3.1127 \(\int e^{2 \tanh ^{-1}(a x)} x^m (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=42 \[ \frac {a^2 c x^{m+3}}{m+3}+\frac {2 a c x^{m+2}}{m+2}+\frac {c x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 42, 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 = {6150, 43} \[ \frac {a^2 c x^{m+3}}{m+3}+\frac {2 a c x^{m+2}}{m+2}+\frac {c x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 6150

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

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Mathematica [A]  time = 0.06, size = 34, normalized size = 0.81 \[ c x^{m+1} \left (\frac {a^2 x^2}{m+3}+\frac {2 a x}{m+2}+\frac {1}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 82, normalized size = 1.95 \[ \frac {{\left ({\left (a^{2} c m^{2} + 3 \, a^{2} c m + 2 \, a^{2} c\right )} x^{3} + 2 \, {\left (a c m^{2} + 4 \, a c m + 3 \, a c\right )} x^{2} + {\left (c m^{2} + 5 \, c m + 6 \, c\right )} x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*c*m^2 + 3*a^2*c*m + 2*a^2*c)*x^3 + 2*(a*c*m^2 + 4*a*c*m + 3*a*c)*x^2 + (c*m^2 + 5*c*m + 6*c)*x)*x^m/(m^3
+ 6*m^2 + 11*m + 6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} c x^{2} - c\right )} {\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 74, normalized size = 1.76 \[ \frac {c \,x^{1+m} \left (a^{2} m^{2} x^{2}+3 a^{2} m \,x^{2}+2 a^{2} x^{2}+2 a \,m^{2} x +8 a m x +6 a x +m^{2}+5 m +6\right )}{\left (3+m \right ) \left (2+m \right ) \left (1+m \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x^(1+m)*(a^2*m^2*x^2+3*a^2*m*x^2+2*a^2*x^2+2*a*m^2*x+8*a*m*x+6*a*x+m^2+5*m+6)/(3+m)/(2+m)/(1+m)

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maxima [A]  time = 0.38, size = 62, normalized size = 1.48 \[ \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{2} c x^{3} + 2 \, {\left (m^{2} + 4 \, m + 3\right )} a c x^{2} + {\left (m^{2} + 5 \, m + 6\right )} c x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((m^2 + 3*m + 2)*a^2*c*x^3 + 2*(m^2 + 4*m + 3)*a*c*x^2 + (m^2 + 5*m + 6)*c*x)*x^m/(m^3 + 6*m^2 + 11*m + 6)

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mupad [B]  time = 0.94, size = 92, normalized size = 2.19 \[ x^m\,\left (\frac {c\,x\,\left (m^2+5\,m+6\right )}{m^3+6\,m^2+11\,m+6}+\frac {2\,a\,c\,x^2\,\left (m^2+4\,m+3\right )}{m^3+6\,m^2+11\,m+6}+\frac {a^2\,c\,x^3\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^m*((c*x*(5*m + m^2 + 6))/(11*m + 6*m^2 + m^3 + 6) + (2*a*c*x^2*(4*m + m^2 + 3))/(11*m + 6*m^2 + m^3 + 6) + (
a^2*c*x^3*(3*m + m^2 + 2))/(11*m + 6*m^2 + m^3 + 6))

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sympy [A]  time = 1.10, size = 299, normalized size = 7.12 \[ \begin {cases} a^{2} c \log {\relax (x )} - \frac {2 a c}{x} - \frac {c}{2 x^{2}} & \text {for}\: m = -3 \\a^{2} c x + 2 a c \log {\relax (x )} - \frac {c}{x} & \text {for}\: m = -2 \\\frac {a^{2} c x^{2}}{2} + 2 a c x + c \log {\relax (x )} & \text {for}\: m = -1 \\\frac {a^{2} c m^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {3 a^{2} c m x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {2 a^{2} c x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {2 a c m^{2} x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {8 a c m x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 a c x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {c m^{2} x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {5 c m x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 c x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c*log(x) - 2*a*c/x - c/(2*x**2), Eq(m, -3)), (a**2*c*x + 2*a*c*log(x) - c/x, Eq(m, -2)), (a**2
*c*x**2/2 + 2*a*c*x + c*log(x), Eq(m, -1)), (a**2*c*m**2*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*a**2*c*m*x**
3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*a**2*c*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*a*c*m**2*x**2*x**m/(m**3
+ 6*m**2 + 11*m + 6) + 8*a*c*m*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 6*a*c*x**2*x**m/(m**3 + 6*m**2 + 11*m +
6) + c*m**2*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 5*c*m*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 6*c*x*x**m/(m**3 +
6*m**2 + 11*m + 6), True))

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