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

Optimal. Leaf size=249 \[ \frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}+\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - (3*e^3*(c + d*x)^2)/(2*b^2*d*(a +
b*ArcSin[c + d*x])) + (2*e^3*(c + d*x)^4)/(b^2*d*(a + b*ArcSin[c + d*x])) + (e^3*CosIntegral[(2*(a + b*ArcSin
[c + d*x]))/b]*Sin[(2*a)/b])/(2*b^3*d) - (e^3*CosIntegral[(4*(a + b*ArcSin[c + d*x]))/b]*Sin[(4*a)/b])/(b^3*d)
- (e^3*Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c + d*x]))/b])/(2*b^3*d) + (e^3*Cos[(4*a)/b]*SinIntegral[(4*
(a + b*ArcSin[c + d*x]))/b])/(b^3*d)

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Rubi [A]  time = 0.659939, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302} \[ \frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^3 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{b^3 d}-\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^3 d}+\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{b^3 d}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^3,x]

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - (3*e^3*(c + d*x)^2)/(2*b^2*d*(a +
b*ArcSin[c + d*x])) + (2*e^3*(c + d*x)^4)/(b^2*d*(a + b*ArcSin[c + d*x])) + (e^3*CosIntegral[(2*a)/b + 2*ArcS
in[c + d*x]]*Sin[(2*a)/b])/(2*b^3*d) - (e^3*CosIntegral[(4*a)/b + 4*ArcSin[c + d*x]]*Sin[(4*a)/b])/(b^3*d) - (
e^3*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c + d*x]])/(2*b^3*d) + (e^3*Cos[(4*a)/b]*SinIntegral[(4*a)/b +
4*ArcSin[c + d*x]])/(b^3*d)

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4633

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] + (Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSin[c*x])^(n + 1))
/Sqrt[1 - c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^
2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4635

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
in[x]^m*Cos[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 (a+b x)} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 (a+b x)}-\frac{\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (2 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (e^3 \cos \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (3 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (2 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (e^3 \sin \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{2 a}{b}\right )}{2 b^3 d}-\frac{e^3 \text{Ci}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{4 a}{b}\right )}{b^3 d}-\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^3 d}+\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.705064, size = 181, normalized size = 0.73 \[ \frac{e^3 \left (-\frac{b^2 \sqrt{1-(c+d x)^2} (c+d x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-2 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac{b \left (4 (c+d x)^4-3 (c+d x)^2\right )}{a+b \sin ^{-1}(c+d x)}\right )}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^3,x]

[Out]

(e^3*(-((b^2*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x])^2) + (b*(-3*(c + d*x)^2 + 4*(c + d*x)^
4))/(a + b*ArcSin[c + d*x]) + CosIntegral[2*(a/b + ArcSin[c + d*x])]*Sin[(2*a)/b] - 2*CosIntegral[4*(a/b + Arc
Sin[c + d*x])]*Sin[(4*a)/b] - Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c + d*x])] + 2*Cos[(4*a)/b]*SinIntegral
[4*(a/b + ArcSin[c + d*x])]))/(2*b^3*d)

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Maple [B]  time = 0.044, size = 506, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^3,x)

[Out]

1/16/d*e^3*(16*arcsin(d*x+c)^2*Si(4*arcsin(d*x+c)+4*a/b)*cos(4*a/b)*b^2-16*arcsin(d*x+c)^2*Ci(4*arcsin(d*x+c)+
4*a/b)*sin(4*a/b)*b^2-8*arcsin(d*x+c)^2*Si(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*b^2+8*arcsin(d*x+c)^2*Ci(2*arcsin
(d*x+c)+2*a/b)*sin(2*a/b)*b^2+32*arcsin(d*x+c)*Si(4*arcsin(d*x+c)+4*a/b)*cos(4*a/b)*a*b-32*arcsin(d*x+c)*Ci(4*
arcsin(d*x+c)+4*a/b)*sin(4*a/b)*a*b-16*arcsin(d*x+c)*Si(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*a*b+16*arcsin(d*x+c)
*Ci(2*arcsin(d*x+c)+2*a/b)*sin(2*a/b)*a*b+4*arcsin(d*x+c)*cos(4*arcsin(d*x+c))*b^2-4*arcsin(d*x+c)*cos(2*arcsi
n(d*x+c))*b^2+16*Si(4*arcsin(d*x+c)+4*a/b)*cos(4*a/b)*a^2-16*Ci(4*arcsin(d*x+c)+4*a/b)*sin(4*a/b)*a^2-8*Si(2*a
rcsin(d*x+c)+2*a/b)*cos(2*a/b)*a^2+8*Ci(2*arcsin(d*x+c)+2*a/b)*sin(2*a/b)*a^2+sin(4*arcsin(d*x+c))*b^2+4*cos(4
*arcsin(d*x+c))*a*b-2*sin(2*arcsin(d*x+c))*b^2-4*cos(2*arcsin(d*x+c))*a*b)/(a+b*arcsin(d*x+c))^2/b^3

