3.108 \(\int \frac {x}{(a+b \sinh ^{-1}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=269 \[ \frac {\sqrt {\pi } c e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}-\frac {\sqrt {\pi } c e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {\sqrt {\frac {\pi }{2}} e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {2 c \sqrt {(c+d x)^2+1}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {(c+d x)^2+1}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.54, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5865, 5803, 5655, 5779, 3308, 2180, 2204, 2205, 5665, 3307} \[ \frac {\sqrt {\pi } c e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}-\frac {\sqrt {\pi } c e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {\sqrt {\frac {\pi }{2}} e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {2 c \sqrt {(c+d x)^2+1}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {(c+d x)^2+1}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 + (c + d*x)^2])/(b*d^2*Sqrt[a + b*ArcSinh[c + d*x]]) - (2*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(b*d^2*
Sqrt[a + b*ArcSinh[c + d*x]]) + (c*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(b^(3/2)*d^2) +
(E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(b^(3/2)*d^2) - (c*Sqrt[Pi]*Erfi
[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(b^(3/2)*d^2*E^(a/b)) + (Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c
+ d*x]])/Sqrt[b]])/(b^(3/2)*d^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 3308

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 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5665

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

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])

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-\frac {c}{d}+\frac {x}{d}}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {c}{d \left (a+b \sinh ^{-1}(x)\right )^{3/2}}+\frac {x}{d \left (a+b \sinh ^{-1}(x)\right )^{3/2}}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d^2}-\frac {c \operatorname {Subst}\left (\int \frac {1}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 c \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}-\frac {(2 c) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{b d^2}\\ &=\frac {2 c \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}+\frac {\operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}-\frac {(2 c) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}\\ &=\frac {2 c \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {2 \operatorname {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^2 d^2}+\frac {2 \operatorname {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^2 d^2}+\frac {c \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}-\frac {c \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d^2}\\ &=\frac {2 c \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {(2 c) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^2 d^2}-\frac {(2 c) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^2 d^2}\\ &=\frac {2 c \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {c e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}-\frac {c e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}+\frac {e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d^2}\\ \end {align*}

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Mathematica [A]  time = 2.69, size = 301, normalized size = 1.12 \[ \frac {\sqrt {2 \pi } \left (\sinh \left (\frac {2 a}{b}\right )+\cosh \left (\frac {2 a}{b}\right )\right ) \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+\sqrt {2 \pi } \left (\cosh \left (\frac {2 a}{b}\right )-\sinh \left (\frac {2 a}{b}\right )\right ) \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )-\frac {2 \sqrt {b} \sinh \left (2 \sinh ^{-1}(c+d x)\right )}{\sqrt {a+b \sinh ^{-1}(c+d x)}}}{2 b^{3/2} d^2}-\frac {c e^{-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \left (-e^{a/b} \left (e^{2 \sinh ^{-1}(c+d x)}+1\right )+e^{\frac {2 a}{b}+\sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+e^{\sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )\right )}{b d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((c*(-(E^(a/b)*(1 + E^(2*ArcSinh[c + d*x]))) + E^((2*a)/b + ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*Ga
mma[1/2, a/b + ArcSinh[c + d*x]] + E^ArcSinh[c + d*x]*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, -((a + b*
ArcSinh[c + d*x])/b)]))/(b*d^2*E^((a + b*ArcSinh[c + d*x])/b)*Sqrt[a + b*ArcSinh[c + d*x]])) + (Sqrt[2*Pi]*Erf
i[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]]*(Cosh[(2*a)/b] - Sinh[(2*a)/b]) + Sqrt[2*Pi]*Erf[(Sqrt[2]*Sq
rt[a + b*ArcSinh[c + d*x]])/Sqrt[b]]*(Cosh[(2*a)/b] + Sinh[(2*a)/b]) - (2*Sqrt[b]*Sinh[2*ArcSinh[c + d*x]])/Sq
rt[a + b*ArcSinh[c + d*x]])/(2*b^(3/2)*d^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(d*x+c))^(3/2),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 \[ \int \frac {x}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asinh(c + d*x))^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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