3.705 \(\int \frac{1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=284 \[ \frac{d \left (-a^2 d^2 \left (2 c^2+d^2\right )+6 a b c^3 d-b^2 \left (-5 c^2 d^2+6 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2} (b c-a d)^3}+\frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2} (b c-a d)^3}-\frac{d^2 \left (-3 a c d+5 b c^2-2 b d^2\right ) \cos (e+f x)}{2 f \left (c^2-d^2\right )^2 (b c-a d)^2 (c+d \sin (e+f x))}-\frac{d^2 \cos (e+f x)}{2 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^2} \]
[Out]
(2*b^3*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*(b*c - a*d)^3*f) + (d*(6*a*b*c^3*d -
a^2*d^2*(2*c^2 + d^2) - b^2*(6*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)^3*(c^2 - d^2)^(5/2)*f) - (d^2*Cos[e + f*x])/(2*(b*c - a*d)*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) -
(d^2*(5*b*c^2 - 3*a*c*d - 2*b*d^2)*Cos[e + f*x])/(2*(b*c - a*d)^2*(c^2 - d^2)^2*f*(c + d*Sin[e + f*x]))
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Rubi [A] time = 1.07979, antiderivative size = 284, 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} \[ \frac{d \left (-a^2 d^2 \left (2 c^2+d^2\right )+6 a b c^3 d-b^2 \left (-5 c^2 d^2+6 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2} (b c-a d)^3}+\frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2} (b c-a d)^3}-\frac{d^2 \left (-3 a c d+5 b c^2-2 b d^2\right ) \cos (e+f x)}{2 f \left (c^2-d^2\right )^2 (b c-a d)^2 (c+d \sin (e+f x))}-\frac{d^2 \cos (e+f x)}{2 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]
[Out]
(2*b^3*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*(b*c - a*d)^3*f) + (d*(6*a*b*c^3*d -
a^2*d^2*(2*c^2 + d^2) - b^2*(6*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)^3*(c^2 - d^2)^(5/2)*f) - (d^2*Cos[e + f*x])/(2*(b*c - a*d)*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) -
(d^2*(5*b*c^2 - 3*a*c*d - 2*b*d^2)*Cos[e + f*x])/(2*(b*c - a*d)^2*(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)) (c+d \sin (e+f x))^3} \, dx &=-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\int \frac{-2 \left (a c d-b \left (c^2-d^2\right )\right )-d (2 b c-a d) \sin (e+f x)+b d^2 \sin ^2(e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{2 (b c-a d) \left (c^2-d^2\right )}\\ &=-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{d^2 \left (5 b c^2-3 a c d-2 b d^2\right ) \cos (e+f x)}{2 (b c-a d)^2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\int \frac{2 b^2 \left (c^2-d^2\right )^2-a b c d \left (4 c^2-d^2\right )+a^2 d^2 \left (2 c^2+d^2\right )-b d \left (4 b c^3-2 a c^2 d-b c d^2-a d^3\right ) \sin (e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{2 (b c-a d)^2 \left (c^2-d^2\right )^2}\\ &=-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{d^2 \left (5 b c^2-3 a c d-2 b d^2\right ) \cos (e+f x)}{2 (b c-a d)^2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{b^3 \int \frac{1}{a+b \sin (e+f x)} \, dx}{(b c-a d)^3}+\frac{\left (d \left (6 a b c^3 d-a^2 d^2 \left (2 c^2+d^2\right )-b^2 \left (6 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)^3 \left (c^2-d^2\right )^2}\\ &=-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{d^2 \left (5 b c^2-3 a c d-2 b d^2\right ) \cos (e+f x)}{2 (b c-a d)^2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\left (2 b^3\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 )}{(b c-a d)^3 f}+\frac{\left (d \left (6 a b c^3 d-a^2 d^2 \left (2 c^2+d^2\right )-b^2 \left (6 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)^3 \left (c^2-d^2\right )^2 f}\\ &=-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{d^2 \left (5 b c^2-3 a c d-2 b d^2\right ) \cos (e+f x)}{2 (b c-a d)^2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac{\left (4 b^3\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 )}{(b c-a d)^3 f}-\frac{\left (2 d \left (6 a b c^3 d-a^2 d^2 \left (2 c^2+d^2\right )-b^2 \left (6 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)^3 \left (c^2-d^2\right )^2 f}\\ &=\frac{2 b^3 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} (b c-a d)^3 f}+\frac{d \left (6 a b c^3 d-a^2 d^2 \left (2 c^2+d^2\right )-b^2 \left (6 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)^3 \left (c^2-d^2\right )^{5/2} f}-\frac{d^2 \cos (e+f x)}{2 (b c-a d) \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{d^2 \left (5 b c^2-3 a c d-2 b d^2\right ) \cos (e+f x)}{2 (b c-a d)^2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 2.13891, size = 263, normalized size = 0.93 \[ \frac{-\frac{2 d \left (a^2 d^2 \left (2 c^2+d^2\right )-6 a b c^3 d+b^2 \left (-5 c^2 d^2+6 c^4+2 d^4\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}+\frac{4 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{d^2 (b c-a d) \left (3 a c d-5 b c^2+2 b d^2\right ) \cos (e+f x)}{(c-d)^2 (c+d)^2 (c+d \sin (e+f x))}-\frac{d^2 (b c-a d)^2 \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))^2}}{2 f (b c-a d)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]