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Maxima [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((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3)/(b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x
+ c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{3} \left (\int \frac{c^{3}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{d^{3} x^{3}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{3 c d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{3 c^{2} d x}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asin(d*x+c))**3,x)

[Out]

e**3*(Integral(c**3/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x) +
Integral(d**3*x**3/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x) + I
ntegral(3*c*d**2*x**2/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x)
+ Integral(3*c**2*d*x/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x))

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Giac [B]  time = 2.01396, size = 2928, normalized size = 11.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^3,x, algorithm="giac")

[Out]

-8*b^2*arcsin(d*x + c)^2*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c
)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 8*b^2*arcsin(d*x + c)^2*cos(a/b)^4*e^3*sin_integral(4*a/b + 4*a
rcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 16*a*b*arcsin(d*x + c)*cos
(a/b)^3*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x +
c) + a^2*b^3*d) + 16*a*b*arcsin(d*x + c)*cos(a/b)^4*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin
(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 4*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(4*a/b + 4
*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 8*a^2*cos(a
/b)^3*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c
) + a^2*b^3*d) + b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*ar
csin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 8*b^2*arcsin(d*x + c)^2*cos(a/b)^2*e^3*sin_integral
(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 8*a^2*cos(a/b)
^4*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*
d) - b^2*arcsin(d*x + c)^2*cos(a/b)^2*e^3*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2
*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 8*a*b*arcsin(d*x + c)*cos(a/b)*cos_integral(4*a/b + 4*arcsin(d*x + c))
*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 2*a*b*arcsin(d*x + c)*cos(a/
b)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d) - 16*a*b*arcsin(d*x + c)*cos(a/b)^2*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d*x
+ c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 2*a*b*arcsin(d*x + c)*cos(a/b)^2*e^3*sin_integral(2*a/b + 2*
arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 2*((d*x + c)^2 - 1)^2*b^2
*arcsin(d*x + c)*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 4*a^2*cos(a/b)*cos_in
tegral(4*a/b + 4*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*
d) + a^2*cos(a/b)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*ar
csin(d*x + c) + a^2*b^3*d) + b^2*arcsin(d*x + c)^2*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d
*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 8*a^2*cos(a/b)^2*e^3*sin_integral(4*a/b + 4*arcsin(d*x +
c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*b^2*arcsin(d*x + c)^2*e^3*sin_inte
gral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - a^2*cos(a/
b)^2*e^3*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^
3*d) + 1/2*(-(d*x + c)^2 + 1)^(3/2)*(d*x + c)*b^2*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a
^2*b^3*d) + 2*((d*x + c)^2 - 1)^2*a*b*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) +
5/2*((d*x + c)^2 - 1)*b^2*arcsin(d*x + c)*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d
) + 2*a*b*arcsin(d*x + c)*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arc
sin(d*x + c) + a^2*b^3*d) + a*b*arcsin(d*x + c)*e^3*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x
+ c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/2*sqrt(-(d*x + c)^2 + 1)*(d*x + c)*b^2*e^3/(b^5*d*arcsin(d
*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 5/2*((d*x + c)^2 - 1)*a*b*e^3/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*b^2*arcsin(d*x + c)*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcs
in(d*x + c) + a^2*b^3*d) + a^2*e^3*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*
d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*a^2*e^3*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2
+ 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*a*b*e^3/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d)