3.2.55 \(\int \frac {x}{(a+b \sinh ^{-1}(c x))^{7/2}} \, dx\) [155]

Optimal. Leaf size=219 \[ -\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {32 x \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {8 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2}+\frac {8 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2} \]

[Out]

-4/15/b^2/c^2/(a+b*arcsinh(c*x))^(3/2)-8/15*x^2/b^2/(a+b*arcsinh(c*x))^(3/2)+8/15*exp(2*a/b)*erf(2^(1/2)*(a+b*
arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(7/2)/c^2+8/15*erfi(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*
2^(1/2)*Pi^(1/2)/b^(7/2)/c^2/exp(2*a/b)-2/5*x*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(c*x))^(5/2)-32/15*x*(c^2*x^2+
1)^(1/2)/b^3/c/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 219, 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 = {5779, 5818, 5778, 3388, 2211, 2236, 2235, 5783} \begin {gather*} \frac {8 \sqrt {2 \pi } e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2}+\frac {8 \sqrt {2 \pi } e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2}-\frac {32 x \sqrt {c^2 x^2+1}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {2 x \sqrt {c^2 x^2+1}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcSinh[c*x])^(7/2),x]

[Out]

(-2*x*Sqrt[1 + c^2*x^2])/(5*b*c*(a + b*ArcSinh[c*x])^(5/2)) - 4/(15*b^2*c^2*(a + b*ArcSinh[c*x])^(3/2)) - (8*x
^2)/(15*b^2*(a + b*ArcSinh[c*x])^(3/2)) - (32*x*Sqrt[1 + c^2*x^2])/(15*b^3*c*Sqrt[a + b*ArcSinh[c*x]]) + (8*E^
((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(15*b^(7/2)*c^2) + (8*Sqrt[2*Pi]*Erfi[(S
qrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(15*b^(7/2)*c^2*E^((2*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 5778

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^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*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_.), 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[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 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] && LtQ[n, -2]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5818

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

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \sinh ^{-1}(c x)\right )^{7/2}} \, dx &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}+\frac {2 \int \frac {1}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \, dx}{5 b c}+\frac {(4 c) \int \frac {x^2}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \, dx}{5 b}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}+\frac {16 \int \frac {x}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {32 x \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {32 \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {32 x \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {16 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^2}+\frac {16 \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {32 x \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {32 \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c^2}+\frac {32 \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c^2}\\ &=-\frac {2 x \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {4}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {8 x^2}{15 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {32 x \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {8 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2}+\frac {8 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^2}\\ \end {align*}

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Mathematica [A]
time = 0.79, size = 208, normalized size = 0.95 \begin {gather*} -\frac {\left (a+b \sinh ^{-1}(c x)\right ) \left (e^{-\frac {2 a}{b}} \left (2 e^{2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \left (4 a+b+4 b \sinh ^{-1}(c x)\right )+8 \sqrt {2} b \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )+e^{-2 \sinh ^{-1}(c x)} \left (-8 a+2 b-8 b \sinh ^{-1}(c x)+8 \sqrt {2} e^{2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \left (a+b \sinh ^{-1}(c x)\right ) \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )\right )+3 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right )}{15 b^3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((a + b*ArcSinh[c*x])*((2*E^(2*(a/b + ArcSinh[c*x]))*(4*a + b + 4*b*ArcSinh[c*x]) + 8*Sqrt[2]*b*(-((a +
b*ArcSinh[c*x])/b))^(3/2)*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b])/E^((2*a)/b) + (-8*a + 2*b - 8*b*ArcSinh[c*x
] + 8*Sqrt[2]*E^(2*(a/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])*Gamma[1/2, (2*(a + b*Ar
cSinh[c*x]))/b])/E^(2*ArcSinh[c*x])) + 3*b^2*Sinh[2*ArcSinh[c*x]])/(b^3*c^2*(a + b*ArcSinh[c*x])^(5/2))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x/(b*arcsinh(c*x) + a)^(7/2), 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/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asinh(c*x))^(7/2),x)

[Out]

int(x/(a + b*asinh(c*x))^(7/2), x)

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