3.11.20 \(\int e^{2 \tanh ^{-1}(a x)} x^3 (c-a^2 c x^2) \, dx\) [1020]

Optimal. Leaf size=29 \[ \frac {c x^4}{4}+\frac {2}{5} a c x^5+\frac {1}{6} a^2 c x^6 \]

[Out]

1/4*c*x^4+2/5*a*c*x^5+1/6*a^2*c*x^6

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Rubi [A]
time = 0.04, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6285, 45} \begin {gather*} \frac {1}{6} a^2 c x^6+\frac {2}{5} a c x^5+\frac {c x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x^3*(c - a^2*c*x^2),x]

[Out]

(c*x^4)/4 + (2*a*c*x^5)/5 + (a^2*c*x^6)/6

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 6285

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, 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)} x^3 \left (c-a^2 c x^2\right ) \, dx &=c \int x^3 (1+a x)^2 \, dx\\ &=c \int \left (x^3+2 a x^4+a^2 x^5\right ) \, dx\\ &=\frac {c x^4}{4}+\frac {2}{5} a c x^5+\frac {1}{6} a^2 c x^6\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 0.76 \begin {gather*} \frac {1}{60} c x^4 \left (15+24 a x+10 a^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*x^3*(c - a^2*c*x^2),x]

[Out]

(c*x^4*(15 + 24*a*x + 10*a^2*x^2))/60

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Maple [A]
time = 0.06, size = 23, normalized size = 0.79

method	result	size
gosper	\(\frac {c \,x^{4} \left (10 a^{2} x^{2}+24 a x +15\right )}{60}\)	\(21\)
default	\(c \left (\frac {1}{6} a^{2} x^{6}+\frac {2}{5} a \,x^{5}+\frac {1}{4} x^{4}\right )\)	\(23\)
norman	\(\frac {1}{4} c \,x^{4}+\frac {2}{5} a c \,x^{5}+\frac {1}{6} a^{2} c \,x^{6}\)	\(24\)
risch	\(\frac {1}{4} c \,x^{4}+\frac {2}{5} a c \,x^{5}+\frac {1}{6} a^{2} c \,x^{6}\)	\(24\)
meijerg	\(-\frac {c \left (-\frac {x^{2} a^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{12}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a^{4}}+\frac {c \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \arctanh \left (a x \right )}{a^{7}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {c \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {c \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a^{4}}\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(1/6*a^2*x^6+2/5*a*x^5+1/4*x^4)

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Maxima [A]
time = 0.25, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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Fricas [A]
time = 0.35, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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Sympy [A]
time = 0.02, size = 26, normalized size = 0.90 \begin {gather*} \frac {a^{2} c x^{6}}{6} + \frac {2 a c x^{5}}{5} + \frac {c x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*x**3*(-a**2*c*x**2+c),x)

[Out]

a**2*c*x**6/6 + 2*a*c*x**5/5 + c*x**4/4

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Giac [A]
time = 0.44, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, a^{2} c x^{6} + \frac {2}{5} \, a c x^{5} + \frac {1}{4} \, c x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

1/6*a^2*c*x^6 + 2/5*a*c*x^5 + 1/4*c*x^4

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Mupad [B]
time = 0.04, size = 20, normalized size = 0.69 \begin {gather*} \frac {c\,x^4\,\left (10\,a^2\,x^2+24\,a\,x+15\right )}{60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^4*(24*a*x + 10*a^2*x^2 + 15))/60

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