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

Optimal. Leaf size=95 \[ \frac{c^2 \sqrt{c-\frac{c}{a x}}}{a}-\frac{c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{c \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+x \left (c-\frac{c}{a x}\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.20145, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6167, 6133, 25, 514, 375, 78, 50, 63, 208} \[ \frac{c^2 \sqrt{c-\frac{c}{a x}}}{a}-\frac{c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{c \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+x \left (c-\frac{c}{a x}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6133

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

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 78

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.0648904, size = 75, normalized size = 0.79 \[ \frac{c^2 \left (3 a^2 x^2-2 a x+2\right ) \sqrt{c-\frac{c}{a x}}-3 a c^{5/2} x \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{3 a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.164, size = 108, normalized size = 1.1 \begin{align*} -{\frac{{c}^{2}}{6\,{x}^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -6\,\sqrt{a{x}^{2}-x}{a}^{5/2}{x}^{3}+3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{3}{a}^{2}+4\, \left ( a{x}^{2}-x \right ) ^{3/2}\sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}{\left (c - \frac{c}{a x}\right )}^{\frac{5}{2}}}{a x - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.86239, size = 397, normalized size = 4.18 \begin{align*} \left [\frac{3 \, a c^{\frac{5}{2}} x \log \left (-2 \, a c x + 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right ) + 2 \,{\left (3 \, a^{2} c^{2} x^{2} - 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{\frac{a c x - c}{a x}}}{6 \, a^{2} x}, \frac{3 \, a \sqrt{-c} c^{2} x \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) +{\left (3 \, a^{2} c^{2} x^{2} - 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{\frac{a c x - c}{a x}}}{3 \, a^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError