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

Optimal. Leaf size=232 \[ -\frac{\text{Unintegrable}\left (\frac{\left (c^2 x^2+1\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )},x\right )}{b c}+\frac{25 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2}+\frac{25 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{25 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2}-\frac{25 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{\left (c^2 x^2+1\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-((1 + c^2*x^2)^3/(b*c*x*(a + b*ArcSinh[c*x]))) + (25*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2)
+ (25*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2) + (5*Cosh[(5*a)/b]*CoshIntegral[(5*(a +
b*ArcSinh[c*x]))/b])/(16*b^2) - (25*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2) - (25*Sinh[(3*a)/
b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c*x]))
/b])/(16*b^2) - Unintegrable[(1 + c^2*x^2)^2/(x^2*(a + b*ArcSinh[c*x])), x]/(b*c)

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Rubi [A]  time = 0.568258, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

-((1 + c^2*x^2)^3/(b*c*x*(a + b*ArcSinh[c*x]))) + (25*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(8*b^2) + (2
5*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2) + (5*Cosh[(5*a)/b]*CoshIntegral[(5*a)/b + 5*A
rcSinh[c*x]])/(16*b^2) - (25*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(8*b^2) - (25*Sinh[(3*a)/b]*SinhInteg
ral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c*x]])/(16*b^2) -
Defer[Int][(1 + c^2*x^2)^2/(x^2*(a + b*ArcSinh[c*x])), x]/(b*c)

Rubi steps

\begin{align*} \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}+\frac{(5 c) \int \frac{\left (1+c^2 x^2\right )^2}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh ^5(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \left (\frac{5 \cosh (x)}{8 (a+b x)}+\frac{5 \cosh (3 x)}{16 (a+b x)}+\frac{\cosh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}+\frac{\left (25 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b}+\frac{\left (25 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}+\frac{\left (5 \cosh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}-\frac{\left (25 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b}-\frac{\left (25 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}-\frac{\left (5 \sinh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}\\ &=-\frac{\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}+\frac{25 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2}+\frac{25 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2}-\frac{25 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2}-\frac{25 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2}-\frac{\int \frac{\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b c}\\ \end{align*}

Mathematica [A]  time = 9.04237, size = 0, normalized size = 0. \[ \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.267, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}} \left ({c}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (c^{6} x^{6} + 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} + 1\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (c^{7} x^{7} + 3 \, c^{5} x^{5} + 3 \, c^{3} x^{3} + c x\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{3} x^{3} + \sqrt{c^{2} x^{2} + 1} a b c^{2} x^{2} + a b c x +{\left (b^{2} c^{3} x^{3} + \sqrt{c^{2} x^{2} + 1} b^{2} c^{2} x^{2} + b^{2} c x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )} + \int \frac{{\left (5 \, c^{7} x^{7} + 8 \, c^{5} x^{5} + c^{3} x^{3} - 2 \, c x\right )}{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} +{\left (10 \, c^{8} x^{8} + 23 \, c^{6} x^{6} + 15 \, c^{4} x^{4} + c^{2} x^{2} - 1\right )}{\left (c^{2} x^{2} + 1\right )} + 5 \,{\left (c^{9} x^{9} + 3 \, c^{7} x^{7} + 3 \, c^{5} x^{5} + c^{3} x^{3}\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{5} x^{6} +{\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{4} + 2 \, a b c^{3} x^{4} + a b c x^{2} +{\left (b^{2} c^{5} x^{6} +{\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{4} + 2 \, b^{2} c^{3} x^{4} + b^{2} c x^{2} + 2 \,{\left (b^{2} c^{4} x^{5} + b^{2} c^{2} x^{3}\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (a b c^{4} x^{5} + a b c^{2} x^{3}\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c^6*x^6 + 3*c^4*x^4 + 3*c^2*x^2 + 1)*(c^2*x^2 + 1) + (c^7*x^7 + 3*c^5*x^5 + 3*c^3*x^3 + c*x)*sqrt(c^2*x^2 +
1))/(a*b*c^3*x^3 + sqrt(c^2*x^2 + 1)*a*b*c^2*x^2 + a*b*c*x + (b^2*c^3*x^3 + sqrt(c^2*x^2 + 1)*b^2*c^2*x^2 + b
^2*c*x)*log(c*x + sqrt(c^2*x^2 + 1))) + integrate(((5*c^7*x^7 + 8*c^5*x^5 + c^3*x^3 - 2*c*x)*(c^2*x^2 + 1)^(3/
2) + (10*c^8*x^8 + 23*c^6*x^6 + 15*c^4*x^4 + c^2*x^2 - 1)*(c^2*x^2 + 1) + 5*(c^9*x^9 + 3*c^7*x^7 + 3*c^5*x^5 +
c^3*x^3)*sqrt(c^2*x^2 + 1))/(a*b*c^5*x^6 + (c^2*x^2 + 1)*a*b*c^3*x^4 + 2*a*b*c^3*x^4 + a*b*c*x^2 + (b^2*c^5*x
^6 + (c^2*x^2 + 1)*b^2*c^3*x^4 + 2*b^2*c^3*x^4 + b^2*c*x^2 + 2*(b^2*c^4*x^5 + b^2*c^2*x^3)*sqrt(c^2*x^2 + 1))*
log(c*x + sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4*x^5 + a*b*c^2*x^3)*sqrt(c^2*x^2 + 1)), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{4} x^{4} + 2 \, c^{2} x^{2} + 1\right )} \sqrt{c^{2} x^{2} + 1}}{b^{2} x \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x \operatorname{arsinh}\left (c x\right ) + a^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^4*x^4 + 2*c^2*x^2 + 1)*sqrt(c^2*x^2 + 1)/(b^2*x*arcsinh(c*x)^2 + 2*a*b*x*arcsinh(c*x) + a^2*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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