 - (2*c^5*d - 12*c^4*d^2 + 41*c^3*d^3 + 8
4*c^2*d^4 - 43*c*d^5 - 72*d^6)*cos(f*x + e)^4 - (2*c^6 - 6*c^5*d + 5*c^4*d^2 + 147*c^3*d^3 + 164*c^2*d^4 - 141
*c*d^5 - 171*d^6)*cos(f*x + e)^3 + (4*c^6 - 19*c^5*d + 22*c^4*d^2 + 128*c^3*d^3 + 64*c^2*d^4 - 109*c*d^5 - 90*
d^6)*cos(f*x + e)^2 - 15*(16*c^2*d^3 + 28*c*d^4 + 12*d^5 + (4*c*d^4 + 3*d^5)*cos(f*x + e)^4 - (4*c^2*d^3 + 11*
c*d^4 + 6*d^5)*cos(f*x + e)^3 - (12*c^2*d^3 + 29*c*d^4 + 15*d^5)*cos(f*x + e)^2 + 2*(4*c^2*d^3 + 7*c*d^4 + 3*d
^5)*cos(f*x + e) + (16*c^2*d^3 + 28*c*d^4 + 12*d^5 - (4*c*d^4 + 3*d^5)*cos(f*x + e)^3 - (4*c^2*d^3 + 15*c*d^4
+ 9*d^5)*cos(f*x + e)^2 + 2*(4*c^2*d^3 + 7*c*d^4 + 3*d^5)*cos(f*x + e))*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-
(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + 3*(3*c^6 - 11*c^5*d + 12*c^4*d^2 + 82*c^3*d^3 + 47*c^2*
d^4 - 71*c*d^5 - 62*d^6)*cos(f*x + e) - (3*c^6 - 6*c^5*d - 3*c^4*d^2 + 12*c^3*d^3 - 3*c^2*d^4 - 6*c*d^5 + 3*d^
6 + (2*c^5*d - 12*c^4*d^2 + 41*c^3*d^3 + 84*c^2*d^4 - 43*c*d^5 - 72*d^6)*cos(f*x + e)^3 - (2*c^6 - 8*c^5*d + 1
7*c^4*d^2 + 106*c^3*d^3 + 80*c^2*d^4 - 98*c*d^5 - 99*d^6)*cos(f*x + e)^2 - 3*(2*c^6 - 9*c^5*d + 13*c^4*d^2 + 7
8*c^3*d^3 + 48*c^2*d^4 - 69*c*d^5 - 63*d^6)*cos(f*x + e))*sin(f*x + e))/((a^3*c^7*d - 3*a^3*c^6*d^2 + a^3*c^5*
d^3 + 5*a^3*c^4*d^4 - 5*a^3*c^3*d^5 - a^3*c^2*d^6 + 3*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e)^4 - (a^3*c^8 - a^3*c
^7*d - 5*a^3*c^6*d^2 + 7*a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 11*a^3*c^3*d^5 + a^3*c^2*d^6 + 5*a^3*c*d^7 - 2*a^3*d^8)
*f*cos(f*x + e)^3 - (3*a^3*c^8 - 4*a^3*c^7*d - 12*a^3*c^6*d^2 + 20*a^3*c^5*d^3 + 10*a^3*c^4*d^4 - 28*a^3*c^3*d
^5 + 4*a^3*c^2*d^6 + 12*a^3*c*d^7 - 5*a^3*d^8)*f*cos(f*x + e)^2 + 2*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6
*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e) + 4*(a^3*c^8 - 2*a^3*c^7*
d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f - ((a^3*c^7*d - 3
*a^3*c^6*d^2 + a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 5*a^3*c^3*d^5 - a^3*c^2*d^6 + 3*a^3*c*d^7 - a^3*d^8)*f*cos(f*x +
e)^3 + (a^3*c^8 - 8*a^3*c^6*d^2 + 8*a^3*c^5*d^3 + 10*a^3*c^4*d^4 - 16*a^3*c^3*d^5 + 8*a^3*c*d^7 - 3*a^3*d^8)*f
*cos(f*x + e)^2 - 2*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2
*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e) - 4*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^
5 + 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f)*sin(f*x + e))]

