3.12.86 \(\int e^{4 \tanh ^{-1}(a x)} (c-a^2 c x^2) \, dx\) [1186]

Optimal. Leaf size=36 \[ -7 c x-2 a c x^2-\frac {1}{3} a^2 c x^3-\frac {8 c \log (1-a x)}{a} \]

[Out]

-7*c*x-2*a*c*x^2-1/3*a^2*c*x^3-8*c*ln(-a*x+1)/a

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.28, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6275, 45} \begin {gather*} -\frac {c (a x+1)^3}{3 a}-\frac {c (a x+1)^2}{a}-\frac {8 c \log (1-a x)}{a}-4 c x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*c*x - (c*(1 + a*x)^2)/a - (c*(1 + a*x)^3)/(3*a) - (8*c*Log[1 - a*x])/a

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6275

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 36, normalized size = 1.00 \begin {gather*} -7 c x-2 a c x^2-\frac {1}{3} a^2 c x^3-\frac {8 c \log (1-a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-7*c*x - 2*a*c*x^2 - (a^2*c*x^3)/3 - (8*c*Log[1 - a*x])/a

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Maple [A]
time = 0.06, size = 32, normalized size = 0.89

method	result	size
default	\(c \left (-\frac {a^{2} x^{3}}{3}-2 x^{2} a -7 x -\frac {8 \ln \left (a x -1\right )}{a}\right )\)	\(32\)
risch	\(-\frac {a^{2} c \,x^{3}}{3}-2 a c \,x^{2}-7 c x -\frac {8 c \ln \left (a x -1\right )}{a}\)	\(34\)
norman	\(\frac {2 a c \,x^{2}+7 c x -\frac {20}{3} a^{2} c \,x^{3}-2 a^{3} c \,x^{4}-\frac {1}{3} a^{4} c \,x^{5}}{a^{2} x^{2}-1}-\frac {8 c \ln \left (a x -1\right )}{a}\)	\(65\)
meijerg	\(\frac {c \left (\frac {x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-70 a^{2} x^{2}+105\right )}{21 a^{6} \left (-a^{2} x^{2}+1\right )}-\frac {5 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{2 \sqrt {-a^{2}}}-\frac {5 c \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c \left (-\frac {x^{2} a^{2} \left (-3 a^{2} x^{2}+6\right )}{3 \left (-a^{2} x^{2}+1\right )}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {5 c \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 a c \,x^{2}}{-a^{2} x^{2}+1}+\frac {c \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(-1/3*a^2*x^3-2*x^2*a-7*x-8/a*ln(a*x-1))

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Maxima [A]
time = 0.26, size = 33, normalized size = 0.92 \begin {gather*} -\frac {1}{3} \, a^{2} c x^{3} - 2 \, a c x^{2} - 7 \, c x - \frac {8 \, c \log \left (a x - 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3 - 2*a*c*x^2 - 7*c*x - 8*c*log(a*x - 1)/a

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Fricas [A]
time = 0.36, size = 37, normalized size = 1.03 \begin {gather*} -\frac {a^{3} c x^{3} + 6 \, a^{2} c x^{2} + 21 \, a c x + 24 \, c \log \left (a x - 1\right )}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^3*c*x^3 + 6*a^2*c*x^2 + 21*a*c*x + 24*c*log(a*x - 1))/a

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Sympy [A]
time = 0.07, size = 36, normalized size = 1.00 \begin {gather*} - \frac {a^{2} c x^{3}}{3} - 2 a c x^{2} - 7 c x - \frac {8 c \log {\left (a x - 1 \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c*x**3/3 - 2*a*c*x**2 - 7*c*x - 8*c*log(a*x - 1)/a

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Giac [A]
time = 0.41, size = 44, normalized size = 1.22 \begin {gather*} -\frac {8 \, c \log \left ({\left | a x - 1 \right |}\right )}{a} - \frac {a^{5} c x^{3} + 6 \, a^{4} c x^{2} + 21 \, a^{3} c x}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*c*log(abs(a*x - 1))/a - 1/3*(a^5*c*x^3 + 6*a^4*c*x^2 + 21*a^3*c*x)/a^3

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Mupad [B]
time = 0.05, size = 33, normalized size = 0.92 \begin {gather*} -7\,c\,x-\frac {a^2\,c\,x^3}{3}-\frac {8\,c\,\ln \left (a\,x-1\right )}{a}-2\,a\,c\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 7*c*x - (a^2*c*x^3)/3 - (8*c*log(a*x - 1))/a - 2*a*c*x^2

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