3.460 \(\int \frac{1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=213 \[ -\frac{3 d \left (2 c^2+2 c d+d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{5/2}}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a f (c-d)^3 (c+d)^2 (c+d \sin (e+f x))}-\frac{d (2 c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2} \]
[Out]
(-3*d*(2*c^2 + 2*c*d + d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 - d^2)^(5/2)*f)
- (d*(2*c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])^2) - Cos[e + f*x]/((c - d)*f*(a +
a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - (d*(2*c + d)*(c + 4*d)*Cos[e + f*x])/(2*a*(c - d)^3*(c + d)^2*f*(c
+ d*Sin[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.322788, antiderivative size = 213, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{2768, 2754, 12, 2660, 618, 204} \[ -\frac{3 d \left (2 c^2+2 c d+d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{5/2}}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a f (c-d)^3 (c+d)^2 (c+d \sin (e+f x))}-\frac{d (2 c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]
[Out]
(-3*d*(2*c^2 + 2*c*d + d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 - d^2)^(5/2)*f)
- (d*(2*c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])^2) - Cos[e + f*x]/((c - d)*f*(a +
a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - (d*(2*c + d)*(c + 4*d)*Cos[e + f*x])/(2*a*(c - d)^3*(c + d)^2*f*(c
+ d*Sin[e + f*x]))
Rule 2768
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2754
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx &=-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}+\frac{d \int \frac{-3 a+2 a \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{a^2 (c-d)}\\ &=-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d \int \frac{2 a (3 c+2 d)-a (2 c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 a^2 (c-d)^2 (c+d)}\\ &=-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}+\frac{d \int -\frac{3 a \left (2 c^2+2 c d+d^2\right )}{c+d \sin (e+f x)} \, dx}{2 a^2 (c-d)^3 (c+d)^2}\\ &=-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}-\frac{\left (3 d \left (2 c^2+2 c d+d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 a (c-d)^3 (c+d)^2}\\ &=-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}-\frac{\left (3 d \left (2 c^2+2 c d+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 )}{a (c-d)^3 (c+d)^2 f}\\ &=-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}+\frac{\left (6 d \left (2 c^2+2 c d+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 )}{a (c-d)^3 (c+d)^2 f}\\ &=-\frac{3 d \left (2 c^2+2 c d+d^2\right ) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a (c-d)^3 (c+d)^2 \sqrt{c^2-d^2} f}-\frac{d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.34363, size = 232, normalized size = 1.09 \[ \frac{\cos (e+f x) \left (\frac{\frac{2 c^2+9 c d+4 d^2}{(c-d)^2 (\sin (e+f x)+1)}+\frac{6 d \left (2 c^2+2 c d+d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{\sqrt{-c-d} (d-c)^{5/2} \sqrt{\cos ^2(e+f x)}}}{c+d}-\frac{d (4 c+d)}{(c-d) (c+d) (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac{d}{(\sin (e+f x)+1) (c+d \sin (e+f x))^2}\right )}{2 a f (d-c) (c+d)} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]
[Out]
(Cos[e + f*x]*(-(d/((1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^2)) - (d*(4*c + d))/((c - d)*(c + d)*(1 + Sin[e +
f*x])*(c + d*Sin[e + f*x])) + ((6*d*(2*c^2 + 2*c*d + d^2)*ArcTan[(Sqrt[-c + d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-
c - d]*Sqrt[1 + Sin[e + f*x]])])/(Sqrt[-c - d]*(-c + d)^(5/2)*Sqrt[Cos[e + f*x]^2]) + (2*c^2 + 9*c*d + 4*d^2)/
((c - d)^2*(1 + Sin[e + f*x])))/(c + d)))/(2*a*(-c + d)*(c + d)*f)
________________________________________________________________________________________
Maple [B] time = 0.115, size = 1224, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x)
[Out]
-7/a/f*d^3/(c-d)^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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3-
2/a/f*d^4/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3+2/a
/f*d^5/(c-d)^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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3-6/a/
f*d^2/(c-d)^3/(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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2-2/a
/f*d^3/(c-d)^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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2-11/a
/f*d^4/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2-4/a/f*
d^5/(c-d)^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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2+2/a/f*d
^6/(c-d)^3/(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^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2-17/a/f
*d^3/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c*tan(1/2*f*x+1/2*e)-6/a/f*d^
4/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)+2/a/f*d^5/(c-
d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)-6/a/f*d^2/(c-d)^
3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c^2-2/a/f*d^3/(c-d)^3/(c*tan(1/2*f*x+1/2
*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c+1/a/f*d^4/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/
2*e)*d+c)^2/(c^2+2*c*d+d^2)-6/a/f*d/(c-d)^3/(c^2+2*c*d+d^2)/(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))*c^2-6/a/f*d^2/(c-d)^3/(c^2+2*c*d+d^2)/(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))*c-3/a/f*d^3/(c-d)^3/(c^2+2*c*d+d^2)/(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))-2/a/f/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)
________________________________________________________________________________________
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+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 2.38012, size = 5072, normalized size = 23.81 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
