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

Optimal. Leaf size=458 \[ -\frac{d^2 \left (-a^2 d^2 \left (2 c^2+d^2\right )+2 a b c d \left (4 c^2-d^2\right )-3 b^2 \left (-5 c^2 d^2+4 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2} (b c-a d)^4}-\frac{\left (-a^2 b d^3 \left (7 c^2-4 d^2\right )+3 a^3 c d^4-3 a b^2 c d^4-b^3 \left (-11 c^2 d^3+2 c^4 d+6 d^5\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right ) \left (c^2-d^2\right )^2 (b c-a d)^3 (c+d \sin (e+f x))}+\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right ) \left (c^2-d^2\right ) (b c-a d)^2 (c+d \sin (e+f x))^2}+\frac{2 b^3 \left (-4 a^2 d+a b c+3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^4}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.43866, antiderivative size = 458, 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} \[ -\frac{d^2 \left (-a^2 d^2 \left (2 c^2+d^2\right )+2 a b c d \left (4 c^2-d^2\right )-3 b^2 \left (-5 c^2 d^2+4 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2} (b c-a d)^4}-\frac{\left (-a^2 b d^3 \left (7 c^2-4 d^2\right )+3 a^3 c d^4-3 a b^2 c d^4-b^3 \left (-11 c^2 d^3+2 c^4 d+6 d^5\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right ) \left (c^2-d^2\right )^2 (b c-a d)^3 (c+d \sin (e+f x))}+\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right ) \left (c^2-d^2\right ) (b c-a d)^2 (c+d \sin (e+f x))^2}+\frac{2 b^3 \left (-4 a^2 d+a b c+3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^4}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^3*(a*b*c - 4*a^2*d + 3*b^2*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(3/2)*(b*c -
a*d)^4*f) - (d^2*(2*a*b*c*d*(4*c^2 - d^2) - a^2*d^2*(2*c^2 + d^2) - 3*b^2*(4*c^4 - 5*c^2*d^2 + 2*d^4))*ArcTan
[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((b*c - a*d)^4*(c^2 - d^2)^(5/2)*f) + (d*(a^2*d^2 + b^2*(2*c^2 - 3
*d^2))*Cos[e + f*x])/(2*(a^2 - b^2)*(b*c - a*d)^2*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) + (b^2*Cos[e + f*x])/(
(a^2 - b^2)*(b*c - a*d)*f*(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - ((3*a^3*c*d^4 - 3*a*b^2*c*d^4 - a^2*b
*d^3*(7*c^2 - 4*d^2) - b^3*(2*c^4*d - 11*c^2*d^3 + 6*d^5))*Cos[e + f*x])/(2*(a^2 - b^2)*(b*c - a*d)^3*(c^2 - d
^2)^2*f*(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))^2 (c+d \sin (e+f x))^3} \, dx &=\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\int \frac{-a b c+a^2 d-3 b^2 d-a b d \sin (e+f x)+2 b^2 d \sin ^2(e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx}{\left (a^2-b^2\right ) (b c-a d)}\\ &=\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\int \frac{-2 \left (a^3 c d^2+a b^2 c \left (c^2-2 d^2\right )-2 a^2 b d \left (c^2-d^2\right )+3 b^3 d \left (c^2-d^2\right )\right )-d \left (2 a^2 b c d-2 b^3 c d-a^3 d^2+a b^2 \left (2 c^2-d^2\right )\right ) \sin (e+f x)+b d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\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 ) (b c-a d)^2 \left (c^2-d^2\right )}\\ &=\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\left (3 a^3 c d^4-3 a b^2 c d^4-a^2 b d^3 \left (7 c^2-4 d^2\right )-b^3 \left (2 c^4 d-11 c^2 d^3+6 d^5\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^3 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac{\int \frac{-6 b^4 d \left (c^2-d^2\right )^2-3 a^3 b c d^2 \left (2 c^2-d^2\right )+a^2 b^2 d \left (6 c^4-14 c^2 d^2+5 d^4\right )-a b^3 c \left (2 c^4-10 c^2 d^2+5 d^4\right )+a^4 \left (2 c^2 d^3+d^5\right )-b d \left (3 a^2 b c d \left (2 c^2-d^2\right )-3 b^3 c d \left (2 c^2-d^2\right )-a^3 \left (2 c^2 d^2+d^4\right )+a b^2 \left (2 c^4-2 c^2 d^2+3 d^4\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 ) (b c-a d)^3 \left (c^2-d^2\right )^2}\\ &=\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\left (3 a^3 c d^4-3 a b^2 c d^4-a^2 b d^3 \left (7 c^2-4 d^2\right )-b^3 \left (2 c^4 d-11 c^2 d^3+6 d^5\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^3 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\left (b^3 \left (a b c-4 a^2 d+3 b^2 d\right )\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) (b c-a d)^4}-\frac{\left (d^2 \left (2 a b c d \left (4 c^2-d^2\right )-a^2 d^2 \left (2 c^2+d^2\right )-3 b^2 \left (4 c^4-5 c^2 d^2+2 d^4\right )\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 (b c-a d)^4 \left (c^2-d^2\right )^2}\\ &=\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\left (3 a^3 c d^4-3 a b^2 c d^4-a^2 b d^3 \left (7 c^2-4 d^2\right )-b^3 \left (2 c^4 d-11 c^2 d^3+6 d^5\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^3 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\left (2 b^3 \left (a b c-4 a^2 d+3 b^2 d\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 ) (b c-a d)^4 f}-\frac{\left (d^2 \left (2 a b c d \left (4 c^2-d^2\right )-a^2 d^2 \left (2 c^2+d^2\right )-3 b^2 \left (4 c^4-5 c^2 d^2+2 d^4\right )\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 )^2 f}\\ &=\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\left (3 a^3 c d^4-3 a b^2 c d^4-a^2 b d^3 \left (7 c^2-4 d^2\right )-b^3 \left (2 c^4 d-11 c^2 d^3+6 d^5\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^3 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac{\left (4 b^3 \left (a b c-4 a^2 d+3 b^2 d\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 ) (b c-a d)^4 f}+\frac{\left (2 d^2 \left (2 a b c d \left (4 c^2-d^2\right )-a^2 d^2 \left (2 c^2+d^2\right )-3 b^2 \left (4 c^4-5 c^2 d^2+2 d^4\right )\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 )^2 f}\\ &=\frac{2 b^3 \left (a b c-4 a^2 d+3 b^2 d\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 )^{3/2} (b c-a d)^4 f}-\frac{d^2 \left (2 a b c d \left (4 c^2-d^2\right )-a^2 d^2 \left (2 c^2+d^2\right )-3 b^2 \left (4 c^4-5 c^2 d^2+2 d^4\right )\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 )^{5/2} f}+\frac{d \left (a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\left (3 a^3 c d^4-3 a b^2 c d^4-a^2 b d^3 \left (7 c^2-4 d^2\right )-b^3 \left (2 c^4 d-11 c^2 d^3+6 d^5\right )\right ) \cos (e+f x)}{2 \left (a^2-b^2\right ) (b c-a d)^3 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 6.44137, size = 346, normalized size = 0.76 \[ \frac{\frac{2 d^2 \left (a^2 d^2 \left (2 c^2+d^2\right )+2 a b c d \left (d^2-4 c^2\right )+3 b^2 \left (-5 c^2 d^2+4 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} (b c-a d)^4}+\frac{4 b^3 \left (-4 a^2 d+a b c+3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^4}-\frac{2 b^4 \cos (e+f x)}{(a-b) (a+b) (a d-b c)^3 (a+b \sin (e+f x))}+\frac{d^3 \left (-3 a c d+7 b c^2-4 b d^2\right ) \cos (e+f x)}{(c-d)^2 (c+d)^2 (b c-a d)^3 (c+d \sin (e+f x))}+\frac{d^3 \cos (e+f x)}{(c-d) (c+d) (b c-a d)^2 (c+d \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b^3*(a*b*c - 4*a^2*d + 3*b^2*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(3/2)*(b*c
- a*d)^4) + (2*d^2*(2*a*b*c*d*(-4*c^2 + d^2) + a^2*d^2*(2*c^2 + d^2) + 3*b^2*(4*c^4 - 5*c^2*d^2 + 2*d^4))*ArcT
an[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((b*c - a*d)^4*(c^2 - d^2)^(5/2)) - (2*b^4*Cos[e + f*x])/((a - b
)*(a + b)*(-(b*c) + a*d)^3*(a + b*Sin[e + f*x])) + (d^3*Cos[e + f*x])/((c - d)*(c + d)*(b*c - a*d)^2*(c + d*Si
n[e + f*x])^2) + (d^3*(7*b*c^2 - 3*a*c*d - 4*b*d^2)*Cos[e + f*x])/((c - d)^2*(c + d)^2*(b*c - a*d)^3*(c + d*Si
n[e + f*x])))/(2*f)

