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

Optimal. Leaf size=186 \[ -\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {2 d^3 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt {c^2-d^2}}-\frac {(2 c-7 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]

[Out]

-1/5*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^3-1/15*(2*c-7*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin(f*x+e))^2-1/15*(2*c^
2-9*c*d+22*d^2)*cos(f*x+e)/(c-d)^3/f/(a^3+a^3*sin(f*x+e))-2*d^3*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2
))/a^3/(c-d)^3/f/(c^2-d^2)^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2766, 2978, 12, 2660, 618, 204} \[ -\frac {2 d^3 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt {c^2-d^2}}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 c-7 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^3*Sqrt[c^2 - d^2]*f) - Cos[e + f*x]/(5*
(c - d)*f*(a + a*Sin[e + f*x])^3) - ((2*c - 7*d)*Cos[e + f*x])/(15*a*(c - d)^2*f*(a + a*Sin[e + f*x])^2) - ((2
*c^2 - 9*c*d + 22*d^2)*Cos[e + f*x])/(15*(c - d)^3*f*(a^3 + a^3*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 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))} \, dx &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a (2 c-5 d)-2 a d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx}{5 a^2 (c-d)}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac {\int \frac {a^2 \left (2 c^2-7 c d+15 d^2\right )+a^2 (2 c-7 d) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int \frac {15 a^3 d^3}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {d^3 \int \frac {1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (2 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 )}{a^3 (c-d)^3 f}\\ &=-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (4 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 )}{a^3 (c-d)^3 f}\\ &=-\frac {2 d^3 \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)^3 \sqrt {c^2-d^2} f}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 301, normalized size = 1.62 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 \left (2 c^2-9 c d+22 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4-\frac {30 d^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )+(c-d) (7 d-2 c) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3+2 (2 c-7 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-3 (c-d)^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{15 a^3 f (c-d)^3 (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])),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]) + 2*(2*c - 7*d)*(c - d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (c - d)*(-2*c +
7*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 2*(2*c^2 - 9*c*d + 22*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^4 - (30*d^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2])^5)/Sqrt[c^2 - d^2]))/(15*a^3*(c - d)^3*f*(1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.53, size = 1744, normalized size = 9.38 \[ \text {result too large to display} \]

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)),x, algorithm="fricas")

[Out]

[1/30*(6*c^4 - 12*c^3*d + 12*c*d^3 - 6*d^4 - 2*(2*c^4 - 9*c^3*d + 20*c^2*d^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^
3 + 2*(4*c^4 - 18*c^3*d + 25*c^2*d^2 + 18*c*d^3 - 29*d^4)*cos(f*x + e)^2 + 15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(
f*x + e)^2 - 2*d^3*cos(f*x + e) - 4*d^3 + (d^3*cos(f*x + e)^2 - 2*d^3*cos(f*x + e) - 4*d^3)*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^4 - 11*
c^3*d + 15*c^2*d^2 + 11*c*d^3 - 18*d^4)*cos(f*x + e) - 2*(3*c^4 - 6*c^3*d + 6*c*d^3 - 3*d^4 - (2*c^4 - 9*c^3*d
+ 20*c^2*d^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^2 - 3*(2*c^4 - 9*c^3*d + 15*c^2*d^2 + 9*c*d^3 - 17*d^4)*cos(f*x
+ e))*sin(f*x + e))/((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)*f*cos(f*
x + e)^3 + 3*(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)*f*cos(f*x + e)^2
- 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)*f*cos(f*x + e) - 4*(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)*f + ((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)*f*cos(f*x + e)^2 - 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)*f*cos(f*x + e) - 4*(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)*f)*sin(f*x + e)), 1/15*(3*c^4 - 6*c^3*d + 6*c*d^3 - 3*d^4 - (2*c^4 - 9*c^3*d + 20*
c^2*d^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^3 + (4*c^4 - 18*c^3*d + 25*c^2*d^2 + 18*c*d^3 - 29*d^4)*cos(f*x + e)^
2 + 15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^2 - 2*d^3*cos(f*x + e) - 4*d^3 + (d^3*cos(f*x + e)^2 - 2*d^3*c
os(f*x + e) - 4*d^3)*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^4 - 11*c^3*d + 15*c^2*d^2 + 11*c*d^3 - 18*d^4)*cos(f*x + e) - (3*c^4 - 6*c^3*d + 6*c*d^3 - 3*d^4 -
(2*c^4 - 9*c^3*d + 20*c^2*d^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^2 - 3*(2*c^4 - 9*c^3*d + 15*c^2*d^2 + 9*c*d^3 -
17*d^4)*cos(f*x + e))*sin(f*x + e))/((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)*f*cos(f*x + e)^3 + 3*(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)*f
*cos(f*x + e)^2 - 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)*f*cos(f*x
+ e) - 4*(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)*f + ((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)*f*cos(f*x + e)^2 - 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)*f*cos(f*x + e) - 4*(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)*f)*sin(f*x + e))]

