3.1114 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^3 \sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=109 \[ \frac {2 a^2 (a x+1)}{\sqrt {c-a^2 c x^2}}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}} \]

[Out]

-5/2*a^2*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))/c^(1/2)+2*a^2*(a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/2*(-a^2*c*x^2+c)^(
1/2)/c/x^2-2*a*(-a^2*c*x^2+c)^(1/2)/c/x

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Rubi [A]  time = 0.32, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 a^2 (a x+1)}{\sqrt {c-a^2 c x^2}}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)}}{x^3 \sqrt {c-a^2 c x^2}} \, dx &=c \int \frac {(1+a x)^2}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\int \frac {-1-2 a x-2 a^2 x^2}{x^3 \sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}+\frac {\int \frac {4 a c+5 a^2 c x}{x^2 \sqrt {c-a^2 c x^2}} \, dx}{2 c}\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}+\frac {1}{2} \left (5 a^2\right ) \int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}+\frac {1}{4} \left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}-\frac {5 \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 )}{2 c}\\ &=\frac {2 a^2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{c x}-\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 94, normalized size = 0.86 \[ \frac {\frac {\left (-8 a^2 x^2+3 a x+1\right ) \sqrt {c-a^2 c x^2}}{x^2 (a x-1)}-5 a^2 \sqrt {c} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )+5 a^2 \sqrt {c} \log (x)}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((1 + 3*a*x - 8*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(x^2*(-1 + a*x)) + 5*a^2*Sqrt[c]*Log[x] - 5*a^2*Sqrt[c]*Log[c +
Sqrt[c]*Sqrt[c - a^2*c*x^2]])/(2*c)

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fricas [A]  time = 0.65, size = 208, normalized size = 1.91 \[ \left [\frac {5 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (8 \, a^{2} x^{2} - 3 \, a x - 1\right )}}{4 \, {\left (a c x^{3} - c x^{2}\right )}}, -\frac {5 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (8 \, a^{2} x^{2} - 3 \, a x - 1\right )}}{2 \, {\left (a c x^{3} - c x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 124, normalized size = 1.14 \[ -\frac {5 a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2 \sqrt {c}}-\frac {2 a \sqrt {-a^{2} c \,x^{2}+c}}{c x}-\frac {2 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{c \left (x -\frac {1}{a}\right )}-\frac {\sqrt {-a^{2} c \,x^{2}+c}}{2 c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*a^2/c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-2*a*(-a^2*c*x^2+c)^(1/2)/c/x-2*a/c/(x-1/a)*(-(x-1/
a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)-1/2*(-a^2*c*x^2+c)^(1/2)/c/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a x + 1\right )}^{2}}{\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (a\,x+1\right )}^2}{x^3\,\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a x}{a x^{4} \sqrt {- a^{2} c x^{2} + c} - x^{3} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a x^{4} \sqrt {- a^{2} c x^{2} + c} - x^{3} \sqrt {- a^{2} c x^{2} + c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x/(a*x**4*sqrt(-a**2*c*x**2 + c) - x**3*sqrt(-a**2*c*x**2 + c)), x) - Integral(1/(a*x**4*sqrt(-a**
2*c*x**2 + c) - x**3*sqrt(-a**2*c*x**2 + c)), x)

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