3.651 \(\int \frac{e^{3 \tanh ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^3} \, dx\)

Optimal. Leaf size=155 \[ -\frac{(a x+1)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (a x+1)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (a x+1)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}+\frac{181 a x+245}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{3 \sin ^{-1}(a x)}{a c^3} \]

[Out]

-(1 + a*x)^3/(7*a*c^3*(1 - a^2*x^2)^(7/2)) + (38*(1 + a*x)^2)/(35*a*c^3*(1 - a^2*x^2)^(5/2)) - (137*(1 + a*x))
/(35*a*c^3*(1 - a^2*x^2)^(3/2)) + (245 + 181*a*x)/(35*a*c^3*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^3) - (
3*ArcSin[a*x])/(a*c^3)

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Rubi [A]  time = 0.439783, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6157, 6148, 1635, 1814, 641, 216} \[ -\frac{(a x+1)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (a x+1)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (a x+1)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}+\frac{181 a x+245}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{3 \sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + a*x)^3/(7*a*c^3*(1 - a^2*x^2)^(7/2)) + (38*(1 + a*x)^2)/(35*a*c^3*(1 - a^2*x^2)^(5/2)) - (137*(1 + a*x))
/(35*a*c^3*(1 - a^2*x^2)^(3/2)) + (245 + 181*a*x)/(35*a*c^3*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^3) - (
3*ArcSin[a*x])/(a*c^3)

Rule 6157

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

Rule 6148

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{3 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^3} \, dx &=-\frac{a^6 \int \frac{e^{3 \tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=-\frac{a^6 \int \frac{x^6 (1+a x)^3}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{a^6 \int \frac{(1+a x)^2 \left (\frac{3}{a^6}+\frac{7 x}{a^5}+\frac{7 x^2}{a^4}+\frac{7 x^3}{a^3}+\frac{7 x^4}{a^2}+\frac{7 x^5}{a}\right )}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (1+a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{a^6 \int \frac{(1+a x) \left (\frac{61}{a^6}+\frac{140 x}{a^5}+\frac{105 x^2}{a^4}+\frac{70 x^3}{a^3}+\frac{35 x^4}{a^2}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (1+a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (1+a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^6 \int \frac{\frac{228}{a^6}+\frac{630 x}{a^5}+\frac{315 x^2}{a^4}+\frac{105 x^3}{a^3}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{105 c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (1+a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (1+a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{245+181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{a^6 \int \frac{\frac{315}{a^6}+\frac{105 x}{a^5}}{\sqrt{1-a^2 x^2}} \, dx}{105 c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (1+a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (1+a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{245+181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac{(1+a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{38 (1+a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{137 (1+a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{245+181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{3 \sin ^{-1}(a x)}{a c^3}\\ \end{align*}

Mathematica [A]  time = 0.118628, size = 96, normalized size = 0.62 \[ \frac{-35 a^5 x^5+286 a^4 x^4-368 a^3 x^3-125 a^2 x^2-105 (a x-1)^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)+423 a x-176}{35 a c^3 (a x-1)^3 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-176 + 423*a*x - 125*a^2*x^2 - 368*a^3*x^3 + 286*a^4*x^4 - 35*a^5*x^5 - 105*(-1 + a*x)^3*Sqrt[1 - a^2*x^2]*Ar
cSin[a*x])/(35*a*c^3*(-1 + a*x)^3*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.058, size = 256, normalized size = 1.7 \begin{align*} -{\frac{a{x}^{2}}{{c}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+8\,{\frac{1}{a{c}^{3}\sqrt{-{a}^{2}{x}^{2}+1}}}+13\,{\frac{x}{{c}^{3}\sqrt{-{a}^{2}{x}^{2}+1}}}-3\,{\frac{1}{{c}^{3}\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{1}{7\,{a}^{4}{c}^{3}} \left ( x-{a}^{-1} \right ) ^{-3}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}}+{\frac{38}{35\,{a}^{3}{c}^{3}} \left ( x-{a}^{-1} \right ) ^{-2}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}}+{\frac{137}{35\,{a}^{2}{c}^{3}} \left ( x-{a}^{-1} \right ) ^{-1}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}}-{\frac{274\,x}{35\,{c}^{3}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/c^3*x^2/(-a^2*x^2+1)^(1/2)+8/a/c^3/(-a^2*x^2+1)^(1/2)+13*x/c^3/(-a^2*x^2+1)^(1/2)-3/c^3/(a^2)^(1/2)*arctan(
(a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/7/a^4/c^3/(x-1/a)^3/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+38/35/a^3/c^3/(x-1/
a)^2/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+137/35/a^2/c^3/(x-1/a)/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-274/35/c^3/(
-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.23451, size = 482, normalized size = 3.11 \begin{align*} \frac{176 \, a^{5} x^{5} - 528 \, a^{4} x^{4} + 352 \, a^{3} x^{3} + 352 \, a^{2} x^{2} - 528 \, a x + 210 \,{\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (35 \, a^{5} x^{5} - 286 \, a^{4} x^{4} + 368 \, a^{3} x^{3} + 125 \, a^{2} x^{2} - 423 \, a x + 176\right )} \sqrt{-a^{2} x^{2} + 1} + 176}{35 \,{\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(176*a^5*x^5 - 528*a^4*x^4 + 352*a^3*x^3 + 352*a^2*x^2 - 528*a*x + 210*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 +
2*a^2*x^2 - 3*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (35*a^5*x^5 - 286*a^4*x^4 + 368*a^3*x^3 + 125
*a^2*x^2 - 423*a*x + 176)*sqrt(-a^2*x^2 + 1) + 176)/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 + 2*a^3*c^3*x
^2 - 3*a^2*c^3*x + a*c^3)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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