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

Optimal. Leaf size=176 \[ \frac {\sqrt {1-a^2 x^2}}{2 a^4 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 a^4 c \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c \sqrt {c-a^2 c x^2}} \]

[Out]

x*(-a^2*x^2+1)^(1/2)/a^3/c/(-a^2*c*x^2+c)^(1/2)+1/2*(-a^2*x^2+1)^(1/2)/a^4/c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+5/4
*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/a^4/c/(-a^2*c*x^2+c)^(1/2)-1/4*ln(a*x+1)*(-a^2*x^2+1)^(1/2)/a^4/c/(-a^2*c*x^2+c
)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 88} \[ \frac {x \sqrt {1-a^2 x^2}}{a^3 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 a^4 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 a^4 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 - a^2*x^2])/(a^3*c*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(2*a^4*c*(1 - a*x)*Sqrt[c - a^2*c*x^2])
+ (5*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(4*a^4*c*Sqrt[c - a^2*c*x^2]) - (Sqrt[1 - a^2*x^2]*Log[1 + a*x])/(4*a^4*c
*Sqrt[c - a^2*c*x^2])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

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

Rule 6153

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 71, normalized size = 0.40 \[ \frac {\sqrt {1-a^2 x^2} \left (4 a x+\frac {2}{1-a x}+5 \log (1-a x)-\log (a x+1)\right )}{4 a^4 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(4*a*x + 2/(1 - a*x) + 5*Log[1 - a*x] - Log[1 + a*x]))/(4*a^4*c*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{3}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^3/(a^5*c^2*x^5 - a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 2*a^2*c^2*x
^2 + a*c^2*x - c^2), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.05, size = 101, normalized size = 0.57 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (4 a^{2} x^{2}+5 \ln \left (a x -1\right ) x a -a x \ln \left (a x +1\right )-4 a x -5 \ln \left (a x -1\right )+\ln \left (a x +1\right )-2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a^{4} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(4*a^2*x^2+5*ln(a*x-1)*x*a-a*x*ln(a*x+1)-4*a*x-5*ln(a*x-1)+ln(a
*x+1)-2)/(a^2*x^2-1)/c^2/a^4/(a*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a \int -\frac {x^{4}}{{\left (a^{2} c^{\frac {3}{2}} x^{2} - c^{\frac {3}{2}}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}\,{d x} - \frac {1}{2 \, {\left (a^{6} c^{\frac {3}{2}} x^{2} - a^{4} c^{\frac {3}{2}}\right )}} + \frac {\log \left (-a^{2} c x^{2} + c\right )}{2 \, a^{4} c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*integrate(-x^4/((a^2*c^(3/2)*x^2 - c^(3/2))*(a*x + 1)*(a*x - 1)), x) - 1/2/(a^6*c^(3/2)*x^2 - a^4*c^(3/2))
+ 1/2*log(-a^2*c*x^2 + c)/(a^4*c^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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