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

Optimal. Leaf size=158 \[ \text{Unintegrable}\left (\frac{1}{x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )},x\right )+\frac{c \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b}+\frac{c \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b}-\frac{c \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b}-\frac{c \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b}+\frac{15 c \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b} \]

[Out]

(c*Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/b + (c*Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*ArcSinh
[c*x]))/b])/(8*b) + (15*c*Log[a + b*ArcSinh[c*x]])/(8*b) - (c*Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x
]))/b])/b - (c*Sinh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(8*b) + Unintegrable[1/(x^2*Sqrt[1 + c^
2*x^2]*(a + b*ArcSinh[c*x])), x]

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Rubi [A]  time = 0.966754, 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 = {} \[ \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(c*Cosh[(2*a)/b]*CoshIntegral[(2*a)/b + 2*ArcSinh[c*x]])/b + (c*Cosh[(4*a)/b]*CoshIntegral[(4*a)/b + 4*ArcSinh
[c*x]])/(8*b) + (15*c*Log[a + b*ArcSinh[c*x]])/(8*b) - (c*Sinh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]]
)/b - (c*Sinh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c*x]])/(8*b) + Defer[Int][1/(x^2*Sqrt[1 + c^2*x^2]*(a
+ b*ArcSinh[c*x])), x]

Rubi steps

\begin{align*} \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx &=\int \left (\frac{3 c^2}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}+\frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}+\frac{3 c^4 x^2}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}+\frac{c^6 x^4}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}\right ) \, dx\\ &=\left (3 c^2\right ) \int \frac{1}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx+\left (3 c^4\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx+c^6 \int \frac{x^4}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ &=\frac{3 c \log \left (a+b \sinh ^{-1}(c x)\right )}{b}+c \operatorname{Subst}\left (\int \frac{\sinh ^4(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+(3 c) \operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ &=\frac{3 c \log \left (a+b \sinh ^{-1}(c x)\right )}{b}+c \operatorname{Subst}\left (\int \left (\frac{3}{8 (a+b x)}-\frac{\cosh (2 x)}{2 (a+b x)}+\frac{\cosh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )-(3 c) \operatorname{Subst}\left (\int \left (\frac{1}{2 (a+b x)}-\frac{\cosh (2 x)}{2 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ &=\frac{15 c \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b}+\frac{1}{8} c \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )-\frac{1}{2} c \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\frac{1}{2} (3 c) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ &=\frac{15 c \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b}-\frac{1}{2} \left (c \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\frac{1}{2} \left (3 c \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\frac{1}{8} \left (c \cosh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\frac{1}{2} \left (c \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )-\frac{1}{2} \left (3 c \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )-\frac{1}{8} \left (c \sinh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ &=\frac{c \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b}+\frac{c \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{8 b}+\frac{15 c \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b}-\frac{c \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b}-\frac{c \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{8 b}+\int \frac{1}{x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx\\ \end{align*}

Mathematica [A]  time = 1.30436, size = 0, normalized size = 0. \[ \int \frac{\left (1+c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sinh ^{-1}(c x)\right )} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.171, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \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^2/(a+b*arcsinh(c*x)),x)

[Out]

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

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Maxima [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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c^2*x^2 + 1)^(5/2)/((b*arcsinh(c*x) + a)*x^2), 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 x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^4*x^4 + 2*c^2*x^2 + 1)*sqrt(c^2*x^2 + 1)/(b*x^2*arcsinh(c*x) + a*x^2), 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**2/(a+b*asinh(c*x)),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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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