3.137 \(\int x \sqrt {a+b \sinh ^{-1}(c x)} \, dx\)

Optimal. Leaf size=145 \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^2}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^2}+\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)} \]

[Out]

-1/32*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*2^(1/2)*Pi^(1/2)/c^2-1/32*erfi(2^(1/2)*
(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*2^(1/2)*Pi^(1/2)/c^2/exp(2*a/b)+1/4*(a+b*arcsinh(c*x))^(1/2)/c^2+1/2
*x^2*(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5663, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^2}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^2}+\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

Sqrt[a + b*ArcSinh[c*x]]/(4*c^2) + (x^2*Sqrt[a + b*ArcSinh[c*x]])/2 - (Sqrt[b]*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*c^2) - (Sqrt[b]*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]]
)/Sqrt[b]])/(16*c^2*E^((2*a)/b))

Rule 2180

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] &&  !$UseGamma === True

Rule 2204

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 2205

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 3307

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5663

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

Rule 5779

Int[((a_.) + ArcSinh[(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*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int x \sqrt {a+b \sinh ^{-1}(c x)} \, dx &=\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {1}{4} (b c) \int \frac {x^2}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx\\ &=\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^2}\\ &=\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {b \operatorname {Subst}\left (\int \left (\frac {1}{2 \sqrt {a+b x}}-\frac {\cosh (2 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^2}\\ &=\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2}\\ &=\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}-\frac {b \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {\operatorname {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c^2}-\frac {\operatorname {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c^2}\\ &=\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {\sqrt {b} 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 )}{16 c^2}-\frac {\sqrt {b} 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 )}{16 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 127, normalized size = 0.88 \[ \frac {e^{-\frac {2 a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (\sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {3}{2},-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {3}{2},\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{8 \sqrt {2} c^2 \sqrt {-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

(Sqrt[a + b*ArcSinh[c*x]]*(Sqrt[a/b + ArcSinh[c*x]]*Gamma[3/2, (-2*(a + b*ArcSinh[c*x]))/b] + E^((4*a)/b)*Sqrt
[-((a + b*ArcSinh[c*x])/b)]*Gamma[3/2, (2*(a + b*ArcSinh[c*x]))/b]))/(8*Sqrt[2]*c^2*E^((2*a)/b)*Sqrt[-((a + b*
ArcSinh[c*x])^2/b^2)])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {arsinh}\left (c x\right ) + a} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*arcsinh(c*x) + a)*x, x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int x \sqrt {a +b \arcsinh \left (c x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^(1/2),x)

[Out]

int(x*(a+b*arcsinh(c*x))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {arsinh}\left (c x\right ) + a} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*arcsinh(c*x) + a)*x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\sqrt {a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asinh(c*x))^(1/2),x)

[Out]

int(x*(a + b*asinh(c*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {a + b \operatorname {asinh}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**(1/2),x)

[Out]

Integral(x*sqrt(a + b*asinh(c*x)), x)

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