3.2.39 \(\int x^2 (a+b \sinh ^{-1}(c x))^{3/2} \, dx\) [139]

Optimal. Leaf size=282 \[ \frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {3 b^{3/2} e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {b^{3/2} e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3}-\frac {3 b^{3/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {b^{3/2} e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3} \]

[Out]

1/3*x^3*(a+b*arcsinh(c*x))^(3/2)+1/288*b^(3/2)*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2
)*Pi^(1/2)/c^3+1/288*b^(3/2)*erfi(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/c^3/exp(3*a/b)-3/
32*b^(3/2)*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/c^3-3/32*b^(3/2)*erfi((a+b*arcsinh(c*x))^(1
/2)/b^(1/2))*Pi^(1/2)/c^3/exp(a/b)+1/3*b*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^(1/2)/c^3-1/6*b*x^2*(c^2*x^2+1)^
(1/2)*(a+b*arcsinh(c*x))^(1/2)/c

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Rubi [A]
time = 0.55, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5777, 5812, 5798, 5774, 3388, 2211, 2236, 2235, 5780, 5556} \begin {gather*} -\frac {3 \sqrt {\pi } b^{3/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3}-\frac {3 \sqrt {\pi } b^{3/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3}-\frac {b x^2 \sqrt {c^2 x^2+1} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {b \sqrt {c^2 x^2+1} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(b*Sqrt[1 + c^2*x^2]*Sqrt[a + b*ArcSinh[c*x]])/(3*c^3) - (b*x^2*Sqrt[1 + c^2*x^2]*Sqrt[a + b*ArcSinh[c*x]])/(6
*c) + (x^3*(a + b*ArcSinh[c*x])^(3/2))/3 - (3*b^(3/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/
(32*c^3) + (b^(3/2)*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(96*c^3) - (3*b^(3
/2)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(32*c^3*E^(a/b)) + (b^(3/2)*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt
[a + b*ArcSinh[c*x]])/Sqrt[b]])/(96*c^3*E^((3*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 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 5774

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

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 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -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^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \, dx &=\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {1}{2} (b c) \int \frac {x^3 \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {1}{12} b^2 \int \frac {x^2}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx+\frac {b \int \frac {x \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {1+c^2 x^2}} \, dx}{3 c}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac {b^2 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{12 c^3}-\frac {b^2 \int \frac {1}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{6 c^2}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {b \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{6 c^3}+\frac {b^2 \text {Subst}\left (\int \left (-\frac {\cosh (x)}{4 \sqrt {a+b x}}+\frac {\cosh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{12 c^3}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {b \text {Subst}\left (\int \frac {e^{-i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{12 c^3}-\frac {b \text {Subst}\left (\int \frac {e^{i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{12 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {\cosh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{48 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\cosh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{48 c^3}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {b \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{6 c^3}-\frac {b \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{6 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{96 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{96 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{96 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{96 c^3}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {b^{3/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}-\frac {b^{3/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}+\frac {b \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{48 c^3}-\frac {b \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{48 c^3}-\frac {b \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{48 c^3}+\frac {b \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{48 c^3}\\ &=\frac {b \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{3 c^3}-\frac {b x^2 \sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}}{6 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac {3 b^{3/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {b^{3/2} e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3}-\frac {3 b^{3/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{32 c^3}+\frac {b^{3/2} e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{96 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 215, normalized size = 0.76 \begin {gather*} -\frac {b e^{-\frac {3 a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (-27 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {5}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\sqrt {3} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {5}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-27 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {5}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+\sqrt {3} e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {5}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{216 c^3 \sqrt {-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/216*(b*Sqrt[a + b*ArcSinh[c*x]]*(-27*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[5/2, a/b + ArcSinh[c
*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c*x]]*Gamma[5/2, (-3*(a + b*ArcSinh[c*x]))/b] - 27*E^((2*a)/b)*Sqrt[a/b + Ar
cSinh[c*x]]*Gamma[5/2, -((a + b*ArcSinh[c*x])/b)] + Sqrt[3]*E^((6*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[
5/2, (3*(a + b*ArcSinh[c*x]))/b]))/(c^3*E^((3*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 \[\int x^{2} \left (a +b \arcsinh \left (c x \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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^2*(a+b*arcsinh(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^(3/2)*x^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^2*(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^{2} \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**2*(a+b*asinh(c*x))**(3/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))^(3/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

[Out]

int(x^2*(a + b*asinh(c*x))^(3/2), x)

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