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Maple [B]  time = 0.174, size = 4023, normalized size = 8.8 \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))^2/(c+d*sin(f*x+e))^3,x)

[Out]

-1/f*d^7/(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)^2/(c^4-2*c^
2*d^2+d^4)*a^2+2/f*b^5/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+
a)/(a^2-b^2)*c+6/f*b^5/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2
*e)+2*b)/(a^2-b^2)^(1/2))*d+23/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)^2*c^3/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*b^2-14/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)^2*c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*b^2-12/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)^2/(c^4-2*c^2*d^2+d^
4)*a*b*c^3+6/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)^2
/(c^4-2*c^2*d^2+d^4)*a*b*c+12/f*d^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^4-2*c^2*d^2+d^4)/(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))*b^2*c^4-15/f*d^4/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b
*c)^2/(c^4-2*c^2*d^2+d^4)/(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))*b^2*c^2+2/f
*d^4/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^4-2*c^2*d^2+d^4)/(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))*a^2*c^2+8/f*d^7/(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)^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^3*a*b+16/f*d^7/(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)^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*a*b+5/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)^2/(c^4-2*c^2*d^2+
d^4)*c*tan(1/2*f*x+1/2*e)^3*a^2-2/f*d^8/(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)^2/(c^4-2*c^2*d^2+d^4)/c*tan(1/2*f*x+1/2*e)^3*a^2+1/f*d^6/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-
b*c)^2/(c^4-2*c^2*d^2+d^4)/(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))*a^2+6/f*d^
6/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^4-2*c^2*d^2+d^4)/(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))*b^2+4/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)^2/(c^4-2*c^2*d^2+d^4)*a^2*c^2+8/f*d^3/(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)^2/(c^4-2*c^2*d^2+d^4)*b^2*c^4-5/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)^2/(c^4-2*c^2*d^2+d^4)*b^2*c^2+7/f*d^7/(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)^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e
)^2*a^2-2/f*b^5/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2
-b^2)*tan(1/2*f*x+1/2*e)*d-2/f*b^4/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f
*x+1/2*e)*b+a)/(a^2-b^2)*d*a-8/f*b^3/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*t
an(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*d+2/f*b^4/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a^2-b^2)^(3/2)*
arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a*c-10/f*d^7/(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)^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^2*b^2+11/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)^2/(c^4-2*c^2*d^2+d^4)*c^
2*tan(1/2*f*x+1/2*e)^2*b^2+11/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)^2*c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*a^2+2/f*b^6/(a*d-b*c)^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)
/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/a/(a^2-b^2)*tan(1/2*f*x+1/2*e)*c-2/f*d^8/(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)^2/c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/
2*e)*a^2-8/f*d^3/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^4-2*c^2*d^2+d^4)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*t
an(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*a*b*c^3+2/f*d^5/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)^2/(c^4-2*c^2*d^2
+d^4)/(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))*a*b*c-14/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)^2/(c^4-2*c^2*d^2+d^4)*c^2*tan(1/2*f*
x+1/2*e)^3*a*b-12/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)^2/(c^4-2*c^2*d^2+d^4)*c^3*tan(1/2*f*x+1/2*e)^2*a*b-18/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)^2/(c^4-2*c^2*d^2+d^4)*c*tan(1/2*f*x+1/2*e)^2*a*b+12/f*d^8/(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)^2/(c^4-2*c^2*d^2+d^4)/c*tan(
1/2*f*x+1/2*e)^2*a*b-34/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)^2*c^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*a*b+9/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)^2/(c^4-2*c^2*d^2+d^4)*c^3*tan(1/2*f*x+1/2*e)^3*b^2-6/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)^2/(c^4-2*c^2*d^2+d^4)*c*
tan(1/2*f*x+1/2*e)^3*b^2+4/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)^2/(c^4-2*c^2*d^2+d^4)*c^2*tan(1/2*f*x+1/2*e)^2*a^2-2/f*d^9/(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)^2/(c^4-2*c^2*d^2+d^4)/c^2*tan(1/2*f*x+1/2*e)^2*a^2+8/f*d^3
/(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)^2/(c^4-2*c^2*d^2+d^
4)*c^4*tan(1/2*f*x+1/2*e)^2*b^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))^2/(c+d*sin(f*x+e))^3,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))^2/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [B]  time = 1.46116, size = 1497, normalized size = 3.27 \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))^2/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

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