3.634 \(\int e^{3 \coth ^{-1}(a x)} (c-a^2 c x^2)^{7/2} \, dx\)

Optimal. Leaf size=139 \[ \frac{(a x+1)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{4 (a x+1)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{2 (a x+1)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}} \]

[Out]

(2*(1 + a*x)^6*(c - a^2*c*x^2)^(7/2))/(3*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) - (4*(1 + a*x)^7*(c - a^2*c*x^2)^(7/
2))/(7*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) + ((1 + a*x)^8*(c - a^2*c*x^2)^(7/2))/(8*a^8*(1 - 1/(a^2*x^2))^(7/2)*x
^7)

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Rubi [A]  time = 0.19329, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6192, 6193, 43} \[ \frac{(a x+1)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{4 (a x+1)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}+\frac{2 (a x+1)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 x^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(7/2),x]

[Out]

(2*(1 + a*x)^6*(c - a^2*c*x^2)^(7/2))/(3*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) - (4*(1 + a*x)^7*(c - a^2*c*x^2)^(7/
2))/(7*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) + ((1 + a*x)^8*(c - a^2*c*x^2)^(7/2))/(8*a^8*(1 - 1/(a^2*x^2))^(7/2)*x
^7)

Rule 6192

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

Rule 6193

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx &=\frac{\left (c-a^2 c x^2\right )^{7/2} \int e^{3 \coth ^{-1}(a x)} \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7 \, dx}{\left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{\left (c-a^2 c x^2\right )^{7/2} \int (-1+a x)^2 (1+a x)^5 \, dx}{a^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{\left (c-a^2 c x^2\right )^{7/2} \int \left (4 (1+a x)^5-4 (1+a x)^6+(1+a x)^7\right ) \, dx}{a^7 \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{2 (1+a x)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}-\frac{4 (1+a x)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}+\frac{(1+a x)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 \left (1-\frac{1}{a^2 x^2}\right )^{7/2} x^7}\\ \end{align*}

Mathematica [A]  time = 0.0490649, size = 63, normalized size = 0.45 \[ -\frac{c^3 (a x+1)^6 \left (21 a^2 x^2-54 a x+37\right ) \sqrt{c-a^2 c x^2}}{168 a^2 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(7/2),x]

[Out]

-(c^3*(1 + a*x)^6*(37 - 54*a*x + 21*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(168*a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.133, size = 100, normalized size = 0.7 \begin{align*}{\frac{x \left ( 21\,{a}^{7}{x}^{7}+72\,{x}^{6}{a}^{6}+28\,{x}^{5}{a}^{5}-168\,{x}^{4}{a}^{4}-210\,{x}^{3}{a}^{3}+56\,{a}^{2}{x}^{2}+252\,ax+168 \right ) }{168\, \left ( ax-1 \right ) ^{2} \left ( ax+1 \right ) ^{5}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(7/2),x)

[Out]

1/168*x*(21*a^7*x^7+72*a^6*x^6+28*a^5*x^5-168*a^4*x^4-210*a^3*x^3+56*a^2*x^2+252*a*x+168)*(-a^2*c*x^2+c)^(7/2)
/(a*x-1)^2/(a*x+1)^5/((a*x-1)/(a*x+1))^(3/2)

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Maxima [A]  time = 1.08362, size = 232, normalized size = 1.67 \begin{align*} -\frac{{\left (21 \, a^{9} \sqrt{-c} c^{3} x^{9} + 51 \, a^{8} \sqrt{-c} c^{3} x^{8} - 44 \, a^{7} \sqrt{-c} c^{3} x^{7} - 196 \, a^{6} \sqrt{-c} c^{3} x^{6} - 42 \, a^{5} \sqrt{-c} c^{3} x^{5} + 266 \, a^{4} \sqrt{-c} c^{3} x^{4} + 196 \, a^{3} \sqrt{-c} c^{3} x^{3} - 84 \, a^{2} \sqrt{-c} c^{3} x^{2} - 168 \, \sqrt{-c} c^{3}\right )}{\left (a x + 1\right )}^{2}}{168 \,{\left (a^{3} x^{2} + 2 \, a^{2} x + a\right )}{\left (a x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

-1/168*(21*a^9*sqrt(-c)*c^3*x^9 + 51*a^8*sqrt(-c)*c^3*x^8 - 44*a^7*sqrt(-c)*c^3*x^7 - 196*a^6*sqrt(-c)*c^3*x^6
- 42*a^5*sqrt(-c)*c^3*x^5 + 266*a^4*sqrt(-c)*c^3*x^4 + 196*a^3*sqrt(-c)*c^3*x^3 - 84*a^2*sqrt(-c)*c^3*x^2 - 1
68*sqrt(-c)*c^3)*(a*x + 1)^2/((a^3*x^2 + 2*a^2*x + a)*(a*x - 1))

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Fricas [A]  time = 1.51507, size = 209, normalized size = 1.5 \begin{align*} -\frac{{\left (21 \, a^{7} c^{3} x^{8} + 72 \, a^{6} c^{3} x^{7} + 28 \, a^{5} c^{3} x^{6} - 168 \, a^{4} c^{3} x^{5} - 210 \, a^{3} c^{3} x^{4} + 56 \, a^{2} c^{3} x^{3} + 252 \, a c^{3} x^{2} + 168 \, c^{3} x\right )} \sqrt{-a^{2} c}}{168 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

-1/168*(21*a^7*c^3*x^8 + 72*a^6*c^3*x^7 + 28*a^5*c^3*x^6 - 168*a^4*c^3*x^5 - 210*a^3*c^3*x^4 + 56*a^2*c^3*x^3
+ 252*a*c^3*x^2 + 168*c^3*x)*sqrt(-a^2*c)/a

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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))**(3/2)*(-a**2*c*x**2+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{7}{2}}}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(7/2)/((a*x - 1)/(a*x + 1))^(3/2), x)