\(\int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx\) [509]

Optimal result

Integrand size = 23, antiderivative size = 42 \[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=-\frac {2 \sqrt {1+a^2 x^2}}{3 a \sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{3/2}} \]

[Out]

-2/3*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(3/2)/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {5783} \[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=-\frac {2 \sqrt {a^2 x^2+1}}{3 a \text {arcsinh}(a x)^{3/2} \sqrt {a^2 c x^2+c}} \]

[In]

Int[1/(Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^(5/2)),x]

[Out]

(-2*Sqrt[1 + a^2*x^2])/(3*a*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^(3/2))

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {1+a^2 x^2}}{3 a \sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=-\frac {2 \sqrt {1+a^2 x^2}}{3 a \sqrt {c \left (1+a^2 x^2\right )} \text {arcsinh}(a x)^{3/2}} \]

[In]

Integrate[1/(Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^(5/2)),x]

[Out]

(-2*Sqrt[1 + a^2*x^2])/(3*a*Sqrt[c*(1 + a^2*x^2)]*ArcSinh[a*x]^(3/2))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 \operatorname {arcsinh}\left (a x \right )^{\frac {3}{2}} a \sqrt {c \left (a^{2} x^{2}+1\right )}}\)	\(36\)

[In]

int(1/arcsinh(a*x)^(5/2)/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/arcsinh(a*x)^(3/2)/a/(c*(a^2*x^2+1))^(1/2)*(a^2*x^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=-\frac {2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1}}{3 \, {\left (a^{3} c x^{2} + a c\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-2/3*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)/((a^3*c*x^2 + a*c)*log(a*x + sqrt(a^2*x^2 + 1))^(3/2))

Sympy [F]

\[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=\int \frac {1}{\sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {asinh}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]

[In]

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

[Out]

Integral(1/(sqrt(c*(a**2*x**2 + 1))*asinh(a*x)**(5/2)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/(sqrt(a^2*c*x^2 + c)*arcsinh(a*x)^(5/2)), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/(sqrt(a^2*c*x^2 + c)*arcsinh(a*x)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {c+a^2 c x^2} \text {arcsinh}(a x)^{5/2}} \, dx=\int \frac {1}{{\mathrm {asinh}\left (a\,x\right )}^{5/2}\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

[In]

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

[Out]

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