3.2.89 \(\int e^{4 \tanh ^{-1}(a x)} (c-a c x)^4 \, dx\) [189]

Optimal. Leaf size=32 \[ c^4 x-\frac {2}{3} a^2 c^4 x^3+\frac {1}{5} a^4 c^4 x^5 \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6264, 41, 200} \begin {gather*} \frac {1}{5} a^4 c^4 x^5-\frac {2}{3} a^2 c^4 x^3+c^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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)^4 \, dx &=c^4 \int (1-a x)^2 (1+a x)^2 \, dx\\ &=c^4 \int \left (1-a^2 x^2\right )^2 \, dx\\ &=c^4 \int \left (1-2 a^2 x^2+a^4 x^4\right ) \, dx\\ &=c^4 x-\frac {2}{3} a^2 c^4 x^3+\frac {1}{5} a^4 c^4 x^5\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 32, normalized size = 1.00 \begin {gather*} c^4 x-\frac {2}{3} a^2 c^4 x^3+\frac {1}{5} a^4 c^4 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.10, size = 23, normalized size = 0.72

method result size
default \(c^{4} \left (\frac {1}{5} a^{4} x^{5}-\frac {2}{3} a^{2} x^{3}+x \right )\) \(23\)
gosper \(\frac {\left (3 a^{4} x^{4}-10 a^{2} x^{2}+15\right ) c^{4} x}{15}\) \(25\)
risch \(c^{4} x -\frac {2}{3} a^{2} c^{4} x^{3}+\frac {1}{5} a^{4} c^{4} x^{5}\) \(29\)
norman \(\frac {-c^{4} x +\frac {5}{3} a^{2} c^{4} x^{3}-\frac {13}{15} a^{4} c^{4} x^{5}+\frac {1}{5} a^{6} c^{4} x^{7}}{a^{2} x^{2}-1}\) \(53\)
meijerg \(\frac {c^{4} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-70 a^{2} x^{2}+105\right )}{21 a^{6} \left (-a^{2} x^{2}+1\right )}-\frac {5 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{\sqrt {-a^{2}}}+\frac {3 c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+\frac {c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {9}{2}} \left (-18 a^{6} x^{6}-42 a^{4} x^{4}-210 a^{2} x^{2}+315\right )}{45 a^{8} \left (-a^{2} x^{2}+1\right )}-\frac {7 \left (-a^{2}\right )^{\frac {9}{2}} \arctanh \left (a x \right )}{a^{9}}\right )}{2 \sqrt {-a^{2}}}\) \(321\)

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)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 28, normalized size = 0.88 \begin {gather*} \frac {1}{5} \, a^{4} c^{4} x^{5} - \frac {2}{3} \, a^{2} c^{4} x^{3} + c^{4} 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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 28, normalized size = 0.88 \begin {gather*} \frac {1}{5} \, a^{4} c^{4} x^{5} - \frac {2}{3} \, a^{2} c^{4} x^{3} + c^{4} 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)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.03, size = 29, normalized size = 0.91 \begin {gather*} \frac {a^{4} c^{4} x^{5}}{5} - \frac {2 a^{2} c^{4} x^{3}}{3} + c^{4} 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)**4,x)

[Out]

a**4*c**4*x**5/5 - 2*a**2*c**4*x**3/3 + c**4*x

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Giac [A]
time = 0.43, size = 28, normalized size = 0.88 \begin {gather*} \frac {1}{5} \, a^{4} c^{4} x^{5} - \frac {2}{3} \, a^{2} c^{4} x^{3} + c^{4} 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)^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.80, size = 24, normalized size = 0.75 \begin {gather*} \frac {c^4\,x\,\left (3\,a^4\,x^4-10\,a^2\,x^2+15\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^4*x*(3*a^4*x^4 - 10*a^2*x^2 + 15))/15

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