[1/4*(4*c^6 - 12*c^4*d^2 + 12*c^2*d^4 - 4*d^6 - 2*(2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*
x + e)^3 + 2*(4*c^5*d + 12*c^4*d^2 - 2*c^3*d^3 - 15*c^2*d^4 - 2*c*d^5 + 3*d^6)*cos(f*x + e)^2 - 3*(2*c^4*d + 6
*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^3 - (4*c^3*d^2 + 6*c^2*d^3 + 4
*c*d^4 + d^5)*cos(f*x + e)^2 + (2*c^4*d + 2*c^3*d^2 + 3*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e) + (2*c^4*d + 6*c
^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^2 + 2*(2*c^3*d^2 + 2*c^2*d^3 + c
*d^4)*cos(f*x + e))*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)) + 2*(2*c^6 + 4*c^5*d + 8*c^4*d^2 + 7*c^3*d^3 - 7*c^2*d^4 - 11*c*d^5 - 3*d^6)*cos(f*x + e
) - 2*(2*c^6 - 6*c^4*d^2 + 6*c^2*d^4 - 2*d^6 - (2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x +
e)^2 - (4*c^5*d + 14*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^7*d
^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^3 + (
2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5 - 3*a*c^3*d^6 - 5*a*c^2*d^7 + a*c*d^8 + a*d^9)
*f*cos(f*x + e)^2 - (a*c^9 - a*c^8*d - 2*a*c^7*d^2 + 2*a*c^6*d^3 + 2*a*c^3*d^6 - 2*a*c^2*d^7 - a*c*d^8 + a*d^9
)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*
a*c^2*d^7 + a*c*d^8 + a*d^9)*f + ((a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d
^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - 2*(a*c^8*d - a*c^7*d^2 - 3*a*c^6*d^3 + 3*a*c^5*d^4 + 3*a*c^4*d^5 - 3*
a*c^3*d^6 - a*c^2*d^7 + a*c*d^8)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 +
6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f)*sin(f*x + e)), 1/2*(2*c^6 - 6*c^4*d^2 + 6*c^2*d
^4 - 2*d^6 - (2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x + e)^3 + (4*c^5*d + 12*c^4*d^2 - 2*
c^3*d^3 - 15*c^2*d^4 - 2*c*d^5 + 3*d^6)*cos(f*x + e)^2 - 3*(2*c^4*d + 6*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 -
(2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^3 - (4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*cos(f*x + e)^2 + (2*c^4*d
+ 2*c^3*d^2 + 3*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e) + (2*c^4*d + 6*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2
*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^2 + 2*(2*c^3*d^2 + 2*c^2*d^3 + c*d^4)*cos(f*x + e))*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))) + (2*c^6 + 4*c^5*d + 8*c^4*d^2 + 7*c^3
*d^3 - 7*c^2*d^4 - 11*c*d^5 - 3*d^6)*cos(f*x + e) - (2*c^6 - 6*c^4*d^2 + 6*c^2*d^4 - 2*d^6 - (2*c^4*d^2 + 9*c^
3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x + e)^2 - (4*c^5*d + 14*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^
5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c
^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^3 + (2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5
- 3*a*c^3*d^6 - 5*a*c^2*d^7 + a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - (a*c^9 - a*c^8*d - 2*a*c^7*d^2 + 2*a*c^6*d^3
+ 2*a*c^3*d^6 - 2*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3
+ 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f + ((a*c^7*d^2 - a*c^6*d^3 - 3*a*c
^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - 2*(a*c^8*d - a*c^7*d^2
- 3*a*c^6*d^3 + 3*a*c^5*d^4 + 3*a*c^4*d^5 - 3*a*c^3*d^6 - a*c^2*d^7 + a*c*d^8)*f*cos(f*x + e) - (a*c^9 + a*c^8
*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f)*s
in(f*x + e))]
________________________________________________________________________________________
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+a*sin(f*x+e))/(c+d*sin(f*x+e))**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.47471, size = 651, normalized size = 3.06 \begin{align*} -\frac{\frac{3 \,{\left (2 \, c^{2} d + 2 \, c d^{2} + d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{{\left (a c^{5} - a c^{4} d - 2 \, a c^{3} d^{2} + 2 \, a c^{2} d^{3} + a c d^{4} - a d^{5}\right )} \sqrt{c^{2} - d^{2}}} + \frac{7 \, c^{3} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, c^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, c d^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, c^{4} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, c^{3} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 11 \, c^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 4 \, c d^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, d^{6} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 17 \, c^{3} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, c^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, c d^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, c^{4} d^{2} + 2 \, c^{3} d^{3} - c^{2} d^{4}}{{\left (a c^{7} - a c^{6} d - 2 \, a c^{5} d^{2} + 2 \, a c^{4} d^{3} + a c^{3} d^{4} - a c^{2} d^{5}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}^{2}} + \frac{2}{{\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-(3*(2*c^2*d + 2*c*d^2 + d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/s
qrt(c^2 - d^2)))/((a*c^5 - a*c^4*d - 2*a*c^3*d^2 + 2*a*c^2*d^3 + a*c*d^4 - a*d^5)*sqrt(c^2 - d^2)) + (7*c^3*d^
3*tan(1/2*f*x + 1/2*e)^3 + 2*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 - 2*c*d^5*tan(1/2*f*x + 1/2*e)^3 + 6*c^4*d^2*tan(1
/2*f*x + 1/2*e)^2 + 2*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 + 11*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 + 4*c*d^5*tan(1/2*f*x
+ 1/2*e)^2 - 2*d^6*tan(1/2*f*x + 1/2*e)^2 + 17*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*c^2*d^4*tan(1/2*f*x + 1/2*e)
- 2*c*d^5*tan(1/2*f*x + 1/2*e) + 6*c^4*d^2 + 2*c^3*d^3 - c^2*d^4)/((a*c^7 - a*c^6*d - 2*a*c^5*d^2 + 2*a*c^4*d^
3 + a*c^3*d^4 - a*c^2*d^5)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2) + 2/((a*c^3 - 3*a*c^2*
d + 3*a*c*d^2 - a*d^3)*(tan(1/2*f*x + 1/2*e) + 1)))/f