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

Optimal. Leaf size=454 \[ -\frac{b^2 \left (-a^2 b^2 \left (2 c^2-15 d^2\right )+8 a^3 b c d-12 a^4 d^2-2 a b^3 c d-b^4 \left (c^2+6 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2} (b c-a d)^4}-\frac{d \left (a^2 b^2 d \left (7 c^2-11 d^2\right )+2 a^4 d^3-3 a b^3 c \left (c^2-d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 \left (c^2-d^2\right ) (b c-a d)^3 (c+d \sin (e+f x))}+\frac{3 b^2 \left (-2 a^2 d+a b c+b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (b c-a d)^2 (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{2 f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{2 d^3 \left (-a c d+4 b c^2-3 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^4} \]

[Out]

-((b^2*(8*a^3*b*c*d - 2*a*b^3*c*d - 12*a^4*d^2 - a^2*b^2*(2*c^2 - 15*d^2) - b^4*(c^2 + 6*d^2))*ArcTan[(b + a*T
an[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*(b*c - a*d)^4*f)) - (2*d^3*(4*b*c^2 - a*c*d - 3*b*d^2)*A
rcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((b*c - a*d)^4*(c^2 - d^2)^(3/2)*f) - (d*(2*a^4*d^3 + a^2*b^2
*d*(7*c^2 - 11*d^2) - 2*b^4*d*(2*c^2 - 3*d^2) - 3*a*b^3*c*(c^2 - d^2))*Cos[e + f*x])/(2*(a^2 - b^2)^2*(b*c - a
*d)^3*(c^2 - d^2)*f*(c + d*Sin[e + f*x])) + (b^2*Cos[e + f*x])/(2*(a^2 - b^2)*(b*c - a*d)*f*(a + b*Sin[e + f*x
])^2*(c + d*Sin[e + f*x])) + (3*b^2*(a*b*c - 2*a^2*d + 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])*(c + d*Sin[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 2.43873, antiderivative size = 454, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{b^2 \left (-a^2 b^2 \left (2 c^2-15 d^2\right )+8 a^3 b c d-12 a^4 d^2-2 a b^3 c d-b^4 \left (c^2+6 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2} (b c-a d)^4}-\frac{d \left (a^2 b^2 d \left (7 c^2-11 d^2\right )+2 a^4 d^3-3 a b^3 c \left (c^2-d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 \left (c^2-d^2\right ) (b c-a d)^3 (c+d \sin (e+f x))}+\frac{3 b^2 \left (-2 a^2 d+a b c+b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (b c-a d)^2 (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{2 f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{2 d^3 \left (-a c d+4 b c^2-3 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(8*a^3*b*c*d - 2*a*b^3*c*d - 12*a^4*d^2 - a^2*b^2*(2*c^2 - 15*d^2) - b^4*(c^2 + 6*d^2))*ArcTan[(b + a*T
an[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*(b*c - a*d)^4*f)) - (2*d^3*(4*b*c^2 - a*c*d - 3*b*d^2)*A
rcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((b*c - a*d)^4*(c^2 - d^2)^(3/2)*f) - (d*(2*a^4*d^3 + a^2*b^2
*d*(7*c^2 - 11*d^2) - 2*b^4*d*(2*c^2 - 3*d^2) - 3*a*b^3*c*(c^2 - d^2))*Cos[e + f*x])/(2*(a^2 - b^2)^2*(b*c - a
*d)^3*(c^2 - d^2)*f*(c + d*Sin[e + f*x])) + (b^2*Cos[e + f*x])/(2*(a^2 - b^2)*(b*c - a*d)*f*(a + b*Sin[e + f*x
])^2*(c + d*Sin[e + f*x])) + (3*b^2*(a*b*c - 2*a^2*d + 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])*(c + d*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))^2} \, 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 (c+d \sin (e+f x))}-\frac{\int \frac{-2 a b c+2 a^2 d-3 b^2 d+b (b c-2 a d) \sin (e+f x)+2 b^2 d \sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, 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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}+\frac{\int \frac{-4 a^3 b c d+4 a b^3 c d+2 a^4 d^2+a^2 b^2 \left (2 c^2-11 d^2\right )+b^4 \left (c^2+6 d^2\right )+b d \left (a^2 b c+2 b^3 c-4 a^3 d+a b^2 d\right ) \sin (e+f x)-3 b^2 d \left (a b c-2 a^2 d+b^2 d\right ) \sin ^2(e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{2 \left (a^2-b^2\right )^2 (b c-a d)^2}\\ &=-\frac{d \left (2 a^4 d^3+a^2 b^2 d \left (7 c^2-11 d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )-3 a b^3 c \left (c^2-d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 (b c-a d)^3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}+\frac{\int \frac{-2 a^5 c d^3-2 a^3 b^2 c d \left (3 c^2-5 d^2\right )+a b^4 c d \left (3 c^2-5 d^2\right )+6 