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

Optimal. Leaf size=285 \[ -\frac{b \left (-a^2 b^2 \left (2 c^2-5 d^2\right )+6 a^3 b c d-6 a^4 d^2-b^4 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2} (b c-a d)^3}+\frac{b^2 \left (-5 a^2 d+3 a b c+2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (b c-a d)^2 (a+b \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{2 f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))^2}-\frac{2 d^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \sqrt{c^2-d^2} (b c-a d)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.03696, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2802, 3055, 3001, 2660, 618, 204} \[ -\frac{b \left (-a^2 b^2 \left (2 c^2-5 d^2\right )+6 a^3 b c d-6 a^4 d^2-b^4 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2} (b c-a d)^3}+\frac{b^2 \left (-5 a^2 d+3 a b c+2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (b c-a d)^2 (a+b \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{2 f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))^2}-\frac{2 d^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \sqrt{c^2-d^2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[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] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rubi steps

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

Mathematica [A]  time = 2.2313, size = 275, normalized size = 0.96 \[ \frac{-\frac{2 b \left (a^2 b^2 \left (2 c^2-5 d^2\right )-6 a^3 b c d+6 a^4 d^2+b^4 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} (a d-b c)^3}+\frac{b^2 \left (-5 a^2 d+3 a b c+2 b^2 d\right ) \cos (e+f x)}{(a-b)^2 (a+b)^2 (b c-a d)^2 (a+b \sin (e+f x))}-\frac{b^2 \cos (e+f x)}{(a-b) (a+b) (a d-b c) (a+b \sin (e+f x))^2}+\frac{4 d^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2} (a d-b c)^3}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.136, size = 2644, normalized size = 9.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*d^3/(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)^(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*b^6/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*c
^2-2/f*b^5/(a*d-b*c)^3/(a^4-2*a^2*b^2+b^4)/(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))*d^2-6/f*b^2/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^4*d^2-
4/f*b^4/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*c^2*a^2+3/f*b^4/(a
*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*d^2-7/f*b^6/(a*d-b*c)^3/
(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2*c^2+6/f*b^6/(a*d-
b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2*d^2-1/f*b^
5/(a*d-b*c)^3/(a^4-2*a^2*b^2+b^4)/(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))*c^2
+4/f*b^5/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2
*e)^3*d^2+2/f*b^7/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a*tan(1/
2*f*x+1/2*e)^3*c^2-6/f*b^2/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)
*a^4*tan(1/2*f*x+1/2*e)^2*d^2-4/f*b^4/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a
^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2*c^2-9/f*b^4/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)
^2/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2*d^2+2/f*b^8/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+
1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2*c^2-17/f*b^3/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*
tan(1/2*f*x+1/2*e)*b+a)^2*a^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*d^2-11/f*b^5/(a*d-b*c)^3/(tan(1/2*f*x+1/2
*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c^2+8/f*b^5/(a*d-b*c)^3/(tan(1/2*
f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*d^2+2/f*b^7/(a*d-b*c)^3/(t
an(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c^2+10/f*b^3/(a*d-b
*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*c*d-6/f*b/(a*d-b*c)^3/(a^4-2
*a^2*b^2+b^4)/(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))*a^4*d^2-2/f*b^3/(a*d-b*
c)^3/(a^4-2*a^2*b^2+b^4)/(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))*c^2*a^2-4/f*
b^5/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*c*d-6/f*b^6/(a*d-b*c
)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^3*c*d-10/f*b^6/
(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c*d-7/f
*b^3/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*tan(1/2*f*x+1/2*e
)^3*d^2-5/f*b^5/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*
f*x+1/2*e)^3*c^2+5/f*b^3/(a*d-b*c)^3/(a^4-2*a^2*b^2+b^4)/(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))*a^2*d^2+12/f*b^4/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^
2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^3*c*d+6/f*b^2/(a*d-b*c)^3/(a^4-2*a^2*b^2+b^4)/(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))*a^3*c*d+10/f*b^3/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+
1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*tan(1/2*f*x+1/2*e)^2*c*d+16/f*b^5/(a*d-b*c)^3/(tan(1/2*f*x+1/2*e)^2*a+2*
tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^2*c*d-8/f*b^7/(a*d-b*c)^3/(tan(1/2*f*x+1/2*
e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^2*c*d+28/f*b^4/(a*d-b*c)^3/(tan(1/
2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*a^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c*d

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.5606, size = 1064, normalized size = 3.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(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/((b^3
*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) - (2*a^2*b^3*c^2 + b^5*c^2 - 6*a^3*b^2*c*d +
6*a^4*b*d^2 - 5*a^2*b^3*d^2 + 2*b^5*d^2)*(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)))/((a^4*b^3*c^3 - 2*a^2*b^5*c^3 + b^7*c^3 - 3*a^5*b^2*c^2*d + 6*a^3*b^4*c^2*d - 3*a*
b^6*c^2*d + 3*a^6*b*c*d^2 - 6*a^4*b^3*c*d^2 + 3*a^2*b^5*c*d^2 - a^7*d^3 + 2*a^5*b^2*d^3 - a^3*b^4*d^3)*sqrt(a^
2 - b^2)) - (5*a^3*b^4*c*tan(1/2*f*x + 1/2*e)^3 - 2*a*b^6*c*tan(1/2*f*x + 1/2*e)^3 - 7*a^4*b^3*d*tan(1/2*f*x +
1/2*e)^3 + 4*a^2*b^5*d*tan(1/2*f*x + 1/2*e)^3 + 4*a^4*b^3*c*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^5*c*tan(1/2*f*x
+ 1/2*e)^2 - 2*b^7*c*tan(1/2*f*x + 1/2*e)^2 - 6*a^5*b^2*d*tan(1/2*f*x + 1/2*e)^2 - 9*a^3*b^4*d*tan(1/2*f*x + 1
/2*e)^2 + 6*a*b^6*d*tan(1/2*f*x + 1/2*e)^2 + 11*a^3*b^4*c*tan(1/2*f*x + 1/2*e) - 2*a*b^6*c*tan(1/2*f*x + 1/2*e
) - 17*a^4*b^3*d*tan(1/2*f*x + 1/2*e) + 8*a^2*b^5*d*tan(1/2*f*x + 1/2*e) + 4*a^4*b^3*c - a^2*b^5*c - 6*a^5*b^2
*d + 3*a^3*b^4*d)/((a^6*b^2*c^2 - 2*a^4*b^4*c^2 + a^2*b^6*c^2 - 2*a^7*b*c*d + 4*a^5*b^3*c*d - 2*a^3*b^5*c*d +
a^8*d^2 - 2*a^6*b^2*d^2 + a^4*b^4*d^2)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f