3.2.40 \(\int x (a+b \sinh ^{-1}(c x))^{3/2} \, dx\) [140]

Optimal. Leaf size=179 \[ -\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {3 b^{3/2} e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^2}+\frac {3 b^{3/2} e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^2} \]

[Out]

1/4*(a+b*arcsinh(c*x))^(3/2)/c^2+1/2*x^2*(a+b*arcsinh(c*x))^(3/2)-3/128*b^(3/2)*exp(2*a/b)*erf(2^(1/2)*(a+b*ar
csinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/c^2+3/128*b^(3/2)*erfi(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*2
^(1/2)*Pi^(1/2)/c^2/exp(2*a/b)-3/8*b*x*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^(1/2)/c

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Rubi [A]
time = 0.32, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5777, 5812, 5783, 5780, 5556, 12, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {3 \sqrt {\frac {\pi }{2}} b^{3/2} e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^2}+\frac {3 \sqrt {\frac {\pi }{2}} b^{3/2} e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^2}-\frac {3 b x \sqrt {c^2 x^2+1} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(-3*b*x*Sqrt[1 + c^2*x^2]*Sqrt[a + b*ArcSinh[c*x]])/(8*c) + (a + b*ArcSinh[c*x])^(3/2)/(4*c^2) + (x^2*(a + b*A
rcSinh[c*x])^(3/2))/2 - (3*b^(3/2)*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64
*c^2) + (3*b^(3/2)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64*c^2*E^((2*a)/b))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3389

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

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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 5780

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5812

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

Rubi steps

\begin {align*} \int x \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \, dx &=\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {1}{4} (3 b c) \int \frac {x^2 \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {1}{16} \left (3 b^2\right ) \int \frac {x}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx+\frac {(3 b) \int \frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {1+c^2 x^2}} \, dx}{8 c}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{2 \sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {(3 b) \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{32 c^2}+\frac {(3 b) \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{32 c^2}\\ &=-\frac {3 b x \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{8 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {3 b^{3/2} 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 )}{64 c^2}+\frac {3 b^{3/2} 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 )}{64 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 129, normalized size = 0.72 \begin {gather*} \frac {b e^{-\frac {2 a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (-\sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {5}{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 {5}{2},\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{16 \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*(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(b*Sqrt[a + b*ArcSinh[c*x]]*(-(Sqrt[a/b + ArcSinh[c*x]]*Gamma[5/2, (-2*(a + b*ArcSinh[c*x]))/b]) + E^((4*a)/b)
*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[5/2, (2*(a + b*ArcSinh[c*x]))/b]))/(16*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 \[\int x \left (a +b \arcsinh \left (c x \right )\right )^{\frac {3}{2}}\, dx\]

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)

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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))^(3/2),x, algorithm="maxima")

[Out]

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

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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))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int 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)

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