3.11.32 \(\int e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx\) [1032]

Optimal. Leaf size=35 \[ \frac {c^2 (1+a x)^4}{2 a}-\frac {c^2 (1+a x)^5}{5 a} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6275, 45} \begin {gather*} \frac {c^2 (a x+1)^4}{2 a}-\frac {c^2 (a x+1)^5}{5 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 23, normalized size = 0.66 \begin {gather*} -\frac {c^2 (1+a x)^4 (-3+2 a x)}{10 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*(c^2*(1 + a*x)^4*(-3 + 2*a*x))/a

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Maple [A]
time = 0.07, size = 28, normalized size = 0.80

method	result	size
default	\(c^{2} \left (-\frac {1}{5} a^{4} x^{5}-\frac {1}{2} a^{3} x^{4}+x^{2} a +x \right )\)	\(28\)
gosper	\(-\frac {c^{2} x \left (2 a^{4} x^{4}+5 a^{3} x^{3}-10 a x -10\right )}{10}\)	\(29\)
norman	\(c^{2} x +a \,c^{2} x^{2}-\frac {1}{2} a^{3} c^{2} x^{4}-\frac {1}{5} a^{4} c^{2} x^{5}\)	\(37\)
risch	\(c^{2} x +a \,c^{2} x^{2}-\frac {1}{2} a^{3} c^{2} x^{4}-\frac {1}{5} a^{4} c^{2} x^{5}\)	\(37\)
meijerg	\(-\frac {c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{2 \sqrt {-a^{2}}}-\frac {c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-\frac {c^{2} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {2 c^{2} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {c^{2} \ln \left (-a^{2} x^{2}+1\right )}{a}+\frac {c^{2} \arctanh \left (a x \right )}{a}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 36, normalized size = 1.03 \begin {gather*} -\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} + c^{2} 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^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 36, normalized size = 1.03 \begin {gather*} -\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} + c^{2} 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^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.02, size = 36, normalized size = 1.03 \begin {gather*} - \frac {a^{4} c^{2} x^{5}}{5} - \frac {a^{3} c^{2} x^{4}}{2} + a c^{2} x^{2} + c^{2} 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**2*c*x**2+c)**2,x)

[Out]

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

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Giac [A]
time = 0.40, size = 36, normalized size = 1.03 \begin {gather*} -\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} + c^{2} 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^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 36, normalized size = 1.03 \begin {gather*} -\frac {a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}+a\,c^2\,x^2+c^2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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