3.2.49 \(\int \frac {x}{(a+b \sinh ^{-1}(c x))^{3/2}} \, dx\) [149]

Optimal. Leaf size=135 \[ -\frac {2 x \sqrt {1+c^2 x^2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}+\frac {e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2} \]

[Out]

1/2*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c^2+1/2*erfi(2^(1/2)*(a+
b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c^2/exp(2*a/b)-2*x*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(
c*x))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5778, 3388, 2211, 2236, 2235} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}+\frac {\sqrt {\frac {\pi }{2}} e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}-\frac {2 x \sqrt {c^2 x^2+1}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(-2*x*Sqrt[1 + c^2*x^2])/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcS
inh[c*x]])/Sqrt[b]])/(b^(3/2)*c^2) + (Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(b^(3/2)*c^
2*E^((2*a)/b))

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 5778

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

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac {2 x \sqrt {1+c^2 x^2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {2 \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {\text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}+\frac {\text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {2 \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}+\frac {2 \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}+\frac {e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 134, normalized size = 0.99 \begin {gather*} \frac {e^{-\frac {2 a}{b}} \left (\sqrt {2} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-\sqrt {2} e^{\frac {4 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-2 e^{\frac {2 a}{b}} \sinh \left (2 \sinh ^{-1}(c x)\right )\right )}{2 b c^2 \sqrt {a+b \sinh ^{-1}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b] - Sqrt[2]*E^((4*a)/b)*Sqrt[a/
b + ArcSinh[c*x]]*Gamma[1/2, (2*(a + b*ArcSinh[c*x]))/b] - 2*E^((2*a)/b)*Sinh[2*ArcSinh[c*x]])/(2*b*c^2*E^((2*
a)/b)*Sqrt[a + b*ArcSinh[c*x]])

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*arcsinh(c*x))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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*arcsinh(c*x))^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*arcsinh(c*x))^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asinh(c*x))^(3/2),x)

[Out]

int(x/(a + b*asinh(c*x))^(3/2), x)

________________________________________________________________________________________