### 3.471 $$\int e^{-2 \coth ^{-1}(a x)} (c-\frac{c}{a x})^{7/2} \, dx$$

Optimal. Leaf size=163 $-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}-\frac{11 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+x \left (c-\frac{c}{a x}\right )^{7/2}$

[Out]

(-21*c^3*Sqrt[c - c/(a*x)])/a - (5*c^2*(c - c/(a*x))^(3/2))/(3*a) + (3*c*(c - c/(a*x))^(5/2))/(5*a) + (c - c/(
a*x))^(7/2)*x - (11*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a + (32*Sqrt[2]*c^(7/2)*ArcTanh[Sqrt[c - c/(a*
x)]/(Sqrt[2]*Sqrt[c])])/a

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Rubi [A]  time = 0.280299, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.417, Rules used = {6167, 6133, 25, 514, 375, 98, 154, 156, 63, 208} $-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}-\frac{11 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+x \left (c-\frac{c}{a x}\right )^{7/2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-21*c^3*Sqrt[c - c/(a*x)])/a - (5*c^2*(c - c/(a*x))^(3/2))/(3*a) + (3*c*(c - c/(a*x))^(5/2))/(5*a) + (c - c/(
a*x))^(7/2)*x - (11*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a + (32*Sqrt[2]*c^(7/2)*ArcTanh[Sqrt[c - c/(a*
x)]/(Sqrt[2]*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 98

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

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 )^{7/2} \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^{7/2} \, dx\\ &=-\int \frac{\left (c-\frac{c}{a x}\right )^{7/2} (1-a x)}{1+a x} \, dx\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{9/2} x}{1+a x} \, dx}{c}\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{9/2}}{a+\frac{1}{x}} \, dx}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{9/2}}{x^2 (a+x)} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{\operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{5/2} \left (\frac{11 c^2}{2}+\frac{3 c^2 x}{2 a}\right )}{x (a+x)} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{2 \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{3/2} \left (\frac{55 c^3}{4}-\frac{25 c^3 x}{4 a}\right )}{x (a+x)} \, dx,x,\frac{1}{x}\right )}{5 c}\\ &=-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{4 \operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}} \left (\frac{165 c^4}{8}-\frac{315 c^4 x}{8 a}\right )}{x (a+x)} \, dx,x,\frac{1}{x}\right )}{15 c}\\ &=-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{8 \operatorname{Subst}\left (\int \frac{\frac{165 c^5}{16}-\frac{795 c^5 x}{16 a}}{x (a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{15 c}\\ &=-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{\left (11 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{2 a}-\frac{\left (32 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x-\left (11 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )+\left (64 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )\\ &=-\frac{21 c^3 \sqrt{c-\frac{c}{a x}}}{a}-\frac{5 c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{3 a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+\left (c-\frac{c}{a x}\right )^{7/2} x-\frac{11 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.260302, size = 125, normalized size = 0.77 $\frac{c^3 \left (15 a^3 x^3-376 a^2 x^2+52 a x-6\right ) \sqrt{c-\frac{c}{a x}}}{15 a^3 x^2}-\frac{11 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c^3*Sqrt[c - c/(a*x)]*(-6 + 52*a*x - 376*a^2*x^2 + 15*a^3*x^3))/(15*a^3*x^2) - (11*c^(7/2)*ArcTanh[Sqrt[c - c
/(a*x)]/Sqrt[c]])/a + (32*Sqrt[2]*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])])/a

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Maple [B]  time = 0.168, size = 281, normalized size = 1.7 \begin{align*} -{\frac{{c}^{3}}{30\,{x}^{3}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 1110\,{a}^{7/2}\sqrt{{a}^{-1}}\sqrt{a{x}^{2}-x}{x}^{4}-480\,{a}^{7/2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}{x}^{4}+480\,{a}^{5/2}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1}{ax+1}} \right ){x}^{4}-660\,{a}^{5/2}\sqrt{{a}^{-1}} \left ( a{x}^{2}-x \right ) ^{3/2}{x}^{2}-555\,\sqrt{{a}^{-1}}\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{4}{a}^{3}+720\,\sqrt{{a}^{-1}}\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{4}{a}^{3}+92\,{a}^{3/2} \left ( a{x}^{2}-x \right ) ^{3/2}x\sqrt{{a}^{-1}}-12\, \left ( a{x}^{2}-x \right ) ^{3/2}\sqrt{a}\sqrt{{a}^{-1}} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{{a}^{-1}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(c*(a*x-1)/a/x)^(1/2)/x^3*c^3/a^(7/2)*(1110*a^(7/2)*(1/a)^(1/2)*(a*x^2-x)^(1/2)*x^4-480*a^(7/2)*(1/a)^(1
/2)*((a*x-1)*x)^(1/2)*x^4+480*a^(5/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*
x^4-660*a^(5/2)*(1/a)^(1/2)*(a*x^2-x)^(3/2)*x^2-555*(1/a)^(1/2)*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(
1/2))*x^4*a^3+720*(1/a)^(1/2)*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*x^4*a^3+92*a^(3/2)*(a*x^2-
x)^(3/2)*x*(1/a)^(1/2)-12*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/((a*x-1)*x)^(1/2)/(1/a)^(1/2)

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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{7}{2}}}{a x + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(7/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.2818, size = 744, normalized size = 4.56 \begin{align*} \left [\frac{480 \, \sqrt{2} a^{2} c^{\frac{7}{2}} x^{2} \log \left (-\frac{2 \, \sqrt{2} a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + 3 \, a c x - c}{a x + 1}\right ) + 165 \, a^{2} c^{\frac{7}{2}} x^{2} \log \left (-2 \, a c x + 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right ) + 2 \,{\left (15 \, a^{3} c^{3} x^{3} - 376 \, a^{2} c^{3} x^{2} + 52 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt{\frac{a c x - c}{a x}}}{30 \, a^{3} x^{2}}, -\frac{480 \, \sqrt{2} a^{2} \sqrt{-c} c^{3} x^{2} \arctan \left (\frac{\sqrt{2} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{2 \, c}\right ) - 165 \, a^{2} \sqrt{-c} c^{3} x^{2} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) -{\left (15 \, a^{3} c^{3} x^{3} - 376 \, a^{2} c^{3} x^{2} + 52 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt{\frac{a c x - c}{a x}}}{15 \, a^{3} x^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(7/2)*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

[1/30*(480*sqrt(2)*a^2*c^(7/2)*x^2*log(-(2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + 3*a*c*x - c)/(a*x + 1
)) + 165*a^2*c^(7/2)*x^2*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) + 2*(15*a^3*c^3*x^3 - 376*a
^2*c^3*x^2 + 52*a*c^3*x - 6*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2), -1/15*(480*sqrt(2)*a^2*sqrt(-c)*c^3*x^2*a
rctan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - 165*a^2*sqrt(-c)*c^3*x^2*arctan(sqrt(-c)*sqrt((a*c*x -
c)/(a*x))/c) - (15*a^3*c^3*x^3 - 376*a^2*c^3*x^2 + 52*a*c^3*x - 6*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (-1 + \frac{1}{a x}\right )\right )^{\frac{7}{2}} \left (a x - 1\right )}{a x + 1}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((c-c/a/x)^(7/2)*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

Exception raised: TypeError