### 3.454 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^{7/2}} \, dx$$

Optimal. Leaf size=145 $-\frac{11}{a c^3 \sqrt{c-\frac{c}{a x}}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{7/2}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}$

[Out]

-11/(7*a*(c - c/(a*x))^(7/2)) - 11/(5*a*c*(c - c/(a*x))^(5/2)) - 11/(3*a*c^2*(c - c/(a*x))^(3/2)) - 11/(a*c^3*
Sqrt[c - c/(a*x)]) + x/(c - c/(a*x))^(7/2) + (11*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.229653, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 12, 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, 51, 63, 208} $-\frac{11}{a c^3 \sqrt{c-\frac{c}{a x}}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{7/2}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*ArcCoth[a*x])/(c - c/(a*x))^(7/2),x]

[Out]

-11/(7*a*(c - c/(a*x))^(7/2)) - 11/(5*a*c*(c - c/(a*x))^(5/2)) - 11/(3*a*c^2*(c - c/(a*x))^(3/2)) - 11/(a*c^3*
Sqrt[c - c/(a*x)]) + x/(c - c/(a*x))^(7/2) + (11*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*c^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac{e^{2 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^{7/2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^{7/2}} \, dx\\ &=-\int \frac{1+a x}{\left (c-\frac{c}{a x}\right )^{7/2} (1-a x)} \, dx\\ &=\frac{c \int \frac{1+a x}{\left (c-\frac{c}{a x}\right )^{9/2} x} \, dx}{a}\\ &=\frac{c \int \frac{a+\frac{1}{x}}{\left (c-\frac{c}{a x}\right )^{9/2}} \, dx}{a}\\ &=-\frac{c \operatorname{Subst}\left (\int \frac{a+x}{x^2 \left (c-\frac{c x}{a}\right )^{9/2}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{(11 c) \operatorname{Subst}\left (\int \frac{1}{x \left (c-\frac{c x}{a}\right )^{9/2}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{x \left (c-\frac{c x}{a}\right )^{7/2}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{x \left (c-\frac{c x}{a}\right )^{5/2}} \, dx,x,\frac{1}{x}\right )}{2 a c}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{x \left (c-\frac{c x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{2 a c^2}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}-\frac{11}{a c^3 \sqrt{c-\frac{c}{a x}}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{2 a c^3}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}-\frac{11}{a c^3 \sqrt{c-\frac{c}{a x}}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )}{c^4}\\ &=-\frac{11}{7 a \left (c-\frac{c}{a x}\right )^{7/2}}-\frac{11}{5 a c \left (c-\frac{c}{a x}\right )^{5/2}}-\frac{11}{3 a c^2 \left (c-\frac{c}{a x}\right )^{3/2}}-\frac{11}{a c^3 \sqrt{c-\frac{c}{a x}}}+\frac{x}{\left (c-\frac{c}{a x}\right )^{7/2}}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0408748, size = 46, normalized size = 0.32 $\frac{7 x-\frac{11 \text{Hypergeometric2F1}\left (-\frac{7}{2},1,-\frac{5}{2},1-\frac{1}{a x}\right )}{a}}{7 \left (c-\frac{c}{a x}\right )^{7/2}}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - c/(a*x))^(7/2),x]

[Out]

(7*x - (11*Hypergeometric2F1[-7/2, 1, -5/2, 1 - 1/(a*x)])/a)/(7*(c - c/(a*x))^(7/2))

________________________________________________________________________________________

Maple [B]  time = 0.174, size = 396, normalized size = 2.7 \begin{align*}{\frac{x}{210\,{c}^{4} \left ( ax-1 \right ) ^{5}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 2310\,{a}^{11/2}\sqrt{ \left ( ax-1 \right ) x}{x}^{5}+1155\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{5}{a}^{5}-2100\,{a}^{9/2} \left ( \left ( ax-1 \right ) x \right ) ^{3/2}{x}^{3}-11550\,\sqrt{ \left ( ax-1 \right ) x}{a}^{9/2}{x}^{4}-5775\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{4}{a}^{4}+5368\, \left ( \left ( ax-1 \right ) x \right ) ^{3/2}{a}^{7/2}{x}^{2}+23100\,\sqrt{ \left ( ax-1 \right ) x}{a}^{7/2}{x}^{3}+11550\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{3}{a}^{3}-4928\, \left ( \left ( ax-1 \right ) x \right ) ^{3/2}{a}^{5/2}x-23100\,{a}^{5/2}\sqrt{ \left ( ax-1 \right ) x}{x}^{2}-11550\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{2}{a}^{2}+1540\,{a}^{3/2} \left ( \left ( ax-1 \right ) x \right ) ^{3/2}+11550\,{a}^{3/2}\sqrt{ \left ( ax-1 \right ) x}x+5775\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) xa-2310\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}-1155\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(a*x-1)/(c-c/a/x)^(7/2),x)