a^4 b d^2 \left (c^2-d^2\right )+b^5 \left (c^4+5 c^2 d^2-6 d^4\right )+2 a^2 b^3 \left (c^4-7 c^2 d^2+6 d^4\right )-b d \left (2 a^4 c d^2-b^4 c \left (c^2-3 d^2\right )+6 a^3 b d \left (c^2-d^2\right )-3 a b^3 d \left (c^2-d^2\right )-2 a^2 b^2 c \left (c^2+d^2\right )\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)^3 \left (c^2-d^2\right )}\\ &=-\frac{d \left (2 a^4 d^3+a^2 b^2 d \left (7 c^2-11 d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )-3 a b^3 c \left (c^2-d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 (b c-a d)^3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}-\frac{\left (d^3 \left (4 b c^2-a c d-3 b d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{(b c-a d)^4 \left (c^2-d^2\right )}-\frac{\left (b^2 \left (8 a^3 b c d-2 a b^3 c d-12 a^4 d^2-a^2 b^2 \left (2 c^2-15 d^2\right )-b^4 \left (c^2+6 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)^4}\\ &=-\frac{d \left (2 a^4 d^3+a^2 b^2 d \left (7 c^2-11 d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )-3 a b^3 c \left (c^2-d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 (b c-a d)^3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}-\frac{\left (2 d^3 \left (4 b c^2-a c d-3 b d^2\right )\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)^4 \left (c^2-d^2\right ) f}-\frac{\left (b^2 \left (8 a^3 b c d-2 a b^3 c d-12 a^4 d^2-a^2 b^2 \left (2 c^2-15 d^2\right )-b^4 \left (c^2+6 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)^4 f}\\ &=-\frac{d \left (2 a^4 d^3+a^2 b^2 d \left (7 c^2-11 d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )-3 a b^3 c \left (c^2-d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 (b c-a d)^3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}+\frac{\left (4 d^3 \left (4 b c^2-a c d-3 b d^2\right )\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)^4 \left (c^2-d^2\right ) f}+\frac{\left (2 b^2 \left (8 a^3 b c d-2 a b^3 c d-12 a^4 d^2-a^2 b^2 \left (2 c^2-15 d^2\right )-b^4 \left (c^2+6 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)^4 f}\\ &=-\frac{b^2 \left (8 a^3 b c d-2 a b^3 c d-12 a^4 d^2-a^2 b^2 \left (2 c^2-15 d^2\right )-b^4 \left (c^2+6 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)^4 f}-\frac{2 d^3 \left (4 b c^2-a c d-3 b d^2\right ) \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)^4 \left (c^2-d^2\right )^{3/2} f}-\frac{d \left (2 a^4 d^3+a^2 b^2 d \left (7 c^2-11 d^2\right )-2 b^4 d \left (2 c^2-3 d^2\right )-3 a b^3 c \left (c^2-d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 (b c-a d)^3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\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 (c+d \sin (e+f x))}+\frac{3 b^2 \left (a b c-2 a^2 d+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)) (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 5.31827, size = 346, normalized size = 0.76 \[ \frac{\frac{2 b^2 \left (a^2 b^2 \left (2 c^2-15 d^2\right )-8 a^3 b c d+12 a^4 d^2+2 a b^3 c d+b^4 \left (c^2+6 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} (b c-a d)^4}+\frac{b^3 \left (7 a^2 d-3 a b c-4 b^2 d\right ) \cos (e+f x)}{(a-b)^2 (a+b)^2 (a d-b c)^3 (a+b \sin (e+f x))}+\frac{b^3 \cos (e+f x)}{(a-b) (a+b) (b c-a d)^2 (a+b \sin (e+f x))^2}+\frac{4 d^3 \left (a c d-4 b c^2+3 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2} (b c-a d)^4}-\frac{2 d^4 \cos (e+f x)}{(c-d) (c+d) (b c-a d)^3 (c+d \sin (e+f x))}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b^2*(-8*a^3*b*c*d + 2*a*b^3*c*d + 12*a^4*d^2 + a^2*b^2*(2*c^2 - 15*d^2) + b^4*(c^2 + 6*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)^4) + (4*d^3*(-4*b*c^2 + a*c*d + 3*b*d^2)*A
rcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((b*c - a*d)^4*(c^2 - d^2)^(3/2)) + (b^3*Cos[e + f*x])/((a -
b)*(a + b)*(b*c - a*d)^2*(a + b*Sin[e + f*x])^2) + (b^3*(-3*a*b*c + 7*a^2*d - 4*b^2*d)*Cos[e + f*x])/((a - b)^
2*(a + b)^2*(-(b*c) + a*d)^3*(a + b*Sin[e + f*x])) - (2*d^4*Cos[e + f*x])/((c - d)*(c + d)*(b*c - a*d)^3*(c +
d*Sin[e + f*x])))/(2*f)

