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

Optimal. Leaf size=287 \[ -\frac{2 \sqrt{2 \pi } c \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{2 \pi } c \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}} \]

[Out]

(2*c*Sqrt[1 - (c + d*x)^2])/(b*d^2*Sqrt[a + b*ArcSin[c + d*x]]) - (2*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d^2*S
qrt[a + b*ArcSin[c + d*x]]) + (2*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[
Pi])])/(b^(3/2)*d^2) + (2*c*Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(b
^(3/2)*d^2) - (2*c*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(b^(3/2)*d^
2) + (2*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(b^(3/2)*d^2)

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Rubi [A]  time = 0.610907, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4805, 4745, 4621, 4723, 3306, 3305, 3351, 3304, 3352, 4631} \[ -\frac{2 \sqrt{2 \pi } c \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{2 \pi } c \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d^2}+\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 - (c + d*x)^2])/(b*d^2*Sqrt[a + b*ArcSin[c + d*x]]) - (2*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d^2*S
qrt[a + b*ArcSin[c + d*x]]) + (2*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[
Pi])])/(b^(3/2)*d^2) + (2*c*Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(b
^(3/2)*d^2) - (2*c*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(b^(3/2)*d^
2) + (2*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(b^(3/2)*d^2)

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 4745

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

Rule 4621

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

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3306

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

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3304

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

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4631

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

Rubi steps

\begin{align*} \int \frac{x}{\left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-\frac{c}{d}+\frac{x}{d}}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{c}{d \left (a+b \sin ^{-1}(x)\right )^{3/2}}+\frac{x}{d \left (a+b \sin ^{-1}(x)\right )^{3/2}}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d^2}-\frac{c \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sqrt{a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{b d^2}\\ &=\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}+\frac{\left (2 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}+\frac{\left (2 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}\\ &=\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (2 c \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}+\frac{\left (4 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d^2}-\frac{\left (2 c \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d^2}+\frac{\left (4 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d^2}\\ &=\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) C\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{b^{3/2} d^2}+\frac{\left (4 c \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d^2}-\frac{\left (4 c \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d^2}\\ &=\frac{2 c \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 (c+d x) \sqrt{1-(c+d x)^2}}{b d^2 \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) C\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d^2}+\frac{2 c \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d^2}-\frac{2 c \sqrt{2 \pi } C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{b^{3/2} d^2}+\frac{2 \sqrt{\pi } S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{b^{3/2} d^2}\\ \end{align*}

Mathematica [C]  time = 2.2409, size = 287, normalized size = 1. \[ \frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{3/2} \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi }}\right )+\frac{-c e^{-\frac{i a}{b}} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-c e^{\frac{i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 \sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{2 a}{b}\right ) \sqrt{a+b \sin ^{-1}(c+d x)} S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi }}\right )+c e^{-i \sin ^{-1}(c+d x)}+c e^{i \sin ^{-1}(c+d x)}-\sin \left (2 \sin ^{-1}(c+d x)\right )}{b \sqrt{a+b \sin ^{-1}(c+d x)}}}{d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[Pi]] + (c/E
^(I*ArcSin[c + d*x]) + c*E^(I*ArcSin[c + d*x]) - (c*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-I)*(a
+ b*ArcSin[c + d*x]))/b])/E^((I*a)/b) - c*E^((I*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, (I*(a +
b*ArcSin[c + d*x]))/b] + 2*Sqrt[b^(-1)]*Sqrt[Pi]*Sqrt[a + b*ArcSin[c + d*x]]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a +
b*ArcSin[c + d*x]])/Sqrt[Pi]]*Sin[(2*a)/b] - Sin[2*ArcSin[c + d*x]])/(b*Sqrt[a + b*ArcSin[c + d*x]]))/d^2

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Maple [A]  time = 0.105, size = 301, normalized size = 1.1 \begin{align*}{\frac{1}{b{d}^{2}} \left ( 2\,\sqrt{2}\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\cos \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }c-2\,\sqrt{2}\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\sin \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }c+2\,\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\cos \left ( 2\,{\frac{a}{b}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }+2\,\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\sin \left ( 2\,{\frac{a}{b}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }+2\,\cos \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) c-\sin \left ( 2\,{\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-2\,{\frac{a}{b}} \right ) \right ){\frac{1}{\sqrt{a+b\arcsin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2/b*(2*2^(1/2)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*a
rcsin(d*x+c))^(1/2)/b)*Pi^(1/2)*c-2*2^(1/2)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi
^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)*c+2*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*
FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)+2*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*
sin(2*a/b)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)+2*cos((a+b*arcsin(d*x+c))/b-a
/b)*c-sin(2*(a+b*arcsin(d*x+c))/b-2*a/b))/(a+b*arcsin(d*x+c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*asin(d*x+c))**(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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