3.688 \(\int e^{\tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{5/2} \, dx\)

Optimal. Leaf size=219 \[ -\frac {a x^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{3 \left (1-a^2 x^2\right )^{5/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{4 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^5 x^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^4 x^5 \log (x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}+\frac {2 a^3 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}} \]

[Out]

-1/4*(c-c/a^2/x^2)^(5/2)*x/(-a^2*x^2+1)^(5/2)-1/3*a*(c-c/a^2/x^2)^(5/2)*x^2/(-a^2*x^2+1)^(5/2)+a^2*(c-c/a^2/x^
2)^(5/2)*x^3/(-a^2*x^2+1)^(5/2)+2*a^3*(c-c/a^2/x^2)^(5/2)*x^4/(-a^2*x^2+1)^(5/2)+a^5*(c-c/a^2/x^2)^(5/2)*x^6/(
-a^2*x^2+1)^(5/2)+a^4*(c-c/a^2/x^2)^(5/2)*x^5*ln(x)/(-a^2*x^2+1)^(5/2)

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Rubi [A]  time = 0.18, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6160, 6150, 88} \[ \frac {a^5 x^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}+\frac {2 a^3 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}}-\frac {a x^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{3 \left (1-a^2 x^2\right )^{5/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{4 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^4 x^5 \log (x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}{\left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2),x]

[Out]

-((c - c/(a^2*x^2))^(5/2)*x)/(4*(1 - a^2*x^2)^(5/2)) - (a*(c - c/(a^2*x^2))^(5/2)*x^2)/(3*(1 - a^2*x^2)^(5/2))
+ (a^2*(c - c/(a^2*x^2))^(5/2)*x^3)/(1 - a^2*x^2)^(5/2) + (2*a^3*(c - c/(a^2*x^2))^(5/2)*x^4)/(1 - a^2*x^2)^(
5/2) + (a^5*(c - c/(a^2*x^2))^(5/2)*x^6)/(1 - a^2*x^2)^(5/2) + (a^4*(c - c/(a^2*x^2))^(5/2)*x^5*Log[x])/(1 - a
^2*x^2)^(5/2)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6160

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/
(1 + (c*x^2)/d)^p, Int[(u*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[c + a^2*d, 0] &&  !IntegerQ[p] &&  !IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx &=\frac {\left (\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{5/2}}{x^5} \, dx}{\left (1-a^2 x^2\right )^{5/2}}\\ &=\frac {\left (\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {(1-a x)^2 (1+a x)^3}{x^5} \, dx}{\left (1-a^2 x^2\right )^{5/2}}\\ &=\frac {\left (\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (a^5+\frac {1}{x^5}+\frac {a}{x^4}-\frac {2 a^2}{x^3}-\frac {2 a^3}{x^2}+\frac {a^4}{x}\right ) \, dx}{\left (1-a^2 x^2\right )^{5/2}}\\ &=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}{4 \left (1-a^2 x^2\right )^{5/2}}-\frac {a \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}{3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}{\left (1-a^2 x^2\right )^{5/2}}+\frac {2 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^6}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (1-a^2 x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 82, normalized size = 0.37 \[ \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \left (12 a^5 x^5+12 a^4 x^4 \log (x)+24 a^3 x^3+12 a^2 x^2-4 a x-3\right )}{12 a^4 x^3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2),x]

[Out]

(c^2*Sqrt[c - c/(a^2*x^2)]*(-3 - 4*a*x + 12*a^2*x^2 + 24*a^3*x^3 + 12*a^5*x^5 + 12*a^4*x^4*Log[x]))/(12*a^4*x^
3*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 1.71, size = 478, normalized size = 2.18 \[ \left [\frac {6 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} + {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (12 \, a^{5} c^{2} x^{5} + 24 \, a^{3} c^{2} x^{3} - {\left (12 \, a^{5} + 24 \, a^{3} + 12 \, a^{2} - 4 \, a - 3\right )} c^{2} x^{4} + 12 \, a^{2} c^{2} x^{2} - 4 \, a c^{2} x - 3 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{12 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}, -\frac {12 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) + {\left (12 \, a^{5} c^{2} x^{5} + 24 \, a^{3} c^{2} x^{3} - {\left (12 \, a^{5} + 24 \, a^{3} + 12 \, a^{2} - 4 \, a - 3\right )} c^{2} x^{4} + 12 \, a^{2} c^{2} x^{2} - 4 \, a c^{2} x - 3 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{12 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 + (a*x^5 - a*x)*sqrt(-a^2*x^2
+ 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c)/(a^2*x^4 - x^2)) - (12*a^5*c^2*x^5 + 24*a^3*c^2*x^3 - (12*
a^5 + 24*a^3 + 12*a^2 - 4*a - 3)*c^2*x^4 + 12*a^2*c^2*x^2 - 4*a*c^2*x - 3*c^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*
x^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3), -1/12*(12*(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(c)*arctan(sqrt(-a^2*x^2 +
1)*(a*x^3 + a*x)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 - (a^2 + 1)*c*x^2 + c)) + (12*a^5*c^2*x^5
+ 24*a^3*c^2*x^3 - (12*a^5 + 24*a^3 + 12*a^2 - 4*a - 3)*c^2*x^4 + 12*a^2*c^2*x^2 - 4*a*c^2*x - 3*c^2)*sqrt(-a
^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 86, normalized size = 0.39 \[ -\frac {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}} x \sqrt {-a^{2} x^{2}+1}\, \left (12 x^{5} a^{5}+12 a^{4} \ln \relax (x ) x^{4}+24 x^{3} a^{3}+12 a^{2} x^{2}-4 a x -3\right )}{12 \left (a^{2} x^{2}-1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*x/(a^2*x^2-1)^3*(-a^2*x^2+1)^(1/2)*(12*x^5*a^5+12*a^4*ln(x)*x^4+24*x^3*a^3
+12*a^2*x^2-4*a*x-3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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