3.5.95 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^3 \, dx\) [495]

Optimal. Leaf size=55 \[ \frac {c^3}{2 a^3 x^2}-\frac {5 c^3}{a^2 x}-c^3 x-\frac {11 c^3 \log (x)}{a}+\frac {16 c^3 \log (1+a x)}{a} \]

[Out]

1/2*c^3/a^3/x^2-5*c^3/a^2/x-c^3*x-11*c^3*ln(x)/a+16*c^3*ln(a*x+1)/a

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Rubi [A]
time = 0.08, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6266, 6264, 90} \begin {gather*} \frac {c^3}{2 a^3 x^2}-\frac {5 c^3}{a^2 x}-\frac {11 c^3 \log (x)}{a}+\frac {16 c^3 \log (a x+1)}{a}+c^3 (-x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

c^3/(2*a^3*x^2) - (5*c^3)/(a^2*x) - c^3*x - (11*c^3*Log[x])/a + (16*c^3*Log[1 + a*x])/a

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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])

Rule 6266

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*(1 + c*(x/d))^
p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx &=-\frac {c^3 \int \frac {e^{-2 \tanh ^{-1}(a x)} (1-a x)^3}{x^3} \, dx}{a^3}\\ &=-\frac {c^3 \int \frac {(1-a x)^4}{x^3 (1+a x)} \, dx}{a^3}\\ &=-\frac {c^3 \int \left (a^3+\frac {1}{x^3}-\frac {5 a}{x^2}+\frac {11 a^2}{x}-\frac {16 a^3}{1+a x}\right ) \, dx}{a^3}\\ &=\frac {c^3}{2 a^3 x^2}-\frac {5 c^3}{a^2 x}-c^3 x-\frac {11 c^3 \log (x)}{a}+\frac {16 c^3 \log (1+a x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 57, normalized size = 1.04 \begin {gather*} \frac {c^3}{2 a^3 x^2}-\frac {5 c^3}{a^2 x}-c^3 x-\frac {11 c^3 \log (a x)}{a}+\frac {16 c^3 \log (1+a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

c^3/(2*a^3*x^2) - (5*c^3)/(a^2*x) - c^3*x - (11*c^3*Log[a*x])/a + (16*c^3*Log[1 + a*x])/a

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Maple [A]
time = 1.11, size = 44, normalized size = 0.80

method	result	size
default	\(\frac {c^{3} \left (-a^{3} x +16 a^{2} \ln \left (a x +1\right )+\frac {1}{2 x^{2}}-\frac {5 a}{x}-11 a^{2} \ln \left (x \right )\right )}{a^{3}}\)	\(44\)
risch	\(-x \,c^{3}+\frac {-5 a \,c^{3} x +\frac {1}{2} c^{3}}{a^{3} x^{2}}-\frac {11 c^{3} \ln \left (x \right )}{a}+\frac {16 c^{3} \ln \left (-a x -1\right )}{a}\)	\(53\)
norman	\(\frac {4 a^{2} c^{3} x^{3}+\frac {c^{3}}{2 a}-\frac {9 x \,c^{3}}{2}-a^{3} c^{3} x^{4}}{a^{2} x^{2} \left (a x +1\right )}-\frac {11 c^{3} \ln \left (x \right )}{a}+\frac {16 c^{3} \ln \left (a x +1\right )}{a}\)	\(77\)
meijerg	\(-\frac {c^{3} \left (\frac {a x \left (3 a x +6\right )}{3 a x +3}-2 \ln \left (a x +1\right )\right )}{a}-\frac {2 c^{3} x}{a x +1}+\frac {3 c^{3} \left (-\frac {a x}{a x +1}+\ln \left (a x +1\right )\right )}{a}-\frac {2 c^{3} \left (-\frac {2 a x}{2 a x +2}-\ln \left (a x +1\right )+1+\ln \left (x \right )+\ln \left (a \right )\right )}{a}+\frac {3 c^{3} \left (\frac {3 a x}{3 a x +3}+2 \ln \left (a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (a \right )-\frac {1}{a x}\right )}{a}-\frac {c^{3} \left (-\frac {4 a x}{4 a x +4}-3 \ln \left (a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {2}{a x}\right )}{a}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3/a^3*(-a^3*x+16*a^2*ln(a*x+1)+1/2/x^2-5*a/x-11*a^2*ln(x))

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Maxima [A]
time = 0.26, size = 52, normalized size = 0.95 \begin {gather*} -c^{3} x + \frac {16 \, c^{3} \log \left (a x + 1\right )}{a} - \frac {11 \, c^{3} \log \left (x\right )}{a} - \frac {10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^3*x + 16*c^3*log(a*x + 1)/a - 11*c^3*log(x)/a - 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)

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Fricas [A]
time = 0.34, size = 62, normalized size = 1.13 \begin {gather*} -\frac {2 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} \log \left (a x + 1\right ) + 22 \, a^{2} c^{3} x^{2} \log \left (x\right ) + 10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3*c^3*x^3 - 32*a^2*c^3*x^2*log(a*x + 1) + 22*a^2*c^3*x^2*log(x) + 10*a*c^3*x - c^3)/(a^3*x^2)

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Sympy [A]
time = 0.22, size = 44, normalized size = 0.80 \begin {gather*} - c^{3} x - \frac {c^{3} \cdot \left (11 \log {\left (x \right )} - 16 \log {\left (x + \frac {1}{a} \right )}\right )}{a} - \frac {10 a c^{3} x - c^{3}}{2 a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**3*x - c**3*(11*log(x) - 16*log(x + 1/a))/a - (10*a*c**3*x - c**3)/(2*a**3*x**2)

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Giac [A]
time = 0.40, size = 100, normalized size = 1.82 \begin {gather*} -\frac {5 \, c^{3} \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {11 \, c^{3} \log \left ({\left | -\frac {1}{a x + 1} + 1 \right |}\right )}{a} - \frac {{\left (2 \, c^{3} + \frac {7 \, c^{3}}{a x + 1} - \frac {10 \, c^{3}}{{\left (a x + 1\right )}^{2}}\right )} {\left (a x + 1\right )}}{2 \, a {\left (\frac {1}{a x + 1} - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*c^3*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a - 11*c^3*log(abs(-1/(a*x + 1) + 1))/a - 1/2*(2*c^3 + 7*c^3/(a*
x + 1) - 10*c^3/(a*x + 1)^2)*(a*x + 1)/(a*(1/(a*x + 1) - 1)^2)

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Mupad [B]
time = 0.88, size = 51, normalized size = 0.93 \begin {gather*} \frac {\frac {c^3}{2}-5\,a\,c^3\,x}{a^3\,x^2}-c^3\,x-\frac {11\,c^3\,\ln \left (x\right )}{a}+\frac {16\,c^3\,\ln \left (a\,x+1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3/2 - 5*a*c^3*x)/(a^3*x^2) - c^3*x - (11*c^3*log(x))/a + (16*c^3*log(a*x + 1))/a

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