3.250 \(\int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=176 \[ \frac{a (2 A c-A d+B c-2 B d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac{a \left (A d (c-2 d)+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 d f (c-d) (c+d)^2 (c+d \sin (e+f x))}+\frac{a (B c-A d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]
[Out]
(a*(2*A*c + B*c - A*d - 2*B*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*(c^2 - d^2)^(3/2)*f)
+ (a*(B*c - A*d)*Cos[e + f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (a*(A*(c - 2*d)*d + B*(c^2 + 2*c*d -
2*d^2))*Cos[e + f*x])/(2*(c - d)*d*(c + d)^2*f*(c + d*Sin[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.420982, antiderivative size = 176, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used =
{2968, 3021, 2754, 12, 2660, 618, 204} \[ \frac{a (2 A c-A d+B c-2 B d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac{a \left (A d (c-2 d)+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 d f (c-d) (c+d)^2 (c+d \sin (e+f x))}+\frac{a (B c-A d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^3,x]
[Out]
(a*(2*A*c + B*c - A*d - 2*B*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*(c^2 - d^2)^(3/2)*f)
+ (a*(B*c - A*d)*Cos[e + f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (a*(A*(c - 2*d)*d + B*(c^2 + 2*c*d -
2*d^2))*Cos[e + f*x])/(2*(c - d)*d*(c + d)^2*f*(c + d*Sin[e + f*x]))
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 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{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx &=\int \frac{a A+(a A+a B) \sin (e+f x)+a B \sin ^2(e+f x)}{(c+d \sin (e+f x))^3} \, dx\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{\int \frac{-2 a (A+B) (c-d) d-a (c-d) (A d+B (c+2 d)) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 d \left (c^2-d^2\right )}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac{\int \frac{a (c-d) d (2 A c+B c-A d-2 B d)}{c+d \sin (e+f x)} \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac{(a (2 A c+B c-A d-2 B d)) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 (c-d) (c+d)^2}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac{(a (2 A c+B c-A d-2 B d)) \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 )}{(c-d) (c+d)^2 f}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}-\frac{(2 a (2 A c+B c-A d-2 B d)) \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 )}{(c-d) (c+d)^2 f}\\ &=\frac{a (2 A c+B c-A d-2 B d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{(c-d) (c+d)^2 \sqrt{c^2-d^2} f}+\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [C] time = 2.63203, size = 345, normalized size = 1.96 \[ \frac{a (\sin (e+f x)+1) \left (\frac{d \csc (e) \left (\left (A d^2 (d-2 c)+B c \left (2 c^2+2 c d-3 d^2\right )\right ) \sin (2 e+f x)-d \left (A d (c-2 d)+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+2 f x)+\sin (f x) \left (B c \left (2 c^2+6 c d-5 d^2\right )-A d \left (-4 c^2+6 c d+d^2\right )\right )\right )+\left (2 c^2+d^2\right ) \cot (e) \left (A d (c-2 d)+B \left (c^2+2 c d-2 d^2\right )\right )}{d^2 (c+d \sin (e+f x))^2}+\frac{4 (\cos (e)-i \sin (e)) (2 A c-A d+B c-2 B d) \tan ^{-1}\left (\frac{(\cos (e)-i \sin (e)) \sec \left (\frac{f x}{2}\right ) \left (c \sin \left (\frac{f x}{2}\right )+d \cos \left (e+\frac{f x}{2}\right )\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{4 f (c-d) (c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^3,x]
[Out]
(a*(1 + Sin[e + f*x])*((4*(2*A*c + B*c - A*d - 2*B*d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x
)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*S
qrt[(Cos[e] - I*Sin[e])^2]) + ((2*c^2 + d^2)*(A*(c - 2*d)*d + B*(c^2 + 2*c*d - 2*d^2))*Cot[e] + d*Csc[e]*(-(d*
(A*(c - 2*d)*d + B*(c^2 + 2*c*d - 2*d^2))*Cos[e + 2*f*x]) + (B*c*(2*c^2 + 6*c*d - 5*d^2) - A*d*(-4*c^2 + 6*c*d
+ d^2))*Sin[f*x] + (A*d^2*(-2*c + d) + B*c*(2*c^2 + 2*c*d - 3*d^2))*Sin[2*e + f*x]))/(d^2*(c + d*Sin[e + f*x]
)^2)))/(4*(c - d)*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)
________________________________________________________________________________________
Maple [B] time = 0.164, size = 2021, normalized size = 11.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x)
[Out]
1/f*a/(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^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3*B-2/f
*a/(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^2*d-c*d^2-d^3)*c^2*tan(1/2*f*x+1/2*e)^2*A-3/f*a/
(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^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^2*A*d^2-2/f*a/(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^2*d-c*d^2-d^3)*c^2*tan(1/2*f*x+1/2*e)^2*B-2/f*a/(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^2*d-c*d^2-d^3)*A*c^2+1/f*a/(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^2*d-c*d^2-d^3)*c*tan(1/2*f*x+1/2*e)^2*B*d+4/f*a/(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^2*d-c*d^2-d^3)/c*tan(1/2*f*x+1/2*e)^2*A*d^3+2/f*a/(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^2*d-c*d^2-d^3)/c^2*tan(1/2*f*x+1/2*e)^2*A*d^4-3/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2
*f*x+1/2*e)*d+c)^2*c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3*A*d-2/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*
