3.2.37 \(\int x \sqrt {a+b \sinh ^{-1}(c x)} \, dx\) [137]

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

[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 = 0.28, 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 = {5777, 5819, 3393, 3388, 2211, 2236, 2235} \begin {gather*} -\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)} \end {gather*}

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 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 3393

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 5777

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 5819

Int[((a_.) + ArcSinh[(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*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
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 \text {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 \text {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 \text {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 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}-\frac {b \text {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 {\text {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 {\text {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 = 0.07, size = 127, normalized size = 0.88 \begin {gather*} \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}}} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 180.00, 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 \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))^(1/2),x, algorithm="maxima")

[Out]

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

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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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))^(1/2),x, algorithm="giac")

[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 \begin {gather*} \int x\,\sqrt {a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \end {gather*}

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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