3.2.66 \(\int \frac {x}{(a+b \text {ArcSin}(c+d x))^{3/2}} \, dx\) [166]

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

[Out]

2*cos(2*a/b)*FresnelC(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(3/2)/d^2+2*FresnelS(2*(a+b*arc
sin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/b^(3/2)/d^2+2*c*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+
b*arcsin(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/d^2-2*c*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))
^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/b^(3/2)/d^2+2*c*(1-(d*x+c)^2)^(1/2)/b/d^2/(a+b*arcsin(d*x+c))^(1/2)-
2*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b/d^2/(a+b*arcsin(d*x+c))^(1/2)

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Rubi [A]
time = 0.36, antiderivative size = 287, normalized size of antiderivative = 1.00, 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 = {4889, 4829, 4717, 4809, 3387, 3386, 3432, 3385, 3433, 4727} \begin {gather*} -\frac {2 \sqrt {2 \pi } c \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(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 \text {ArcSin}(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 \text {ArcSin}(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 \text {ArcSin}(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 \text {ArcSin}(c+d x)}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{b d^2 \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}

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 3385

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 3386

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 3387

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 3432

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

Rule 3433

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

Rule 4717

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 4727

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

Rule 4809

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

Rule 4829

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 4889

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac {\text {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 {\text {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 {\text {Subst}\left (\int \frac {x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d^2}-\frac {c \text {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 \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.47, size = 287, normalized size = 1.00 \begin {gather*} \frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {\pi }}\right )+\frac {c e^{-i \text {ArcSin}(c+d x)}+c e^{i \text {ArcSin}(c+d x)}-c e^{-\frac {i a}{b}} \sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )-c e^{\frac {i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )+2 \sqrt {\frac {1}{b}} \sqrt {\pi } \sqrt {a+b \text {ArcSin}(c+d x)} S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )-\sin (2 \text {ArcSin}(c+d x))}{b \sqrt {a+b \text {ArcSin}(c+d x)}}}{d^2} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.46, size = 326, normalized size = 1.14

method	result	size
default	\(\frac {-2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, c -2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, c +\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}-\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}+2 \cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) c +\sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right )}{d^{2} b \sqrt {a +b \arcsin \left (d x +c \right )}}\)	\(326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2/b*(-2*Pi^(1/2)*2^(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*ar
csin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)*c-2*Pi^(1/2)*2^(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)*(-1/b)^(1/2)*c+Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos
(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-2/b)^(1/2)-Pi^(1/2)*2^(1/2)*(a
+b*arcsin(d*x+c))^(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-2/b
)^(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asin(c + d*x))^(3/2),x)

[Out]

int(x/(a + b*asin(c + d*x))^(3/2), x)

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