3.2.69 \(\int e^{2 \tanh ^{-1}(a x)} (c-a c x)^4 \, dx\) [169]

Optimal. Leaf size=37 \[ -\frac {c^4 (1-a x)^4}{2 a}+\frac {c^4 (1-a x)^5}{5 a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 37, 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 {c^4 (1-a x)^5}{5 a}-\frac {c^4 (1-a x)^4}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 32, normalized size = 0.86 \begin {gather*} c^4 \left (x-a x^2+\frac {a^3 x^4}{2}-\frac {a^4 x^5}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.08, size = 29, normalized size = 0.78

method	result	size
gosper	\(-\frac {\left (2 a^{4} x^{4}-5 a^{3} x^{3}+10 a x -10\right ) c^{4} x}{10}\)	\(29\)
default	\(c^{4} \left (-\frac {1}{5} a^{4} x^{5}+\frac {1}{2} a^{3} x^{4}-a \,x^{2}+x \right )\)	\(29\)
norman	\(c^{4} x -\frac {1}{5} a^{4} c^{4} x^{5}-c^{4} a \,x^{2}+\frac {1}{2} c^{4} x^{4} a^{3}\)	\(38\)
risch	\(c^{4} x -\frac {1}{5} a^{4} c^{4} x^{5}-c^{4} a \,x^{2}+\frac {1}{2} c^{4} x^{4} a^{3}\)	\(38\)
meijerg	\(-\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{2 \sqrt {-a^{2}}}+\frac {c^{4} \left (\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c^{4} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}+\frac {c^{4} \ln \left (-a^{2} x^{2}+1\right )}{a}+\frac {c^{4} \arctanh \left (a x \right )}{a}\)	\(252\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 37, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, a^{4} c^{4} x^{5} + \frac {1}{2} \, a^{3} c^{4} x^{4} - a c^{4} x^{2} + c^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 37, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, a^{4} c^{4} x^{5} + \frac {1}{2} \, a^{3} c^{4} x^{4} - a c^{4} x^{2} + c^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.02, size = 36, normalized size = 0.97 \begin {gather*} - \frac {a^{4} c^{4} x^{5}}{5} + \frac {a^{3} c^{4} x^{4}}{2} - a c^{4} x^{2} + c^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 37, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, a^{4} c^{4} x^{5} + \frac {1}{2} \, a^{3} c^{4} x^{4} - a c^{4} x^{2} + c^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 37, normalized size = 1.00 \begin {gather*} -\frac {a^4\,c^4\,x^5}{5}+\frac {a^3\,c^4\,x^4}{2}-a\,c^4\,x^2+c^4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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