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Maple [B]  time = 0.186, size = 3241, normalized size = 7.1 \begin{align*} \text{output 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))^2,x)

[Out]

-1/f*b^7/(a*d-b*c)^4/(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^8/(a*d-
b*c)^4/(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+8/f*
b^3/(a*d-b*c)^4/(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^5/(a*d-b*c)^4/(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+11/f*b^5/(a*d-b*c)^4/(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+5/f*b^6/(a*d-b*c)^4/(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+2/f*b^4/(a*d-b*c)^4/(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^2-15/f*b^4/(a*d-b*c)^4/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arc
tan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*d^2-2/f*b^9/(a*d-b*c)^4/(tan(1/2*f*x+1/2*e)^2*a+2*ta
n(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+23/f*b^4/(a*d-b*c)^4/(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^2-2/f*d^5/(a^2*d^2-2*a*b*c*d+
b^2*c^2)/(a*d-b*c)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c^2-d^2)*tan(1/2*f*x+1/2*e)*b-2/f*d^4/
(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c^2-d^2)*c*b+2/f*d^
4/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1
/2))*a*c-8/f*d^3/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*
d)/(c^2-d^2)^(1/2))*c^2*b+11/f*b^6/(a*d-b*c)^4/(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^2-14/f*b^6/(a*d-b*c)^4/(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^8/(a*d-b*c)^4/(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^2-12/f*b^4/(a*d-b*c)^4/(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^6/(a*d-b*c)^4/(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+12/f*b^2/(a*d-b*c)^4/(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+9/f*b^4/(a*d-b*c)^4/(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+8/f*b^7/(a*d-b*c)^4/(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+16/f*b^7/(a*d-b*c)^4/(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-6/f*b^6/(a*d-b*c)^4/(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^2-18/f*b^6/(a*d-b*c)^4/(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+12/f*b^8/(a*d-b*c)^4/(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-34/f*b^5/(a*d-b*c
)^4/(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-8/f*b^3
/(a*d-b*c)^4/(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+2/f*b^5/(a*d-b*c)^4/(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*c*d-14/f*b^5/(a*d-b*c)^4/(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+7/f*b^7/(a*d-b*c)^4/(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+2/f*d^6/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(
1/2*f*x+1/2*e)*d+c)/c/(c^2-d^2)*tan(1/2*f*x+1/2*e)*a-12/f*b^4/(a*d-b*c)^4/(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-10/f*b^7/(a*d-b*c)^4/(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+6/f*b^6/(a*d-b*c)^4/(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+1/f*b^6/(a*d-b*c)^4/(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+2/f*d^5/(a^2*d^2-2*a*b*
c*d+b^2*c^2)/(a*d-b*c)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c^2-d^2)*a+6/f*d^5/(a^2*d^2-2*a*b*
c*d+b^2*c^2)/(a*d-b*c)^2/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*b+8/f*b^3/(a
*d-b*c)^4/(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^5/(a*d-b*c)^4/
(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^2-5/f*b^5/(a*d-b*c)^4/(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

