3.229 \(\int \frac{(c e+d e x)^2}{(a+b \sin ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=248 \[ -\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^3 d}+\frac{9 e^2 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 b^3 d}-\frac{e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^3 d}+\frac{9 e^2 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 b^3 d}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

[Out]

-(e^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - (e^2*(c + d*x))/(b^2*d*(a + b*Arc
Sin[c + d*x])) + (3*e^2*(c + d*x)^3)/(2*b^2*d*(a + b*ArcSin[c + d*x])) - (e^2*Cos[a/b]*CosIntegral[(a + b*ArcS
in[c + d*x])/b])/(8*b^3*d) + (9*e^2*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcSin[c + d*x]))/b])/(8*b^3*d) - (e^2*
Sin[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(8*b^3*d) + (9*e^2*Sin[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[
c + d*x]))/b])/(8*b^3*d)

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Rubi [A]  time = 0.577872, antiderivative size = 306, normalized size of antiderivative = 1.23, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302, 4623} \[ -\frac{9 e^2 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{9 e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2/(a + b*ArcSin[c + d*x])^3,x]

[Out]

-(e^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - (e^2*(c + d*x))/(b^2*d*(a + b*Arc
Sin[c + d*x])) + (3*e^2*(c + d*x)^3)/(2*b^2*d*(a + b*ArcSin[c + d*x])) - (9*e^2*Cos[a/b]*CosIntegral[a/b + Arc
Sin[c + d*x]])/(8*b^3*d) + (9*e^2*Cos[(3*a)/b]*CosIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(8*b^3*d) + (e^2*Cos[
a/b]*CosIntegral[(a + b*ArcSin[c + d*x])/b])/(b^3*d) - (9*e^2*Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(8*
b^3*d) + (9*e^2*Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(8*b^3*d) + (e^2*Sin[a/b]*SinIntegral[(
a + b*ArcSin[c + d*x])/b])/(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]

Rule 4623

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

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac{e^2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}-\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}-\frac{x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{b^3 d}-\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{4 (a+b x)}-\frac{\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (e^2 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{b^3 d}+\frac{\left (e^2 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{b^3 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}+\frac{e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}+\frac{e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{\left (9 e^2 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}-\frac{\left (9 e^2 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{9 e^2 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{9 e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.799369, size = 219, normalized size = 0.88 \[ \frac{e^2 \left (-\frac{4 b^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+8 \left (\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+9 \left (-\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )+\frac{4 b \left (3 (c+d x)^3-2 (c+d x)\right )}{a+b \sin ^{-1}(c+d x)}\right )}{8 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^2/(a + b*ArcSin[c + d*x])^3,x]

[Out]

(e^2*((-4*b^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x])^2 + (4*b*(-2*(c + d*x) + 3*(c + d*x)^
3))/(a + b*ArcSin[c + d*x]) + 8*(Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]] + Sin[a/b]*SinIntegral[a/b + ArcS
in[c + d*x]]) + 9*(-(Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]]) + Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c
+ d*x])] - Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]] + Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])])
))/(8*b^3*d)

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Maple [B]  time = 0.053, size = 474, normalized size = 1.9 \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)^2/(a+b*arcsin(d*x+c))^3,x)

[Out]

-1/8/d*e^2*(arcsin(d*x+c)^2*Si(arcsin(d*x+c)+a/b)*sin(a/b)*b^2+arcsin(d*x+c)^2*Ci(arcsin(d*x+c)+a/b)*cos(a/b)*
b^2-9*arcsin(d*x+c)^2*Si(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*b^2-9*arcsin(d*x+c)^2*Ci(3*arcsin(d*x+c)+3*a/b)*cos
(3*a/b)*b^2+2*arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*sin(a/b)*a*b+2*arcsin(d*x+c)*Ci(arcsin(d*x+c)+a/b)*cos(a/b)*
a*b-18*arcsin(d*x+c)*Si(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*a*b-18*arcsin(d*x+c)*Ci(3*arcsin(d*x+c)+3*a/b)*cos(3
*a/b)*a*b+3*arcsin(d*x+c)*sin(3*arcsin(d*x+c))*b^2-arcsin(d*x+c)*(d*x+c)*b^2+Si(arcsin(d*x+c)+a/b)*sin(a/b)*a^
2+Ci(arcsin(d*x+c)+a/b)*cos(a/b)*a^2-9*Si(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*a^2-9*Ci(3*arcsin(d*x+c)+3*a/b)*co
s(3*a/b)*a^2+(1-(d*x+c)^2)^(1/2)*b^2-cos(3*arcsin(d*x+c))*b^2+3*sin(3*arcsin(d*x+c))*a*b-(d*x+c)*a*b)/(a+b*arc
sin(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)^2/(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^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{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)^2/(a+b*arcsin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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

[Out]

e**2*(Integral(c**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(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) + I
ntegral(2*c*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 = 1.99738, size = 2183, normalized size = 8.8 \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)^2/(a+b*arcsin(d*x+c))^3,x, algorithm="giac")

[Out]

9/2*b^2*arcsin(d*x + c)^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*
a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 9/2*b^2*arcsin(d*x + c)^2*cos(a/b)^2*e^2*sin(a/b)*sin_integral(3*a/b +
3*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) + 9*a*b*arcsin(d*x + c)*c
os(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a
^2*b^3*d) + 9*a*b*arcsin(d*x + c)*cos(a/b)^2*e^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^5*d*arcsi
n(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 27/8*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(3*a/b
+ 3*arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 9/2*a^2*cos(a/b)
^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*
d) - 1/8*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(a/b + arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a
*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 9/8*b^2*arcsin(d*x + c)^2*e^2*sin(a/b)*sin_integral(3*a/b + 3*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) + 9/2*a^2*cos(a/b)^2*e^2*sin(a/b)*sin
_integral(3*a/b + 3*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) - 1/8*b
^2*arcsin(d*x + c)^2*e^2*sin(a/b)*sin_integral(a/b + 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) + 3/2*((d*x + c)^2 - 1)*(d*x + c)*b^2*arcsin(d*x + c)*e^2/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 27/4*a*b*arcsin(d*x + c)*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x
+ c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/4*a*b*arcsin(d*x + c)*cos(a/b)
*cos_integral(a/b + arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 9
/4*a*b*arcsin(d*x + c)*e^2*sin(a/b)*sin_integral(3*a/b + 3*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) - 1/4*a*b*arcsin(d*x + c)*e^2*sin(a/b)*sin_integral(a/b + 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) + 3/2*((d*x + c)^2 - 1)*(d*x + c)*a*b*e^2/(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)*b^2*arcsin(d*x + c)*e^2/(b^5*d*ar
csin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 27/8*a^2*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x
+ c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/8*a^2*cos(a/b)*cos_integral(a
/b + arcsin(d*x + c))*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 9/8*a^2*e^2*sin(
a/b)*sin_integral(3*a/b + 3*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)
- 1/8*a^2*e^2*sin(a/b)*sin_integral(a/b + 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)*b^2*e^2/(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)*a*b*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/2*sqr
t(-(d*x + c)^2 + 1)*b^2*e^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d)