3.2.90 \(\int e^{4 \tanh ^{-1}(a x)} (c-a c x)^3 \, dx\) [190]

Optimal. Leaf size=35 \[ \frac {2 c^3 (1+a x)^3}{3 a}-\frac {c^3 (1+a x)^4}{4 a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6264, 45} \begin {gather*} \frac {2 c^3 (a x+1)^3}{3 a}-\frac {c^3 (a x+1)^4}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^3*(1 + a*x)^3)/(3*a) - (c^3*(1 + a*x)^4)/(4*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 6264

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 30, normalized size = 0.86 \begin {gather*} -\frac {1}{12} c^3 x \left (-12-6 a x+4 a^2 x^2+3 a^3 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(c^3*x*(-12 - 6*a*x + 4*a^2*x^2 + 3*a^3*x^3))

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Maple [A]
time = 1.08, size = 29, normalized size = 0.83

method	result	size
gosper	\(-\frac {\left (3 a^{3} x^{3}+4 a^{2} x^{2}-6 a x -12\right ) x \,c^{3}}{12}\)	\(29\)
default	\(c^{3} \left (-\frac {1}{4} a^{3} x^{4}-\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a \,x^{2}+x \right )\)	\(29\)
risch	\(-\frac {1}{4} a^{3} c^{3} x^{4}-\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{2} a \,c^{3} x^{2}+x \,c^{3}\)	\(38\)
norman	\(\frac {-\frac {1}{2} a \,c^{3} x^{2}-x \,c^{3}+\frac {4}{3} a^{2} c^{3} x^{3}+\frac {3}{4} a^{3} c^{3} x^{4}-\frac {1}{4} a^{5} c^{3} x^{6}-\frac {1}{3} c^{3} a^{4} x^{5}}{a^{2} x^{2}-1}\)	\(73\)
meijerg	\(-\frac {c^{3} \left (\frac {a^{2} x^{2} \left (-2 a^{4} x^{4}-6 a^{2} x^{2}+12\right )}{-4 a^{2} x^{2}+4}+3 \ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}+\frac {c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-70 a^{2} x^{2}+105\right )}{21 a^{6} \left (-a^{2} x^{2}+1\right )}-\frac {5 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{3} \left (-\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{3 \left (-a^{2} x^{2}+1\right )}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}+\frac {3 c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{3} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}+\frac {3 c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}+\frac {a \,c^{3} x^{2}}{-2 a^{2} x^{2}+2}+\frac {c^{3} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}\)	\(413\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 37, normalized size = 1.06 \begin {gather*} -\frac {1}{4} \, a^{3} c^{3} x^{4} - \frac {1}{3} \, a^{2} c^{3} x^{3} + \frac {1}{2} \, a c^{3} x^{2} + c^{3} x \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*c*x+c)^3,x, algorithm="maxima")

[Out]

-1/4*a^3*c^3*x^4 - 1/3*a^2*c^3*x^3 + 1/2*a*c^3*x^2 + c^3*x

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Fricas [A]
time = 0.34, size = 37, normalized size = 1.06 \begin {gather*} -\frac {1}{4} \, a^{3} c^{3} x^{4} - \frac {1}{3} \, a^{2} c^{3} x^{3} + \frac {1}{2} \, a c^{3} x^{2} + c^{3} x \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*c*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*a^3*c^3*x^4 - 1/3*a^2*c^3*x^3 + 1/2*a*c^3*x^2 + c^3*x

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Sympy [A]
time = 0.03, size = 37, normalized size = 1.06 \begin {gather*} - \frac {a^{3} c^{3} x^{4}}{4} - \frac {a^{2} c^{3} x^{3}}{3} + \frac {a c^{3} x^{2}}{2} + c^{3} x \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*c*x+c)**3,x)

[Out]

-a**3*c**3*x**4/4 - a**2*c**3*x**3/3 + a*c**3*x**2/2 + c**3*x

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Giac [A]
time = 0.42, size = 37, normalized size = 1.06 \begin {gather*} -\frac {1}{4} \, a^{3} c^{3} x^{4} - \frac {1}{3} \, a^{2} c^{3} x^{3} + \frac {1}{2} \, a c^{3} x^{2} + c^{3} x \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*c*x+c)^3,x, algorithm="giac")

[Out]

-1/4*a^3*c^3*x^4 - 1/3*a^2*c^3*x^3 + 1/2*a*c^3*x^2 + c^3*x

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Mupad [B]
time = 0.05, size = 37, normalized size = 1.06 \begin {gather*} -\frac {a^3\,c^3\,x^4}{4}-\frac {a^2\,c^3\,x^3}{3}+\frac {a\,c^3\,x^2}{2}+c^3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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