3.2.44 \(\int (a+b \sinh ^{-1}(c x))^{5/2} \, dx\) [144]

Optimal. Leaf size=155 \[ \frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c} \]

[Out]

x*(a+b*arcsinh(c*x))^(5/2)+15/16*b^(5/2)*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/c-15/16*b^(5/
2)*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/c/exp(a/b)-5/2*b*(a+b*arcsinh(c*x))^(3/2)*(c^2*x^2+1)^(1/2)
/c+15/4*b^2*x*(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5772, 5798, 5819, 3389, 2211, 2236, 2235} \begin {gather*} \frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}+\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^(5/2),x]

[Out]

(15*b^2*x*Sqrt[a + b*ArcSinh[c*x]])/4 - (5*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(3/2))/(2*c) + x*(a + b*Ar
cSinh[c*x])^(5/2) + (15*b^(5/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(16*c) - (15*b^(5/2)*S
qrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(16*c*E^(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 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 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 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 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 \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \, dx &=x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {1}{2} (5 b c) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {1}{4} \left (15 b^2\right ) \int \sqrt {a+b \sinh ^{-1}(c x)} \, dx\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {1}{8} \left (15 b^3 c\right ) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (15 b^3\right ) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}-\frac {\left (15 b^3\right ) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^2\right ) \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}\\ \end {align*}

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Mathematica [A]
time = 2.36, size = 282, normalized size = 1.82 \begin {gather*} \frac {\sqrt {b} e^{-\frac {a}{b}} \left (-\left (\left (4 a^2-15 b^2\right ) e^{\frac {2 a}{b}} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )\right )+\left (4 a^2-15 b^2\right ) \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )+\frac {4 \sqrt {b} \left (e^{a/b} \left (a+b \sinh ^{-1}(c x)\right ) \left (5 \left (3 b c x-2 a \sqrt {1+c^2 x^2}\right )+2 \left (4 a c x-5 b \sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)+4 b c x \sinh ^{-1}(c x)^2\right )-2 a^2 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )-2 a^2 \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )\right )}{\sqrt {a+b \sinh ^{-1}(c x)}}\right )}{16 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])^(5/2),x]

[Out]

(Sqrt[b]*(-((4*a^2 - 15*b^2)*E^((2*a)/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]]) + (4*a^2 - 15*b^2)*Sq
rt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]] + (4*Sqrt[b]*(E^(a/b)*(a + b*ArcSinh[c*x])*(5*(3*b*c*x - 2*a*Sqr
t[1 + c^2*x^2]) + 2*(4*a*c*x - 5*b*Sqrt[1 + c^2*x^2])*ArcSinh[c*x] + 4*b*c*x*ArcSinh[c*x]^2) - 2*a^2*E^((2*a)/
b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[3/2, a/b + ArcSinh[c*x]] - 2*a^2*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[3/2,
-((a + b*ArcSinh[c*x])/b)]))/Sqrt[a + b*ArcSinh[c*x]]))/(16*c*E^(a/b))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsinh \left (c x \right )\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^(5/2),x)

[Out]

int((a+b*arcsinh(c*x))^(5/2),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((a+b*arcsinh(c*x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^(5/2), 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((a+b*arcsinh(c*x))^(5/2),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 \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**(5/2),x)

[Out]

Integral((a + b*asinh(c*x))**(5/2), 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((a+b*arcsinh(c*x))^(5/2),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 = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^(5/2),x)

[Out]

int((a + b*asinh(c*x))^(5/2), x)

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