[Out]

1/210*(c*(a*x-1)/a/x)^(1/2)*x*(2310*a^(11/2)*((a*x-1)*x)^(1/2)*x^5+1155*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*
a*x-1)/a^(1/2))*x^5*a^5-2100*a^(9/2)*((a*x-1)*x)^(3/2)*x^3-11550*((a*x-1)*x)^(1/2)*a^(9/2)*x^4-5775*ln(1/2*(2*
((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*x^4*a^4+5368*((a*x-1)*x)^(3/2)*a^(7/2)*x^2+23100*((a*x-1)*x)^(1/2)
*a^(7/2)*x^3+11550*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*x^3*a^3-4928*((a*x-1)*x)^(3/2)*a^(5/2
)*x-23100*a^(5/2)*((a*x-1)*x)^(1/2)*x^2-11550*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*x^2*a^2+15
40*a^(3/2)*((a*x-1)*x)^(3/2)+11550*a^(3/2)*((a*x-1)*x)^(1/2)*x+5775*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-
1)/a^(1/2))*x*a-2310*((a*x-1)*x)^(1/2)*a^(1/2)-1155*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2)))/((a
*x-1)*x)^(1/2)/c^4/(a*x-1)^5/a^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{{\left (a x - 1\right )}{\left (c - \frac{c}{a x}\right )}^{\frac{7}{2}}}\,{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)/((a*x - 1)*(c - c/(a*x))^(7/2)), x)

________________________________________________________________________________________

Fricas [A]  time = 1.80492, size = 774, normalized size = 5.34 \begin{align*} \left [\frac{1155 \,{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \sqrt{c} \log \left (-2 \, a c x - 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right ) + 2 \,{\left (105 \, a^{5} x^{5} - 1936 \, a^{4} x^{4} + 4466 \, a^{3} x^{3} - 3850 \, a^{2} x^{2} + 1155 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{210 \,{\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}}, -\frac{1155 \,{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) -{\left (105 \, a^{5} x^{5} - 1936 \, a^{4} x^{4} + 4466 \, a^{3} x^{3} - 3850 \, a^{2} x^{2} + 1155 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{105 \,{\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}}\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)^(7/2),x, algorithm="fricas")

[Out]

[1/210*(1155*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*sqrt(c)*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x -
c)/(a*x)) + c) + 2*(105*a^5*x^5 - 1936*a^4*x^4 + 4466*a^3*x^3 - 3850*a^2*x^2 + 1155*a*x)*sqrt((a*c*x - c)/(a*x
)))/(a^5*c^4*x^4 - 4*a^4*c^4*x^3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^4), -1/105*(1155*(a^4*x^4 - 4*a^3*x^3 + 6
*a^2*x^2 - 4*a*x + 1)*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - (105*a^5*x^5 - 1936*a^4*x^4 + 4466
*a^3*x^3 - 3850*a^2*x^2 + 1155*a*x)*sqrt((a*c*x - c)/(a*x)))/(a^5*c^4*x^4 - 4*a^4*c^4*x^3 + 6*a^3*c^4*x^2 - 4*
a^2*c^4*x + a*c^4)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.31902, size = 250, normalized size = 1.72 \begin{align*} -\frac{1}{105} \, a c{\left (\frac{2 \,{\left (30 \, c^{3} + \frac{63 \,{\left (a c x - c\right )} c^{2}}{a x} + \frac{140 \,{\left (a c x - c\right )}^{2} c}{a^{2} x^{2}} + \frac{525 \,{\left (a c x - c\right )}^{3}}{a^{3} x^{3}}\right )} a x^{3}}{{\left (a c x - c\right )}^{3} c^{4} \sqrt{\frac{a c x - c}{a x}}} + \frac{1155 \, \arctan \left (\frac{\sqrt{\frac{a c x - c}{a x}}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{4}} - \frac{105 \, \sqrt{\frac{a c x - c}{a x}}}{a^{2}{\left (c - \frac{a c x - c}{a x}\right )} c^{4}}\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)^(7/2),x, algorithm="giac")

[Out]

-1/105*a*c*(2*(30*c^3 + 63*(a*c*x - c)*c^2/(a*x) + 140*(a*c*x - c)^2*c/(a^2*x^2) + 525*(a*c*x - c)^3/(a^3*x^3)
)*a*x^3/((a*c*x - c)^3*c^4*sqrt((a*c*x - c)/(a*x))) + 1155*arctan(sqrt((a*c*x - c)/(a*x))/sqrt(-c))/(a^2*sqrt(
-c)*c^4) - 105*sqrt((a*c*x - c)/(a*x))/(a^2*(c - (a*c*x - c)/(a*x))*c^4))