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

Optimal. Leaf size=118 \[ \frac{3 c^3 \sqrt{c-\frac{c}{a x}}}{a}+\frac{c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{a}-\frac{3 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+x \left (c-\frac{c}{a x}\right )^{7/2} \]

[Out]

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

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Rubi [A]  time = 0.225019, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{3 c^3 \sqrt{c-\frac{c}{a x}}}{a}+\frac{c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{a}-\frac{3 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}+\frac{3 c \left (c-\frac{c}{a x}\right )^{5/2}}{5 a}+x \left (c-\frac{c}{a x}\right )^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c^3*Sqrt[c - c/(a*x)])/a + (c^2*(c - c/(a*x))^(3/2))/a + (3*c*(c - c/(a*x))^(5/2))/(5*a) + (c - c/(a*x))^(7
/2)*x - (3*c^(7/2)*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 )^{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{c \int \frac{\left (c-\frac{c}{a x}\right )^{5/2} (1+a x)}{x} \, dx}{a}\\ &=\frac{c \int \left (a+\frac{1}{x}\right ) \left (c-\frac{c}{a x}\right )^{5/2} \, dx}{a}\\ &=-\frac{c \operatorname{Subst}\left (\int \frac{(a+x) \left (c-\frac{c x}{a}\right )^{5/2}}{x^2} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\left (c-\frac{c}{a x}\right )^{7/2} x+\frac{(3 c) \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{5/2}}{x} \, dx,x,\frac{1}{x}\right )}{2 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 (3 c^2\right ) \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^2 \left (c-\frac{c}{a x}\right )^{3/2}}{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 (3 c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{x} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=\frac{3 c^3 \sqrt{c-\frac{c}{a x}}}{a}+\frac{c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{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 (3 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{3 c^3 \sqrt{c-\frac{c}{a x}}}{a}+\frac{c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{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 (3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )\\ &=\frac{3 c^3 \sqrt{c-\frac{c}{a x}}}{a}+\frac{c^2 \left (c-\frac{c}{a x}\right )^{3/2}}{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{3 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.116848, size = 83, normalized size = 0.7 \[ \frac{c^3 \left (5 a^3 x^3+8 a^2 x^2+4 a x-2\right ) \sqrt{c-\frac{c}{a x}}}{5 a^3 x^2}-\frac{3 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Sqrt[c - c/(a*x)]*(-2 + 4*a*x + 8*a^2*x^2 + 5*a^3*x^3))/(5*a^3*x^2) - (3*c^(7/2)*ArcTanh[Sqrt[c - c/(a*x)
]/Sqrt[c]])/a

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Maple [A]  time = 0.164, size = 144, normalized size = 1.2 \begin{align*} -{\frac{{c}^{3}}{10\,{x}^{3}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -30\,\sqrt{a{x}^{2}-x}{a}^{7/2}{x}^{4}+20\,{a}^{5/2} \left ( a{x}^{2}-x \right ) ^{3/2}{x}^{2}+15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{4}{a}^{3}+4\,{a}^{3/2} \left ( a{x}^{2}-x \right ) ^{3/2}x-4\, \left ( a{x}^{2}-x \right ) ^{3/2}\sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{7}{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)^(7/2),x)

[Out]

-1/10*(c*(a*x-1)/a/x)^(1/2)/x^3*c^3*(-30*(a*x^2-x)^(1/2)*a^(7/2)*x^4+20*a^(5/2)*(a*x^2-x)^(3/2)*x^2+15*ln(1/2*
(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*x^4*a^3+4*a^(3/2)*(a*x^2-x)^(3/2)*x-4*(a*x^2-x)^(3/2)*a^(1/2))/((
a*x-1)*x)^(1/2)/a^(7/2)

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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(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(7/2),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 = 1.87326, size = 460, normalized size = 3.9 \begin{align*} \left [\frac{15 \, 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 (5 \, a^{3} c^{3} x^{3} + 8 \, a^{2} c^{3} x^{2} + 4 \, a c^{3} x - 2 \, c^{3}\right )} \sqrt{\frac{a c x - c}{a x}}}{10 \, a^{3} x^{2}}, \frac{15 \, a^{2} \sqrt{-c} c^{3} x^{2} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) +{\left (5 \, a^{3} c^{3} x^{3} + 8 \, a^{2} c^{3} x^{2} + 4 \, a c^{3} x - 2 \, c^{3}\right )} \sqrt{\frac{a c x - c}{a x}}}{5 \, a^{3} x^{2}}\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]

[1/10*(15*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*(5*a^3*c^3*x^3 + 8*a^2
*c^3*x^2 + 4*a*c^3*x - 2*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2), 1/5*(15*a^2*sqrt(-c)*c^3*x^2*arctan(sqrt(-c)
*sqrt((a*c*x - c)/(a*x))/c) + (5*a^3*c^3*x^3 + 8*a^2*c^3*x^2 + 4*a*c^3*x - 2*c^3)*sqrt((a*c*x - c)/(a*x)))/(a^
3*x^2)]

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Sympy [C]  time = 19.9582, size = 729, normalized size = 6.18 \begin{align*} \text{result too large to display} \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]

c**3*Piecewise((sqrt(a)*sqrt(c)*x**(3/2)/sqrt(a*x - 1) - sqrt(c)*acosh(sqrt(a)*sqrt(x))/a - sqrt(c)*sqrt(x)/(s
qrt(a)*sqrt(a*x - 1)), Abs(a*x) > 1), (I*sqrt(c)*asin(sqrt(a)*sqrt(x))/a + I*sqrt(c)*sqrt(x)*sqrt(-a*x + 1)/sq
rt(a), True)) + 2*c**4*atan(sqrt(c - c/(a*x))/sqrt(-c))/(a*sqrt(-c)) + 2*c**3*sqrt(c - c/(a*x))/a - c**3*Piece
wise((0, Eq(c, 0)), (2*a*(c - c/(a*x))**(3/2)/(3*c), True))/a**2 + c**3*Piecewise((-4*a**(11/2)*sqrt(c)*x**(7/
2)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) + 4*a**(9/2)*sqrt(c)*x**(5/2)/(15*a**(7/2)*x**(7/2) - 15*a**(
5/2)*x**(5/2)) + 4*a**5*sqrt(c)*x**3*sqrt(a*x - 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) - 2*a**4*sqrt
(c)*x**2*sqrt(a*x - 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) - 8*a**3*sqrt(c)*x*sqrt(a*x - 1)/(15*a**(
7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) + 6*a**2*sqrt(c)*sqrt(a*x - 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5
/2)), Abs(a*x) > 1), (-4*a**(11/2)*sqrt(c)*x**(7/2)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) + 4*a**(9/2)
*sqrt(c)*x**(5/2)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) + 4*I*a**5*sqrt(c)*x**3*sqrt(-a*x + 1)/(15*a**
(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) - 2*I*a**4*sqrt(c)*x**2*sqrt(-a*x + 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5
/2)*x**(5/2)) - 8*I*a**3*sqrt(c)*x*sqrt(-a*x + 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)) + 6*I*a**2*sqr
t(c)*sqrt(-a*x + 1)/(15*a**(7/2)*x**(7/2) - 15*a**(5/2)*x**(5/2)), True))/a**3

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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)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError