3.341 \(\int \frac {(1-c^2 x^2)^{5/2}}{x^4 (a+b \cosh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=31 \[ \text {Int}\left (\frac {\left (1-c^2 x^2\right )^{5/2}}{x^4 \left (a+b \cosh ^{-1}(c x)\right )^2},x\right ) \]

[Out]

Unintegrable((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(c*x))^2,x)

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Rubi [A]  time = 0.53, 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 = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^4 \left (a+b \cosh ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(1 - c^2*x^2)^(5/2)/(x^4*(a + b*ArcCosh[c*x])^2),x]

[Out]

(Sqrt[1 - c^2*x^2]*Defer[Int][((-1 + c*x)^(5/2)*(1 + c*x)^(5/2))/(x^4*(a + b*ArcCosh[c*x])^2), x])/(Sqrt[-1 +
c*x]*Sqrt[1 + c*x])

Rubi steps

\begin {align*} \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^4 \left (a+b \cosh ^{-1}(c x)\right )^2} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {(-1+c x)^{5/2} (1+c x)^{5/2}}{x^4 \left (a+b \cosh ^{-1}(c x)\right )^2} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]  time = 179.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^4 \left (a+b \cosh ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(1 - c^2*x^2)^(5/2)/(x^4*(a + b*ArcCosh[c*x])^2),x]

[Out]

Integrate[(1 - c^2*x^2)^(5/2)/(x^4*(a + b*ArcCosh[c*x])^2), x]

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fricas [A]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-c^{2} x^{2} + 1}}{b^{2} x^{4} \operatorname {arcosh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {arcosh}\left (c x\right ) + a^{2} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(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^4*arccosh(c*x)^2 + 2*a*b*x^4*arccosh(c*x) + a^2*x
^4), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}^{2} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(c*x))^2,x, algorithm="giac")

[Out]

integrate((-c^2*x^2 + 1)^(5/2)/((b*arccosh(c*x) + a)^2*x^4), x)

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maple [A]  time = 1.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x^{4} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(c*x))^2,x)

[Out]

int((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(c*x))^2,x)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left ({\left (c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1\right )} {\left (c x + 1\right )} \sqrt {c x - 1} + {\left (c^{7} x^{7} - 3 \, c^{5} x^{5} + 3 \, c^{3} x^{3} - c x\right )} \sqrt {c x + 1}\right )} \sqrt {-c x + 1}}{a b c^{3} x^{6} + \sqrt {c x + 1} \sqrt {c x - 1} a b c^{2} x^{5} - a b c x^{4} + {\left (b^{2} c^{3} x^{6} + \sqrt {c x + 1} \sqrt {c x - 1} b^{2} c^{2} x^{5} - b^{2} c x^{4}\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )} + \int \frac {{\left ({\left (2 \, c^{7} x^{7} + c^{5} x^{5} - 8 \, c^{3} x^{3} + 5 \, c x\right )} {\left (c x + 1\right )}^{\frac {3}{2}} {\left (c x - 1\right )} + 2 \, {\left (2 \, c^{8} x^{8} - c^{6} x^{6} - 6 \, c^{4} x^{4} + 7 \, c^{2} x^{2} - 2\right )} {\left (c x + 1\right )} \sqrt {c x - 1} + {\left (2 \, c^{9} x^{9} - 3 \, c^{7} x^{7} - 3 \, c^{5} x^{5} + 7 \, c^{3} x^{3} - 3 \, c x\right )} \sqrt {c x + 1}\right )} \sqrt {-c x + 1}}{a b c^{5} x^{9} + {\left (c x + 1\right )} {\left (c x - 1\right )} a b c^{3} x^{7} - 2 \, a b c^{3} x^{7} + a b c x^{5} + 2 \, {\left (a b c^{4} x^{8} - a b c^{2} x^{6}\right )} \sqrt {c x + 1} \sqrt {c x - 1} + {\left (b^{2} c^{5} x^{9} + {\left (c x + 1\right )} {\left (c x - 1\right )} b^{2} c^{3} x^{7} - 2 \, b^{2} c^{3} x^{7} + b^{2} c x^{5} + 2 \, {\left (b^{2} c^{4} x^{8} - b^{2} c^{2} x^{6}\right )} \sqrt {c x + 1} \sqrt {c x - 1}\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(5/2)/x^4/(a+b*arccosh(c*x))^2,x, algorithm="maxima")

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x^4\,{\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - c^2*x^2)^(5/2)/(x^4*(a + b*acosh(c*x))^2),x)

[Out]

int((1 - c^2*x^2)^(5/2)/(x^4*(a + b*acosh(c*x))^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*x**2+1)**(5/2)/x**4/(a+b*acosh(c*x))**2,x)

[Out]

Timed out

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