3.5.31 \(\int \frac {(1+c^2 x^2)^{5/2}}{x (a+b \sinh ^{-1}(c x))^2} \, dx\) [431]

Optimal. Leaf size=233 \[ -\frac {\left (1+c^2 x^2\right )^3}{b c x \left (a+b \sinh ^{-1}(c x)\right )}+\frac {25 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2}+\frac {25 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {25 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2}-\frac {25 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {\text {Int}\left (\frac {\left (1+c^2 x^2\right )^2}{x^2 \left (a+b \sinh ^{-1}(c x)\right )},x\right )}{b c} \]

[Out]

-(c^2*x^2+1)^3/b/c/x/(a+b*arcsinh(c*x))+25/8*Chi((a+b*arcsinh(c*x))/b)*cosh(a/b)/b^2+25/16*Chi(3*(a+b*arcsinh(
c*x))/b)*cosh(3*a/b)/b^2+5/16*Chi(5*(a+b*arcsinh(c*x))/b)*cosh(5*a/b)/b^2-25/8*Shi((a+b*arcsinh(c*x))/b)*sinh(
a/b)/b^2-25/16*Shi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b^2-5/16*Shi(5*(a+b*arcsinh(c*x))/b)*sinh(5*a/b)/b^2-Un
integrable((c^2*x^2+1)^2/x^2/(a+b*arcsinh(c*x)),x)/b/c

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Rubi [A]
time = 0.36, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (1+c^2 x^2\right )^{5/2}}{x \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx \end {gather*}

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])/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) - 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 \text {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 \text {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 \text {Subst}\left (\int \frac {\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}+\frac {25 \text {Subst}\left (\int \frac {\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b}+\frac {25 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 6.37, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+c^2 x^2\right )^{5/2}}{x \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x \left (a +b \arcsinh \left (c x \right )\right )^{2}}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((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 [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{x \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

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]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c^2\,x^2+1\right )}^{5/2}}{x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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