3.451 \(\int \frac{(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=207 \[ \frac{5 a^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d)^3 \sqrt{c^2-d^2}}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 f (c+d)^3 (c+d \sin (e+f x))}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 f (c+d)^2 (c+d \sin (e+f x))^2}+\frac{(c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

[Out]

(5*a^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^3*Sqrt[c^2 - d^2]*f) + ((c - d)*Cos[e + f*x]
*(a^3 + a^3*Sin[e + f*x]))/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^3) + (a^3*(c - d)*(2*c + 7*d)*Cos[e + f*x])/(6*
d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^2) - (a^3*(2*c^2 + 9*c*d + 22*d^2)*Cos[e + f*x])/(6*d^2*(c + d)^3*f*(c +
d*Sin[e + f*x]))

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Rubi [A]  time = 0.475085, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2762, 2968, 3021, 2754, 12, 2660, 618, 204} \[ \frac{5 a^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d)^3 \sqrt{c^2-d^2}}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 f (c+d)^3 (c+d \sin (e+f x))}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 f (c+d)^2 (c+d \sin (e+f x))^2}+\frac{(c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^4,x]

[Out]

(5*a^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^3*Sqrt[c^2 - d^2]*f) + ((c - d)*Cos[e + f*x]
*(a^3 + a^3*Sin[e + f*x]))/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^3) + (a^3*(c - d)*(2*c + 7*d)*Cos[e + f*x])/(6*
d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^2) - (a^3*(2*c^2 + 9*c*d + 22*d^2)*Cos[e + f*x])/(6*d^2*(c + d)^3*f*(c +
d*Sin[e + f*x]))

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

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))^3}{(c+d \sin (e+f x))^4} \, dx &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a \int \frac{(a+a \sin (e+f x)) (a (c-7 d)-2 a (c+2 d) \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a \int \frac{a^2 (c-7 d)+\left (a^2 (c-7 d)-2 a^2 (c+2 d)\right ) \sin (e+f x)-2 a^2 (c+2 d) \sin ^2(e+f x)}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}+\frac{a \int \frac{2 a^2 (c-d) d (c+11 d)+a^2 (c-d) \left (2 c^2+7 c d+15 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{6 (c-d) d^2 (c+d)^2}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}-\frac{a \int -\frac{15 a^2 (c-d)^2 d^2}{c+d \sin (e+f x)} \, dx}{6 (c-d)^2 d^2 (c+d)^3}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}+\frac{\left (5 a^3\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 (c+d)^3}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}+\frac{\left (5 a^3\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 )}{(c+d)^3 f}\\ &=\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}-\frac{\left (10 a^3\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 )}{(c+d)^3 f}\\ &=\frac{5 a^3 \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{(c+d)^3 \sqrt{c^2-d^2} f}+\frac{(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac{a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.45778, size = 178, normalized size = 0.86 \[ \frac{a^3 \cos (e+f x) \left (-\frac{(\sin (e+f x)+1)^2}{(c+d \sin (e+f x))^3}-\frac{5 (\sin (e+f x)+1)}{2 (c+d) (c+d \sin (e+f x))^2}-\frac{15}{2 (c+d)^2 (c+d \sin (e+f x))}+\frac{15 \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{(-c-d)^{5/2} \sqrt{d-c} \sqrt{\cos ^2(e+f x)}}\right )}{3 f (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^4,x]

[Out]

(a^3*Cos[e + f*x]*((15*ArcTan[(Sqrt[-c + d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/((
-c - d)^(5/2)*Sqrt[-c + d]*Sqrt[Cos[e + f*x]^2]) - (1 + Sin[e + f*x])^2/(c + d*Sin[e + f*x])^3 - (5*(1 + Sin[e
+ f*x]))/(2*(c + d)*(c + d*Sin[e + f*x])^2) - 15/(2*(c + d)^2*(c + d*Sin[e + f*x]))))/(3*(c + d)*f)

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Maple [B]  time = 0.17, size = 1924, 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((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x)

[Out]

-18/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^2
*d^3-4/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2
*e)^2*d^4-38/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x
+1/2*e)*d-2/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+
1/2*e)*d^3-6/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x
+1/2*e)^5*d-2/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*
x+1/2*e)^5*d^3+3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2
*f*x+1/2*e)^4*d-18/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1
/2*f*x+1/2*e)^4*d^3-4/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*
tan(1/2*f*x+1/2*e)^4*d^4-44/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c*d/(c^3+3*c^2*d+3*c*d^2
+d^3)*tan(1/2*f*x+1/2*e)^3-100/3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c*d^3/(c^3+3*c^2*d+
3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^3-12/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2*d^4/(c^3+3*
c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^3-6/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3+3*c^2
*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5*d^2-30/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3+3*c
^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^4*d^2-18/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*d^2/(c
^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^3-12/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3
+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)*d^2-60/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3
+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^2*d^2+3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2
/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5-6/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^
2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^4-16/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*
c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^2-3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3
*c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)-3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/
(c^3+3*c^2*d+3*c*d^2+d^3)*c*d-22/3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3+3*c^2*d+3*c*
d^2+d^3)*c^2-2/3/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^3+3*c^2*d+3*c*d^2+d^3)*d^2-8/3/f
*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^3*d^5/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^
3-12/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^
2*d+5/f*a^3/(c^3+3*c^2*d+3*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))

