3.1.26 \(\int e^{4 \tanh ^{-1}(a x)} x^3 \, dx\) [26]

Optimal. Leaf size=57 \[ \frac {12 x}{a^3}+\frac {4 x^2}{a^2}+\frac {4 x^3}{3 a}+\frac {x^4}{4}+\frac {4}{a^4 (1-a x)}+\frac {16 \log (1-a x)}{a^4} \]

[Out]

12*x/a^3+4*x^2/a^2+4/3*x^3/a+1/4*x^4+4/a^4/(-a*x+1)+16*ln(-a*x+1)/a^4

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Rubi [A]
time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6261, 90} \begin {gather*} \frac {4}{a^4 (1-a x)}+\frac {16 \log (1-a x)}{a^4}+\frac {12 x}{a^3}+\frac {4 x^2}{a^2}+\frac {4 x^3}{3 a}+\frac {x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*x^3,x]

[Out]

(12*x)/a^3 + (4*x^2)/a^2 + (4*x^3)/(3*a) + x^4/4 + 4/(a^4*(1 - a*x)) + (16*Log[1 - a*x])/a^4

Rule 90

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 6261

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rubi steps

\begin {align*} \int e^{4 \tanh ^{-1}(a x)} x^3 \, dx &=\int \frac {x^3 (1+a x)^2}{(1-a x)^2} \, dx\\ &=\int \left (\frac {12}{a^3}+\frac {8 x}{a^2}+\frac {4 x^2}{a}+x^3+\frac {4}{a^3 (-1+a x)^2}+\frac {16}{a^3 (-1+a x)}\right ) \, dx\\ &=\frac {12 x}{a^3}+\frac {4 x^2}{a^2}+\frac {4 x^3}{3 a}+\frac {x^4}{4}+\frac {4}{a^4 (1-a x)}+\frac {16 \log (1-a x)}{a^4}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 57, normalized size = 1.00 \begin {gather*} \frac {12 x}{a^3}+\frac {4 x^2}{a^2}+\frac {4 x^3}{3 a}+\frac {x^4}{4}+\frac {4}{a^4 (1-a x)}+\frac {16 \log (1-a x)}{a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])*x^3,x]

[Out]

(12*x)/a^3 + (4*x^2)/a^2 + (4*x^3)/(3*a) + x^4/4 + 4/(a^4*(1 - a*x)) + (16*Log[1 - a*x])/a^4

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Maple [A]
time = 0.80, size = 55, normalized size = 0.96

method	result	size
risch	\(\frac {x^{4}}{4}+\frac {4 x^{3}}{3 a}+\frac {4 x^{2}}{a^{2}}+\frac {12 x}{a^{3}}-\frac {4}{a^{4} \left (a x -1\right )}+\frac {16 \ln \left (a x -1\right )}{a^{4}}\)	\(52\)
default	\(\frac {\frac {1}{4} a^{3} x^{4}+\frac {4}{3} a^{2} x^{3}+4 a \,x^{2}+12 x}{a^{3}}-\frac {4}{a^{4} \left (a x -1\right )}+\frac {16 \ln \left (a x -1\right )}{a^{4}}\)	\(55\)
norman	\(\frac {\frac {15 x^{4}}{4}+\frac {4 a \,x^{5}}{3}-\frac {16 x}{a^{3}}+\frac {32 x^{3}}{3 a}+\frac {x^{6} a^{2}}{4}-\frac {8}{a^{4}}}{a^{2} x^{2}-1}+\frac {16 \ln \left (a x -1\right )}{a^{4}}\)	\(64\)
meijerg	\(\frac {\frac {a^{2} x^{2} \left (-2 a^{4} x^{4}-6 a^{2} x^{2}+12\right )}{-4 a^{2} x^{2}+4}+3 \ln \left (-a^{2} x^{2}+1\right )}{2 a^{4}}-\frac {2 \left (\frac {x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-70 a^{2} x^{2}+105\right )}{21 a^{6} \left (-a^{2} x^{2}+1\right )}-\frac {5 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{a^{3} \sqrt {-a^{2}}}-\frac {3 \left (-\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{3 \left (-a^{2} x^{2}+1\right )}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{a^{4}}+\frac {\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {6 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}}{a^{3} \sqrt {-a^{2}}}+\frac {\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )}{2 a^{4}}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*x^3,x,method=_RETURNVERBOSE)

[Out]

1/a^3*(1/4*a^3*x^4+4/3*a^2*x^3+4*a*x^2+12*x)-4/a^4/(a*x-1)+16/a^4*ln(a*x-1)

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Maxima [A]
time = 0.27, size = 58, normalized size = 1.02 \begin {gather*} -\frac {4}{a^{5} x - a^{4}} + \frac {3 \, a^{3} x^{4} + 16 \, a^{2} x^{3} + 48 \, a x^{2} + 144 \, x}{12 \, a^{3}} + \frac {16 \, \log \left (a x - 1\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^3,x, algorithm="maxima")

[Out]

-4/(a^5*x - a^4) + 1/12*(3*a^3*x^4 + 16*a^2*x^3 + 48*a*x^2 + 144*x)/a^3 + 16*log(a*x - 1)/a^4

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Fricas [A]
time = 0.36, size = 66, normalized size = 1.16 \begin {gather*} \frac {3 \, a^{5} x^{5} + 13 \, a^{4} x^{4} + 32 \, a^{3} x^{3} + 96 \, a^{2} x^{2} - 144 \, a x + 192 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 48}{12 \, {\left (a^{5} x - a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^3,x, algorithm="fricas")

[Out]

1/12*(3*a^5*x^5 + 13*a^4*x^4 + 32*a^3*x^3 + 96*a^2*x^2 - 144*a*x + 192*(a*x - 1)*log(a*x - 1) - 48)/(a^5*x - a
^4)

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Sympy [A]
time = 0.09, size = 49, normalized size = 0.86 \begin {gather*} \frac {x^{4}}{4} - \frac {4}{a^{5} x - a^{4}} + \frac {4 x^{3}}{3 a} + \frac {4 x^{2}}{a^{2}} + \frac {12 x}{a^{3}} + \frac {16 \log {\left (a x - 1 \right )}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*x**3,x)

[Out]

x**4/4 - 4/(a**5*x - a**4) + 4*x**3/(3*a) + 4*x**2/a**2 + 12*x/a**3 + 16*log(a*x - 1)/a**4

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Giac [A]
time = 0.42, size = 61, normalized size = 1.07 \begin {gather*} \frac {16 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{4}} - \frac {4}{{\left (a x - 1\right )} a^{4}} + \frac {3 \, a^{8} x^{4} + 16 \, a^{7} x^{3} + 48 \, a^{6} x^{2} + 144 \, a^{5} x}{12 \, a^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^3,x, algorithm="giac")

[Out]

16*log(abs(a*x - 1))/a^4 - 4/((a*x - 1)*a^4) + 1/12*(3*a^8*x^4 + 16*a^7*x^3 + 48*a^6*x^2 + 144*a^5*x)/a^8

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Mupad [B]
time = 0.81, size = 57, normalized size = 1.00 \begin {gather*} \frac {16\,\ln \left (a\,x-1\right )}{a^4}-\frac {4}{a\,\left (a^4\,x-a^3\right )}+\frac {12\,x}{a^3}+\frac {x^4}{4}+\frac {4\,x^3}{3\,a}+\frac {4\,x^2}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)

[Out]

(16*log(a*x - 1))/a^4 - 4/(a*(a^4*x - a^3)) + (12*x)/a^3 + x^4/4 + (4*x^3)/(3*a) + (4*x^2)/a^2

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