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giac [A]  time = 1.31, size = 517, normalized size = 1.73 \[ -\frac {2 \, {\left (\frac {15 \, {\left (4 \, c d^{3} + 3 \, d^{4}\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 )}}{{\left (a^{3} c^{5} - 3 \, a^{3} c^{4} d + 2 \, a^{3} c^{3} d^{2} + 2 \, a^{3} c^{2} d^{3} - 3 \, a^{3} c d^{4} + a^{3} d^{5}\right )} \sqrt {c^{2} - d^{2}}} + \frac {15 \, {\left (d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c d^{4}\right )}}{{\left (a^{3} c^{6} - 3 \, a^{3} c^{5} d + 2 \, a^{3} c^{4} d^{2} + 2 \, a^{3} c^{3} d^{3} - 3 \, a^{3} c^{2} d^{4} + a^{3} c d^{5}\right )} {\left (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 )}} + \frac {15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 90 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 150 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 300 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 190 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 420 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 110 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 270 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} - 34 \, c d + 72 \, d^{2}}{{\left (a^{3} c^{4} - 4 \, a^{3} c^{3} d + 6 \, a^{3} c^{2} d^{2} - 4 \, a^{3} c d^{3} + a^{3} d^{4}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-2/15*(15*(4*c*d^3 + 3*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)/sqr
t(c^2 - d^2)))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*sqrt(c^2 - d^2
)) + 15*(d^5*tan(1/2*f*x + 1/2*e) + c*d^4)/((a^3*c^6 - 3*a^3*c^5*d + 2*a^3*c^4*d^2 + 2*a^3*c^3*d^3 - 3*a^3*c^2
*d^4 + a^3*c*d^5)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) + (15*c^2*tan(1/2*f*x + 1/2*e)^4
- 60*c*d*tan(1/2*f*x + 1/2*e)^4 + 90*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1/2*f*x + 1/2*e)^3 - 150*c*d*tan(
1/2*f*x + 1/2*e)^3 + 300*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*c^2*tan(1/2*f*x + 1/2*e)^2 - 190*c*d*tan(1/2*f*x + 1/
2*e)^2 + 420*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x + 1/2*e) - 110*c*d*tan(1/2*f*x + 1/2*e) + 270*d^2
*tan(1/2*f*x + 1/2*e) + 7*c^2 - 34*c*d + 72*d^2)/((a^3*c^4 - 4*a^3*c^3*d + 6*a^3*c^2*d^2 - 4*a^3*c*d^3 + a^3*d
^4)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

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maple [A]  time = 0.35, size = 511, normalized size = 1.71 \[ -\frac {2 d^{5} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a^{3} f \left (c -d \right )^{4} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right ) c}-\frac {2 d^{4}}{a^{3} f \left (c -d \right )^{4} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}-\frac {8 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) c}{a^{3} f \left (c -d \right )^{4} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {6 d^{4} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{a^{3} f \left (c -d \right )^{4} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {16 c}{3 a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8 d}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8 d}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4 c}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 c^{2}}{a^{3} f \left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {8 c d}{a^{3} f \left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {12 d^{2}}{a^{3} f \left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a^3/f*d^5/(c-d)^4/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)/c*tan(1/2*f*x+1/2*e)-2/a^3/f*d^4/
(c-d)^4/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)-8/a^3/f*d^3/(c-d)^4/(c+d)/(c^2-d^2)^(1/2)*arct
an(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*c-6/a^3/f*d^4/(c-d)^4/(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))-16/3/a^3/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^3*c+8/a^3/f/(c-d)^3/(tan
(1/2*f*x+1/2*e)+1)^3*d-8/a^3/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^2*d+4/a^3/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^2*c-2
/a^3/f/(c-d)^4/(tan(1/2*f*x+1/2*e)+1)*c^2+8/a^3/f/(c-d)^4/(tan(1/2*f*x+1/2*e)+1)*c*d-12/a^3/f/(c-d)^4/(tan(1/2
*f*x+1/2*e)+1)*d^2-8/5/a^3/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^5+4/a^3/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^4