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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((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14736, size = 2369, normalized size = 11.44 \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))^3/(c+d*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

[-1/12*(2*(2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^2*d^2 - 9*a^3*c*d^3 - 22*a^3*d^4)*cos(f*x + e)^3 - 6*(3*a^3*c^4
+ 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x + e)*sin(f*x + e) + 15*(3*a^3*c*d^2*cos(f*x + e)^2 - a^3*c^
3 - 3*a^3*c*d^2 + (a^3*d^3*cos(f*x + e)^2 - 3*a^3*c^2*d - a^3*d^3)*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)) - 12*(4*a^3*c^4 + 3*a^3*c^3*d - 3*a^3*c*d
^3 - 4*a^3*d^4)*cos(f*x + e))/(3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^3*d^5 - 3*c^2*d^6 - c*d^7)*f*cos(f*x +
e)^2 - (c^8 + 3*c^7*d + 5*c^6*d^2 + 7*c^5*d^3 + 3*c^4*d^4 - 7*c^3*d^5 - 9*c^2*d^6 - 3*c*d^7)*f + ((c^5*d^3 +
3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f*cos(f*x + e)^2 - (3*c^7*d + 9*c^6*d^2 + 7*c^5*d^3 - 3*c^4
*d^4 - 7*c^3*d^5 - 5*c^2*d^6 - 3*c*d^7 - d^8)*f)*sin(f*x + e)), -1/6*((2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^2*d^
2 - 9*a^3*c*d^3 - 22*a^3*d^4)*cos(f*x + e)^3 - 3*(3*a^3*c^4 + 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x
+ e)*sin(f*x + e) + 15*(3*a^3*c*d^2*cos(f*x + e)^2 - a^3*c^3 - 3*a^3*c*d^2 + (a^3*d^3*cos(f*x + e)^2 - 3*a^3*
c^2*d - a^3*d^3)*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))) -
6*(4*a^3*c^4 + 3*a^3*c^3*d - 3*a^3*c*d^3 - 4*a^3*d^4)*cos(f*x + e))/(3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^
3*d^5 - 3*c^2*d^6 - c*d^7)*f*cos(f*x + e)^2 - (c^8 + 3*c^7*d + 5*c^6*d^2 + 7*c^5*d^3 + 3*c^4*d^4 - 7*c^3*d^5 -
9*c^2*d^6 - 3*c*d^7)*f + ((c^5*d^3 + 3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f*cos(f*x + e)^2 - (3
*c^7*d + 9*c^6*d^2 + 7*c^5*d^3 - 3*c^4*d^4 - 7*c^3*d^5 - 5*c^2*d^6 - 3*c*d^7 - d^8)*f)*sin(f*x + e))]

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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((a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.48734, size = 900, normalized size = 4.35 \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))^3/(c+d*sin(f*x+e))^4,x, algorithm="giac")

[Out]

1/3*(15*(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)))*a^3/(
(c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sqrt(c^2 - d^2)) + (9*a^3*c^5*tan(1/2*f*x + 1/2*e)^5 - 18*a^3*c^4*d*tan(1/2*f*
x + 1/2*e)^5 - 18*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 6*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 - 18*a^3*c^5*tan(1
/2*f*x + 1/2*e)^4 + 9*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^4 - 90*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^4 - 54*a^3*c^2*d^
3*tan(1/2*f*x + 1/2*e)^4 - 12*a^3*c*d^4*tan(1/2*f*x + 1/2*e)^4 - 132*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 54*a^3
*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 100*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 36*a^3*c*d^4*tan(1/2*f*x + 1/2*e)^3
- 8*a^3*d^5*tan(1/2*f*x + 1/2*e)^3 - 48*a^3*c^5*tan(1/2*f*x + 1/2*e)^2 - 36*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^2
- 180*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 54*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 - 12*a^3*c*d^4*tan(1/2*f*x +
1/2*e)^2 - 9*a^3*c^5*tan(1/2*f*x + 1/2*e) - 114*a^3*c^4*d*tan(1/2*f*x + 1/2*e) - 36*a^3*c^3*d^2*tan(1/2*f*x +
1/2*e) - 6*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e) - 22*a^3*c^5 - 9*a^3*c^4*d - 2*a^3*c^3*d^2)/((c^6 + 3*c^5*d + 3*c^
4*d^2 + c^3*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^3))/f