3.220 \(\int \frac {1}{(a+b \sinh ^{-1}(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ \frac {2 \sqrt {\pi } e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 \sqrt {\pi } e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5863, 5655, 5774, 5657, 3307, 2180, 2205, 2204} \[ \frac {2 \sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 \sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (4*(c + d*x))/(3*b^2*d*Sqrt[a + b*ArcSinh[
c + d*x]]) + (2*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(3*b^(5/2)*d) + (2*Sqrt[Pi]*Erfi[S
qrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(3*b^(5/2)*d*E^(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 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 5657

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 5774

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

Rule 5863

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {2 \operatorname {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+d x)\right )}{3 b^3 d}+\frac {2 \operatorname {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+d x)\right )}{3 b^3 d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}+\frac {4 \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}\\ &=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 207, normalized size = 1.31 \[ \frac {e^{-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \left (-e^{a/b} \left (2 a \left (e^{2 \sinh ^{-1}(c+d x)}-1\right )-2 b \sinh ^{-1}(c+d x)+b e^{2 \sinh ^{-1}(c+d x)} \left (2 \sinh ^{-1}(c+d x)+1\right )+b\right )-2 b e^{\sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )-2 e^{\frac {2 a}{b}+\sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(E^(a/b)*(b + 2*a*(-1 + E^(2*ArcSinh[c + d*x])) - 2*b*ArcSinh[c + d*x] + b*E^(2*ArcSinh[c + d*x])*(1 + 2*Arc
Sinh[c + d*x]))) - 2*E^((2*a)/b + ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamm
a[1/2, a/b + ArcSinh[c + d*x]] - 2*b*E^ArcSinh[c + d*x]*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, -((a
+ b*ArcSinh[c + d*x])/b)])/(3*b^2*d*E^((a + b*ArcSinh[c + d*x])/b)*(a + b*ArcSinh[c + d*x])^(3/2))

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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(1/(a+b*arcsinh(d*x+c))^(5/2),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 \[ \int \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(-5/2), x)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(d*x+c))^(5/2),x)

[Out]

int(1/(a+b*arcsinh(d*x+c))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x))^(5/2),x)

[Out]

int(1/(a + b*asinh(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asinh(d*x+c))**(5/2),x)

[Out]

Integral((a + b*asinh(c + d*x))**(-5/2), x)

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