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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(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,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 = 10.30, size = 987, normalized size = 3.31 \[ -\frac {\frac {2\,\left (7\,c^4-27\,c^3\,d+38\,c^2\,d^2+72\,c\,d^3+15\,d^4\right )}{15\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (5\,c^4-18\,c^3\,d+19\,c^2\,d^2+84\,c\,d^3+15\,d^4\right )}{3\,c\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (20\,c^5-76\,c^4\,d+106\,c^3\,d^2+346\,c^2\,d^3+219\,c\,d^4+15\,d^5\right )}{15\,c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (c^5-3\,c^4\,d+2\,c^3\,d^2+6\,c^2\,d^3+d^5\right )}{c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (2\,c^5-6\,c^4\,d+4\,c^3\,d^2+24\,c^2\,d^3+13\,c\,d^4+5\,d^5\right )}{c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (11\,c^5-27\,c^4\,d+4\,c^3\,d^2+162\,c^2\,d^3+135\,c\,d^4+30\,d^5\right )}{3\,c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (47\,c^5-137\,c^4\,d+88\,c^3\,d^2+812\,c^2\,d^3+690\,c\,d^4+75\,d^5\right )}{15\,c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )}}{f\,\left (a^3\,c+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,a^3\,c+2\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (5\,a^3\,c+2\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (11\,a^3\,c+10\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (11\,a^3\,c+10\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (15\,a^3\,c+20\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (15\,a^3\,c+20\,a^3\,d\right )+a^3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\right )}-\frac {2\,d^3\,\mathrm {atan}\left (\frac {\frac {d^3\,\left (4\,c+3\,d\right )\,\left (2\,a^3\,c^5\,d-6\,a^3\,c^4\,d^2+4\,a^3\,c^3\,d^3+4\,a^3\,c^2\,d^4-6\,a^3\,c\,d^5+2\,a^3\,d^6\right )}{a^3\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{9/2}}+\frac {2\,c\,d^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c+3\,d\right )\,\left (a^3\,c^5-3\,a^3\,c^4\,d+2\,a^3\,c^3\,d^2+2\,a^3\,c^2\,d^3-3\,a^3\,c\,d^4+a^3\,d^5\right )}{a^3\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{9/2}}}{6\,d^4+8\,c\,d^3}\right )\,\left (4\,c+3\,d\right )}{a^3\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*(72*c*d^3 - 27*c^3*d + 7*c^4 + 15*d^4 + 38*c^2*d^2))/(15*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)
) + (4*tan(e/2 + (f*x)/2)^3*(84*c*d^3 - 18*c^3*d + 5*c^4 + 15*d^4 + 19*c^2*d^2))/(3*c*(c - d)*(3*c*d^2 - 3*c^2
*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)*(219*c*d^4 - 76*c^4*d + 20*c^5 + 15*d^5 + 346*c^2*d^3 + 106*c^3*d^2))
/(15*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^6*(c^5 - 3*c^4*d + d^5 + 6*c^2
*d^3 + 2*c^3*d^2))/(c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^5*(13*c*d^4 - 6
*c^4*d + 2*c^5 + 5*d^5 + 24*c^2*d^3 + 4*c^3*d^2))/(c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan
(e/2 + (f*x)/2)^4*(135*c*d^4 - 27*c^4*d + 11*c^5 + 30*d^5 + 162*c^2*d^3 + 4*c^3*d^2))/(3*c*(c + d)*(c - d)*(3*
c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^2*(690*c*d^4 - 137*c^4*d + 47*c^5 + 75*d^5 + 812*c^2*d^3
+ 88*c^3*d^2))/(15*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)))/(f*(a^3*c + tan(e/2 + (f*x)/2)*(5*a^3*
c + 2*a^3*d) + tan(e/2 + (f*x)/2)^6*(5*a^3*c + 2*a^3*d) + tan(e/2 + (f*x)/2)^2*(11*a^3*c + 10*a^3*d) + tan(e/2
+ (f*x)/2)^5*(11*a^3*c + 10*a^3*d) + tan(e/2 + (f*x)/2)^3*(15*a^3*c + 20*a^3*d) + tan(e/2 + (f*x)/2)^4*(15*a^
3*c + 20*a^3*d) + a^3*c*tan(e/2 + (f*x)/2)^7)) - (2*d^3*atan(((d^3*(4*c + 3*d)*(2*a^3*d^6 - 6*a^3*c*d^5 + 2*a^
3*c^5*d + 4*a^3*c^2*d^4 + 4*a^3*c^3*d^3 - 6*a^3*c^4*d^2))/(a^3*(c + d)^(3/2)*(c - d)^(9/2)) + (2*c*d^3*tan(e/2
+ (f*x)/2)*(4*c + 3*d)*(a^3*c^5 + a^3*d^5 - 3*a^3*c*d^4 - 3*a^3*c^4*d + 2*a^3*c^2*d^3 + 2*a^3*c^3*d^2))/(a^3*
(c + d)^(3/2)*(c - d)^(9/2)))/(8*c*d^3 + 6*d^4))*(4*c + 3*d))/(a^3*f*(c + d)^(3/2)*(c - d)^(9/2))

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

[Out]

Timed out

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