3.156 \(\int \frac {1}{(a+b \sinh ^{-1}(c x))^{7/2}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {4 \sqrt {\pi } e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 \sqrt {\pi } e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}-\frac {8 \sqrt {c^2 x^2+1}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {2 \sqrt {c^2 x^2+1}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \]

[Out]

-4/15*x/b^2/(a+b*arcsinh(c*x))^(3/2)-4/15*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c+4/
15*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c/exp(a/b)-2/5*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(c
*x))^(5/2)-8/15*(c^2*x^2+1)^(1/2)/b^3/c/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 178, 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 = {5655, 5774, 5779, 3308, 2180, 2204, 2205} \[ -\frac {8 \sqrt {c^2 x^2+1}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {4 \sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 \sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {2 \sqrt {c^2 x^2+1}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^(-7/2),x]

[Out]

(-2*Sqrt[1 + c^2*x^2])/(5*b*c*(a + b*ArcSinh[c*x])^(5/2)) - (4*x)/(15*b^2*(a + b*ArcSinh[c*x])^(3/2)) - (8*Sqr
t[1 + c^2*x^2])/(15*b^3*c*Sqrt[a + b*ArcSinh[c*x]]) - (4*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]
])/(15*b^(7/2)*c) + (4*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(15*b^(7/2)*c*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 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 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 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])

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sinh ^{-1}(c x)\right )^{7/2}} \, dx &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}+\frac {(2 c) \int \frac {x}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \, dx}{5 b}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}+\frac {4 \int \frac {1}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {(8 c) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{15 b^3}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {4 \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c}+\frac {4 \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {8 \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c}+\frac {8 \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c}\\ &=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4 x}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 210, normalized size = 1.18 \[ \frac {-2 e^{-\sinh ^{-1}(c x)} \left (4 a^2+2 a b \left (4 \sinh ^{-1}(c x)-1\right )+b^2 \left (4 \sinh ^{-1}(c x)^2-2 \sinh ^{-1}(c x)+3\right )\right )+8 e^{a/b} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \left (a+b \sinh ^{-1}(c x)\right )^2 \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )-4 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c x)\right ) \left (e^{\frac {a}{b}+\sinh ^{-1}(c x)} \left (2 a+2 b \sinh ^{-1}(c x)+b\right )+2 b \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )\right )-6 b^2 e^{\sinh ^{-1}(c x)}}{30 b^3 c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^(-7/2),x]

[Out]

(-6*b^2*E^ArcSinh[c*x] - (2*(4*a^2 + 2*a*b*(-1 + 4*ArcSinh[c*x]) + b^2*(3 - 2*ArcSinh[c*x] + 4*ArcSinh[c*x]^2)
))/E^ArcSinh[c*x] + 8*E^(a/b)*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])^2*Gamma[1/2, a/b + ArcSinh[c*x]] -
(4*(a + b*ArcSinh[c*x])*(E^(a/b + ArcSinh[c*x])*(2*a + b + 2*b*ArcSinh[c*x]) + 2*b*(-((a + b*ArcSinh[c*x])/b)
)^(3/2)*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)]))/E^(a/b))/(30*b^3*c*(a + b*ArcSinh[c*x])^(5/2))

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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(c*x))^(7/2),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 \[ \int \frac {1}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(c*x))^(7/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^(-7/2), x)

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(c*x))^(7/2),x)

[Out]

int(1/(a+b*arcsinh(c*x))^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(c*x))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^(-7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c*x))^(7/2),x)

[Out]

int(1/(a + b*asinh(c*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asinh(c*x))**(7/2),x)

[Out]

Integral((a + b*asinh(c*x))**(-7/2), x)

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