[Out]
((4*b^3*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (2*d*(-6*a*b*c^3*d + a^2*d^2*(2*c^
2 + d^2) + b^2*(6*c^4 - 5*c^2*d^2 + 2*d^4))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(5/2
) - (d^2*(b*c - a*d)^2*Cos[e + f*x])/((c - d)*(c + d)*(c + d*Sin[e + f*x])^2) + (d^2*(b*c - a*d)*(-5*b*c^2 + 3
*a*c*d + 2*b*d^2)*Cos[e + f*x])/((c - d)^2*(c + d)^2*(c + d*Sin[e + f*x])))/(2*(b*c - a*d)^3*f)
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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))/(c+d*sin(f*x+e))^3,x)
[Out]
6/f*d^6/(a*d-b*c)^3/(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-2/f*b^3/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/(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))+7/f*d^3/(a*d-b*c)^3/(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^2-4/f*d^5/(a*d-b*c)^3/(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^4/(a*d-b*c)^3/(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^8/(a*d-b*c)^3/(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+6/f*d^2/(a*d-b*c)^3/(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+9/f*d^4/(a*d-b*c)^3/(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^5/(a*d-b*c)^3/(
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*d^7/(a*d-b
*c)^3/(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+17/f*d^
3/(a*d-b*c)^3/(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
^2-8/f*d^5/(a*d-b*c)^3/(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^2-10/f*d^3/(a*d-b*c)^3/(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
+4/f*d^5/(a*d-b*c)^3/(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+10/f*d^6/(a
*d-b*c)^3/(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
^5/(a*d-b*c)^3/(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^2-5/f*d^3/(a*d-b*c)^3/(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+6/f*d/(a*d-b*c)^3/(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+2/f*d^3/(a*d-b*c)^3/(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^2-2/f*d^7/(a*d-b*c)^3/(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+7/f*d^6/(a*d-b*c)^3/(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^2-6/f*d^6/(a*d-b*c)^3/(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+4/f*d^4/(a*d-b*c)^3/(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+6/f*d^2/(a*d-b*c)^3/(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^4-3/f*d^4/(a*d-b*c)^3/(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+1/f*d^5/(a*d-b*c)^3/(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+2/f*d^5/(a*d-b*c)^3/(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^2-12/f*d^4/(a*d-b*c)^3/(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-10/f*d^3/(a*d-b*c)^3/(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-16/f*d^5/(a*d-b*c)^3/(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+8/f*d^7/(a*d-b*c)^3/(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-28/f*d^4/(a*d-b*c)^3/(c*t
an(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-6/f*d^2/(a*d-b*
c)^3/(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^3-1/f*
d^6/(a*d-b*c)^3/(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
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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))/(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))/(c+d*sin(f*x+e))^3,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))/(c+d*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] time = 1.34678, size = 1061, normalized size = 3.74 \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))/(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
(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)))*b^3/((b^3*
c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a^2 - b^2)) - (6*b^2*c^4*d - 6*a*b*c^3*d^2 + 2*a^2*c^2*d^3
- 5*b^2*c^2*d^3 + a^2*d^5 + 2*b^2*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^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5*d^2 - 2*b^3*c^5*d^2 - a^3*c^4*d^3 + 6*a*b^2
*c^4*d^3 - 6*a^2*b*c^3*d^4 + b^3*c^3*d^4 + 2*a^3*c^2*d^5 - 3*a*b^2*c^2*d^5 + 3*a^2*b*c*d^6 - a^3*d^7)*sqrt(c^2
- d^2)) - (7*b*c^4*d^3*tan(1/2*f*x + 1/2*e)^3 - 5*a*c^3*d^4*tan(1/2*f*x + 1/2*e)^3 - 4*b*c^2*d^5*tan(1/2*f*x
+ 1/2*e)^3 + 2*a*c*d^6*tan(1/2*f*x + 1/2*e)^3 + 6*b*c^5*d^2*tan(1/2*f*x + 1/2*e)^2 - 4*a*c^4*d^3*tan(1/2*f*x +
1/2*e)^2 + 9*b*c^3*d^4*tan(1/2*f*x + 1/2*e)^2 - 7*a*c^2*d^5*tan(1/2*f*x + 1/2*e)^2 - 6*b*c*d^6*tan(1/2*f*x +
1/2*e)^2 + 2*a*d^7*tan(1/2*f*x + 1/2*e)^2 + 17*b*c^4*d^3*tan(1/2*f*x + 1/2*e) - 11*a*c^3*d^4*tan(1/2*f*x + 1/2
*e) - 8*b*c^2*d^5*tan(1/2*f*x + 1/2*e) + 2*a*c*d^6*tan(1/2*f*x + 1/2*e) + 6*b*c^5*d^2 - 4*a*c^4*d^3 - 3*b*c^3*
d^4 + a*c^2*d^5)/((b^2*c^8 - 2*a*b*c^7*d + a^2*c^6*d^2 - 2*b^2*c^6*d^2 + 4*a*b*c^5*d^3 - 2*a^2*c^4*d^4 + b^2*c
^4*d^4 - 2*a*b*c^3*d^5 + a^2*c^2*d^6)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f