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.50065, size = 1500, normalized size = 3.3 \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))^2,x, algorithm="giac")

[Out]

((2*a^2*b^4*c^2 + b^6*c^2 - 8*a^3*b^3*c*d + 2*a*b^5*c*d + 12*a^4*b^2*d^2 - 15*a^2*b^4*d^2 + 6*b^6*d^2)*(pi*flo
or(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^4*c^4 - 2*a^
2*b^6*c^4 + b^8*c^4 - 4*a^5*b^3*c^3*d + 8*a^3*b^5*c^3*d - 4*a*b^7*c^3*d + 6*a^6*b^2*c^2*d^2 - 12*a^4*b^4*c^2*d
^2 + 6*a^2*b^6*c^2*d^2 - 4*a^7*b*c*d^3 + 8*a^5*b^3*c*d^3 - 4*a^3*b^5*c*d^3 + a^8*d^4 - 2*a^6*b^2*d^4 + a^4*b^4
*d^4)*sqrt(a^2 - b^2)) - 2*(4*b*c^2*d^3 - a*c*d^4 - 3*b*d^5)*(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)))/((b^4*c^6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*d^2 - b^4*c^4*d^2 -
4*a^3*b*c^3*d^3 + 4*a*b^3*c^3*d^3 + a^4*c^2*d^4 - 6*a^2*b^2*c^2*d^4 + 4*a^3*b*c*d^5 - a^4*d^6)*sqrt(c^2 - d^2)
) - 2*(d^5*tan(1/2*f*x + 1/2*e) + c*d^4)/((b^3*c^6 - 3*a*b^2*c^5*d + 3*a^2*b*c^4*d^2 - b^3*c^4*d^2 - a^3*c^3*d
^3 + 3*a*b^2*c^3*d^3 - 3*a^2*b*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))
+ (5*a^3*b^5*c*tan(1/2*f*x + 1/2*e)^3 - 2*a*b^7*c*tan(1/2*f*x + 1/2*e)^3 - 9*a^4*b^4*d*tan(1/2*f*x + 1/2*e)^3
+ 6*a^2*b^6*d*tan(1/2*f*x + 1/2*e)^3 + 4*a^4*b^4*c*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^6*c*tan(1/2*f*x + 1/2*e)^
2 - 2*b^8*c*tan(1/2*f*x + 1/2*e)^2 - 8*a^5*b^3*d*tan(1/2*f*x + 1/2*e)^2 - 11*a^3*b^5*d*tan(1/2*f*x + 1/2*e)^2
+ 10*a*b^7*d*tan(1/2*f*x + 1/2*e)^2 + 11*a^3*b^5*c*tan(1/2*f*x + 1/2*e) - 2*a*b^7*c*tan(1/2*f*x + 1/2*e) - 23*
a^4*b^4*d*tan(1/2*f*x + 1/2*e) + 14*a^2*b^6*d*tan(1/2*f*x + 1/2*e) + 4*a^4*b^4*c - a^2*b^6*c - 8*a^5*b^3*d + 5
*a^3*b^5*d)/((a^6*b^3*c^3 - 2*a^4*b^5*c^3 + a^2*b^7*c^3 - 3*a^7*b^2*c^2*d + 6*a^5*b^4*c^2*d - 3*a^3*b^6*c^2*d
+ 3*a^8*b*c*d^2 - 6*a^6*b^3*c*d^2 + 3*a^4*b^5*c*d^2 - a^9*d^3 + 2*a^7*b^2*d^3 - a^5*b^4*d^3)*(a*tan(1/2*f*x +
1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f