x+1/2*e)*d+c)^2*c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3*B*d+2/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1
/2*e)*d+c)^2/c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)*A*d^3-6/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*
e)*d+c)^2*c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)*B*d+2/f*a/(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^2*d-c*d^2-d^3)*c*tan(1/2*f*x+1/2*e)^2*A*d+2/f*a/(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^2*d-c*d^2-d^3)/c*tan(1/2*f*x+1/2*e)^2*B*d^3+2/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2
/c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3*A*d^3-5/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*
c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)*A*d+1/f*a/(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^2*d-c*d^2-d^3)*A*d^2-2/f*a/(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^2*d-c*d^2-d^3)*B*c^2-2
/f*a/(c^3+c^2*d-c*d^2-d^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))*B*d+1/f*a/
(c^3+c^2*d-c*d^2-d^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))*B*c+1/f*a/(c*ta
n(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*B*c*d+2/f*a/(c^3+c^2*d-c*d^2-d^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))*A*c+2/f*a/(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^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3*A*d^2-1/f*a/(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^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)*B+2/f*a/(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^2*d-c*d^2-d^3)*A*c*d-4/f*a/(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^2*d-c*d
^2-d^3)*tan(1/2*f*x+1/2*e)^2*B*d^2-1/f*a/(c^3+c^2*d-c*d^2-d^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))*A*d+6/f*a/(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^2*d-c*d^2-d^3)*
tan(1/2*f*x+1/2*e)*A*d^2+4/f*a/(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^2*d-c*d^2-d^3)*tan(1
/2*f*x+1/2*e)*B*d^2
________________________________________________________________________________________
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((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 2.47855, size = 2151, normalized size = 12.22 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
[1/4*(2*(B*a*c^4 + (A + 2*B)*a*c^3*d - (2*A + 3*B)*a*c^2*d^2 - (A + 2*B)*a*c*d^3 + 2*(A + B)*a*d^4)*cos(f*x +
e)*sin(f*x + e) + ((2*A + B)*a*c^3 - (A + 2*B)*a*c^2*d + (2*A + B)*a*c*d^2 - (A + 2*B)*a*d^3 - ((2*A + B)*a*c*
d^2 - (A + 2*B)*a*d^3)*cos(f*x + e)^2 + 2*((2*A + B)*a*c^2*d - (A + 2*B)*a*c*d^2)*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*(A + B)*a*c^4 - (2*
A + B)*a*c^3*d - (3*A + 2*B)*a*c^2*d^2 + (2*A + B)*a*c*d^3 + A*a*d^4)*cos(f*x + e))/((c^5*d^2 + c^4*d^3 - 2*c^
3*d^4 - 2*c^2*d^5 + c*d^6 + d^7)*f*cos(f*x + e)^2 - 2*(c^6*d + c^5*d^2 - 2*c^4*d^3 - 2*c^3*d^4 + c^2*d^5 + c*d
^6)*f*sin(f*x + e) - (c^7 + c^6*d - c^5*d^2 - c^4*d^3 - c^3*d^4 - c^2*d^5 + c*d^6 + d^7)*f), 1/2*((B*a*c^4 + (
A + 2*B)*a*c^3*d - (2*A + 3*B)*a*c^2*d^2 - (A + 2*B)*a*c*d^3 + 2*(A + B)*a*d^4)*cos(f*x + e)*sin(f*x + e) + ((
2*A + B)*a*c^3 - (A + 2*B)*a*c^2*d + (2*A + B)*a*c*d^2 - (A + 2*B)*a*d^3 - ((2*A + B)*a*c*d^2 - (A + 2*B)*a*d^
3)*cos(f*x + e)^2 + 2*((2*A + B)*a*c^2*d - (A + 2*B)*a*c*d^2)*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*(A + B)*a*c^4 - (2*A + B)*a*c^3*d - (3*A + 2*B)*a*c^2*d^2 + (2
*A + B)*a*c*d^3 + A*a*d^4)*cos(f*x + e))/((c^5*d^2 + c^4*d^3 - 2*c^3*d^4 - 2*c^2*d^5 + c*d^6 + d^7)*f*cos(f*x
+ e)^2 - 2*(c^6*d + c^5*d^2 - 2*c^4*d^3 - 2*c^3*d^4 + c^2*d^5 + c*d^6)*f*sin(f*x + e) - (c^7 + c^6*d - c^5*d^2
- c^4*d^3 - c^3*d^4 - c^2*d^5 + c*d^6 + d^7)*f)]
________________________________________________________________________________________
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((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.29728, size = 802, normalized size = 4.56 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
((2*A*a*c + B*a*c - A*a*d - 2*B*a*d)*(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)))/((c^3 + c^2*d - c*d^2 - d^3)*sqrt(c^2 - d^2)) + (B*a*c^4*tan(1/2*f*x + 1/2*e)^3 - 3*A*
a*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 2*B*a*c^3*d*tan(1/2*f*x + 1/2*e)^3 + 2*A*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 2
*A*a*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 2*A*a*c^4*tan(1/2*f*x + 1/2*e)^2 - 2*B*a*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*A*
a*c^3*d*tan(1/2*f*x + 1/2*e)^2 + B*a*c^3*d*tan(1/2*f*x + 1/2*e)^2 - 3*A*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 - 4*B
*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 + 4*A*a*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*B*a*c*d^3*tan(1/2*f*x + 1/2*e)^2 +
2*A*a*d^4*tan(1/2*f*x + 1/2*e)^2 - B*a*c^4*tan(1/2*f*x + 1/2*e) - 5*A*a*c^3*d*tan(1/2*f*x + 1/2*e) - 6*B*a*c^3
*d*tan(1/2*f*x + 1/2*e) + 6*A*a*c^2*d^2*tan(1/2*f*x + 1/2*e) + 4*B*a*c^2*d^2*tan(1/2*f*x + 1/2*e) + 2*A*a*c*d^
3*tan(1/2*f*x + 1/2*e) - 2*A*a*c^4 - 2*B*a*c^4 + 2*A*a*c^3*d + B*a*c^3*d + A*a*c^2*d^2)/((c^5 + c^4*d - c^3*d^
2 - c^2*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f