3.457 \(\int e^{3 \coth ^{-1}(a x)} (c-\frac {c}{a x})^{5/2} \, dx\)

Optimal. Leaf size=156 \[ \frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a}+\frac {c^5 x \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (c-\frac {c}{a x}\right )^{5/2}}-\frac {c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 a \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {c^3 \sqrt {1-\frac {1}{a^2 x^2}}}{a \sqrt {c-\frac {c}{a x}}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6177, 879, 865, 875, 208} \[ \frac {c^5 x \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (c-\frac {c}{a x}\right )^{5/2}}-\frac {c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 a \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {c^3 \sqrt {1-\frac {1}{a^2 x^2}}}{a \sqrt {c-\frac {c}{a x}}}+\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])*(c - c/(a*x))^(5/2),x]

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 865

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*
x)^m*(f + g*x)^(n + 1)*(a + c*x^2)^p)/(g*(m - n - 1)), x] - Dist[(c*m*(e*f + d*g))/(e^2*g*(m - n - 1)), Int[(d
+ e*x)^(m + 1)*(f + g*x)^n*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0
] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[p, 0] && NeQ[m - n - 1, 0] &&  !IGtQ[n, 0]
&&  !(IntegerQ[n + p] && LtQ[n + p + 2, 0]) && RationalQ[n]

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
EqQ[m + p - 1, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 6177

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 89, normalized size = 0.57 \[ \frac {c^2 \sqrt {c-\frac {c}{a x}} \left (\sqrt {\frac {1}{a x}+1} \left (3 a^2 x^2+2 a x+2\right )+3 a x \tanh ^{-1}\left (\sqrt {\frac {1}{a x}+1}\right )\right )}{3 a^2 x \sqrt {1-\frac {1}{a x}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - c/(a*x))^(5/2),x]

[Out]

(c^2*Sqrt[c - c/(a*x)]*(Sqrt[1 + 1/(a*x)]*(2 + 2*a*x + 3*a^2*x^2) + 3*a*x*ArcTanh[Sqrt[1 + 1/(a*x)]]))/(3*a^2*
Sqrt[1 - 1/(a*x)]*x)

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fricas [A]  time = 0.53, size = 381, normalized size = 2.44 \[ \left [\frac {3 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (3 \, a^{3} c^{2} x^{3} + 5 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{12 \, {\left (a^{3} x^{2} - a^{2} x\right )}}, -\frac {3 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (3 \, a^{3} c^{2} x^{3} + 5 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{6 \, {\left (a^{3} x^{2} - a^{2} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a^2*c^2*x^2 - a*c^2*x)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x + 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)
*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(3*a^3*c^2*x^3 + 5*a^2*c^2*x^2 + 4*a*c^
2*x + 2*c^2)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2 - a^2*x), -1/6*(3*(a^2*c^2*x^2 - a*c^
2*x)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2
- a*c*x - c)) - 2*(3*a^3*c^2*x^3 + 5*a^2*c^2*x^2 + 4*a*c^2*x + 2*c^2)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x -
c)/(a*x)))/(a^3*x^2 - a^2*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^(5/2),x, algorithm="giac")

[Out]

integrate((c - c/(a*x))^(5/2)/((a*x - 1)/(a*x + 1))^(3/2), x)

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maple [A]  time = 0.07, size = 144, normalized size = 0.92 \[ \frac {\left (a x -1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{2} \left (6 a^{\frac {5}{2}} x^{2} \sqrt {\left (a x +1\right ) x}+3 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) x^{2} a^{2}+4 a^{\frac {3}{2}} x \sqrt {\left (a x +1\right ) x}+4 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}\right )}{6 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) x \,a^{\frac {5}{2}} \sqrt {\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^(5/2),x, algorithm="maxima")

[Out]

integrate((c - c/(a*x))^(5/2)/((a*x - 1)/(a*x + 1))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a\,x}\right )}^{5/2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/(a*x))^(5/2)/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

int((c - c/(a*x))^(5/2)/((a*x - 1)/(a*x + 1))^(3/2), 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(1/((a*x-1)/(a*x+1))**(3/2)*(c-c/a/x)**(5/2),x)

[Out]

Timed out

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