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giac [B]  time = 0.34, size = 364, normalized size = 1.96 \[ -\frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d^{3}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 105 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 135 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 135 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 185 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 75 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 115 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} - 24 \, c d + 32 \, d^{2}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \]

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)),x, algorithm="giac")

[Out]

-2/15*(15*(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)))*d^3
/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) + (15*c^2*tan(1/2*f*x + 1/2*e)^4 - 45*c*d*t
an(1/2*f*x + 1/2*e)^4 + 45*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1/2*f*x + 1/2*e)^3 - 105*c*d*tan(1/2*f*x +
1/2*e)^3 + 135*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*c^2*tan(1/2*f*x + 1/2*e)^2 - 135*c*d*tan(1/2*f*x + 1/2*e)^2 + 1
85*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x + 1/2*e) - 75*c*d*tan(1/2*f*x + 1/2*e) + 115*d^2*tan(1/2*f*
x + 1/2*e) + 7*c^2 - 24*c*d + 32*d^2)/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(tan(1/2*f*x + 1/2*e) +
1)^5))/f

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maple [A]  time = 0.33, size = 325, normalized size = 1.75 \[ -\frac {2 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{a^{3} f \left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}+\frac {4 c}{a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {6 d}{a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16 c}{3 a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {20 d}{3 a^{3} f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 c^{2}}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {6 c d}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {6 d^{2}}{a^{3} f \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 a^{3} f \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{a^{3} f \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}} \]

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)),x)

[Out]

-2/a^3/f*d^3/(c-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))+4/a^3/f/(c-d)^2/
(tan(1/2*f*x+1/2*e)+1)^2*c-6/a^3/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*d-16/3/a^3/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1
)^3*c+20/3/a^3/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*d-2/a^3/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*c^2+6/a^3/f/(c-d)^3
/(tan(1/2*f*x+1/2*e)+1)*c*d-6/a^3/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*d^2-8/5/a^3/f/(c-d)/(tan(1/2*f*x+1/2*e)+1)^
5+4/a^3/f/(c-d)/(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)),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 = 10.07, size = 466, normalized size = 2.51 \[ \frac {2\,d^3\,\mathrm {atan}\left (\frac {\frac {d^3\,\left (-2\,a^3\,c^3\,d+6\,a^3\,c^2\,d^2-6\,a^3\,c\,d^3+2\,a^3\,d^4\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^3\,c^3-3\,a^3\,c^2\,d+3\,a^3\,c\,d^2-a^3\,d^3\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}}{2\,d^3}\right )}{a^3\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (7\,c^2-24\,c\,d+32\,d^2\right )}{15\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c^2-15\,c\,d+23\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-3\,c\,d+3\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^2-7\,c\,d+9\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,c^2-27\,c\,d+37\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]

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))),x)

[Out]

(2*d^3*atan(((d^3*(2*a^3*d^4 - 6*a^3*c*d^3 - 2*a^3*c^3*d + 6*a^3*c^2*d^2))/(a^3*(c + d)^(1/2)*(c - d)^(7/2)) -
(2*c*d^3*tan(e/2 + (f*x)/2)*(a^3*c^3 - a^3*d^3 + 3*a^3*c*d^2 - 3*a^3*c^2*d))/(a^3*(c + d)^(1/2)*(c - d)^(7/2)
))/(2*d^3)))/(a^3*f*(c + d)^(1/2)*(c - d)^(7/2)) - ((2*(7*c^2 - 24*c*d + 32*d^2))/(15*(c - d)*(c^2 - 2*c*d + d
^2)) + (2*tan(e/2 + (f*x)/2)*(4*c^2 - 15*c*d + 23*d^2))/(3*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2
)^4*(c^2 - 3*c*d + 3*d^2))/((c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^3*(2*c^2 - 7*c*d + 9*d^2))/((
c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^2*(8*c^2 - 27*c*d + 37*d^2))/(3*(c - d)*(c^2 - 2*c*d + d^2
)))/(f*(10*a^3*tan(e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^4 + a^3*tan(e/2 +
(f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x)/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)),x)

[Out]

Timed out

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