3.1115 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x^4 \sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{2 a^3 (a x+1)}{\sqrt{c-a^2 c x^2}}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )}{\sqrt{c}} \]

[Out]

(2*a^3*(1 + a*x))/Sqrt[c - a^2*c*x^2] - Sqrt[c - a^2*c*x^2]/(3*c*x^3) - (a*Sqrt[c - a^2*c*x^2])/(c*x^2) - (8*a
^2*Sqrt[c - a^2*c*x^2])/(3*c*x) - (3*a^3*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c]

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Rubi [A]  time = 0.406531, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {6151, 1805, 1807, 807, 266, 63, 208} \[ \frac{2 a^3 (a x+1)}{\sqrt{c-a^2 c x^2}}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/(x^4*Sqrt[c - a^2*c*x^2]),x]

[Out]

(2*a^3*(1 + a*x))/Sqrt[c - a^2*c*x^2] - Sqrt[c - a^2*c*x^2]/(3*c*x^3) - (a*Sqrt[c - a^2*c*x^2])/(c*x^2) - (8*a
^2*Sqrt[c - a^2*c*x^2])/(3*c*x) - (3*a^3*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c]

Rule 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[x^m*(c
+ d*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] ||
GtQ[c, 0]) && IGtQ[n/2, 0]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*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[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 \tanh ^{-1}(a x)}}{x^4 \sqrt{c-a^2 c x^2}} \, dx &=c \int \frac{(1+a x)^2}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\int \frac{-1-2 a x-2 a^2 x^2-2 a^3 x^3}{x^4 \sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}+\frac{\int \frac{6 a c+8 a^2 c x+6 a^3 c x^2}{x^3 \sqrt{c-a^2 c x^2}} \, dx}{3 c}\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{\int \frac{-16 a^2 c^2-18 a^3 c^2 x}{x^2 \sqrt{c-a^2 c x^2}} \, dx}{6 c^2}\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}+\left (3 a^3\right ) \int \frac{1}{x \sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}+\frac{1}{2} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-a^2 c x}} \, dx,x,x^2\right )\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2 c}} \, dx,x,\sqrt{c-a^2 c x^2}\right )}{c}\\ &=\frac{2 a^3 (1+a x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{3 c x^3}-\frac{a \sqrt{c-a^2 c x^2}}{c x^2}-\frac{8 a^2 \sqrt{c-a^2 c x^2}}{3 c x}-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )}{\sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.165517, size = 101, normalized size = 0.75 \[ \frac{\left (-14 a^3 x^3+5 a^2 x^2+2 a x+1\right ) \sqrt{c-a^2 c x^2}}{3 c x^3 (a x-1)}-\frac{3 a^3 \log \left (\sqrt{c} \sqrt{c-a^2 c x^2}+c\right )}{\sqrt{c}}+\frac{3 a^3 \log (x)}{\sqrt{c}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcTanh[a*x])/(x^4*Sqrt[c - a^2*c*x^2]),x]

[Out]

(Sqrt[c - a^2*c*x^2]*(1 + 2*a*x + 5*a^2*x^2 - 14*a^3*x^3))/(3*c*x^3*(-1 + a*x)) + (3*a^3*Log[x])/Sqrt[c] - (3*
a^3*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]])/Sqrt[c]

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Maple [A]  time = 0.045, size = 150, normalized size = 1.1 \begin{align*} -{\frac{8\,{a}^{2}}{3\,cx}\sqrt{-{a}^{2}c{x}^{2}+c}}-3\,{\frac{{a}^{3}}{\sqrt{c}}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{-{a}^{2}c{x}^{2}+c}}{x}} \right ) }-2\,{\frac{{a}^{2}}{c}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}-{\frac{a}{c{x}^{2}}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{1}{3\,c{x}^{3}}\sqrt{-{a}^{2}c{x}^{2}+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^(1/2),x)

[Out]

-8/3*a^2*(-a^2*c*x^2+c)^(1/2)/c/x-3*a^3/c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-2*a^2/c/(x-1/a)*(-c
*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)-a*(-a^2*c*x^2+c)^(1/2)/c/x^2-1/3*(-a^2*c*x^2+c)^(1/2)/c/x^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a x + 1\right )}^{2}}{\sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} - 1\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

-integrate((a*x + 1)^2/(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 1)*x^4), x)

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Fricas [A]  time = 2.86264, size = 477, normalized size = 3.53 \begin{align*} \left [\frac{9 \,{\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt{c} \log \left (-\frac{a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{c} - 2 \, c}{x^{2}}\right ) - 2 \,{\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt{-a^{2} c x^{2} + c}}{6 \,{\left (a c x^{4} - c x^{3}\right )}}, -\frac{9 \,{\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) +{\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt{-a^{2} c x^{2} + c}}{3 \,{\left (a c x^{4} - c x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(9*(a^4*x^4 - a^3*x^3)*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 2*(14*a^3*x
^3 - 5*a^2*x^2 - 2*a*x - 1)*sqrt(-a^2*c*x^2 + c))/(a*c*x^4 - c*x^3), -1/3*(9*(a^4*x^4 - a^3*x^3)*sqrt(-c)*arct
an(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) + (14*a^3*x^3 - 5*a^2*x^2 - 2*a*x - 1)*sqrt(-a^2*c*x^2 + c))
/(a*c*x^4 - c*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{a x}{a x^{5} \sqrt{- a^{2} c x^{2} + c} - x^{4} \sqrt{- a^{2} c x^{2} + c}}\, dx - \int \frac{1}{a x^{5} \sqrt{- a^{2} c x^{2} + c} - x^{4} \sqrt{- a^{2} c x^{2} + c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)/x**4/(-a**2*c*x**2+c)**(1/2),x)

[Out]

-Integral(a*x/(a*x**5*sqrt(-a**2*c*x**2 + c) - x**4*sqrt(-a**2*c*x**2 + c)), x) - Integral(1/(a*x**5*sqrt(-a**
2*c*x**2 + c) - x**4*sqrt(-a**2*c*x**2 + c)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

undef