3.6.68 \(\int \frac {a^x b^x}{x^3} \, dx\) [568]

Optimal. Leaf size=51 \[ -\frac {a^x b^x}{2 x^2}-\frac {a^x b^x (\log (a)+\log (b))}{2 x}+\frac {1}{2} \text {Ei}(x (\log (a)+\log (b))) (\log (a)+\log (b))^2 \]

[Out]

-1/2*a^x*b^x/x^2-1/2*a^x*b^x*(ln(a)+ln(b))/x+1/2*Ei(x*(ln(a)+ln(b)))*(ln(a)+ln(b))^2

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2325, 2208, 2209} \begin {gather*} -\frac {a^x b^x}{2 x^2}-\frac {a^x b^x (\log (a)+\log (b))}{2 x}+\frac {1}{2} (\log (a)+\log (b))^2 \text {Ei}(x (\log (a)+\log (b))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^x*b^x)/x^3,x]

[Out]

-1/2*(a^x*b^x)/x^2 - (a^x*b^x*(Log[a] + Log[b]))/(2*x) + (ExpIntegralEi[x*(Log[a] + Log[b])]*(Log[a] + Log[b])
^2)/2

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {align*} \int \frac {a^x b^x}{x^3} \, dx &=\int \frac {e^{x (\log (a)+\log (b))}}{x^3} \, dx\\ &=-\frac {a^x b^x}{2 x^2}-\frac {1}{2} (-\log (a)-\log (b)) \int \frac {e^{x (\log (a)+\log (b))}}{x^2} \, dx\\ &=-\frac {a^x b^x}{2 x^2}-\frac {a^x b^x (\log (a)+\log (b))}{2 x}+\frac {1}{2} (\log (a)+\log (b))^2 \int \frac {e^{x (\log (a)+\log (b))}}{x} \, dx\\ &=-\frac {a^x b^x}{2 x^2}-\frac {a^x b^x (\log (a)+\log (b))}{2 x}+\frac {1}{2} \text {Ei}(x (\log (a)+\log (b))) (\log (a)+\log (b))^2\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^x b^x}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a^x*b^x)/x^3,x]

[Out]

Integrate[(a^x*b^x)/x^3, x]

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.03, size = 225, normalized size = 4.41

method	result	size
meijerg	\(\ln \left (b \right )^{2} \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )^{2} \left (\frac {9 x^{2} \ln \left (b \right )^{2} \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )^{2}+12 x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )+6}{12 x^{2} \ln \left (b \right )^{2} \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )^{2}}-\frac {\left (3+3 x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )\right ) {\mathrm e}^{x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )}}{6 x^{2} \ln \left (b \right )^{2} \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )^{2}}-\frac {\ln \left (-x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )\right )}{2}-\frac {\expIntegral \left (1, -x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )\right )}{2}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {i \pi }{2}+\frac {\ln \left (\ln \left (b \right )\right )}{2}+\frac {\ln \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )}{2}-\frac {1}{2 x^{2} \ln \left (b \right )^{2} \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )^{2}}-\frac {1}{x \ln \left (b \right ) \left (1+\frac {\ln \left (a \right )}{\ln \left (b \right )}\right )}\right )\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x*b^x/x^3,x,method=_RETURNVERBOSE)

[Out]

ln(b)^2*(1+ln(a)/ln(b))^2*(1/12/x^2/ln(b)^2/(1+ln(a)/ln(b))^2*(9*x^2*ln(b)^2*(1+ln(a)/ln(b))^2+12*x*ln(b)*(1+l
n(a)/ln(b))+6)-1/6/x^2/ln(b)^2/(1+ln(a)/ln(b))^2*(3+3*x*ln(b)*(1+ln(a)/ln(b)))*exp(x*ln(b)*(1+ln(a)/ln(b)))-1/
2*ln(-x*ln(b)*(1+ln(a)/ln(b)))-1/2*Ei(1,-x*ln(b)*(1+ln(a)/ln(b)))-3/4+1/2*ln(x)+1/2*I*Pi+1/2*ln(ln(b))+1/2*ln(
1+ln(a)/ln(b))-1/2/x^2/ln(b)^2/(1+ln(a)/ln(b))^2-1/x/ln(b)/(1+ln(a)/ln(b)))

________________________________________________________________________________________

Maxima [A]
time = 0.32, size = 19, normalized size = 0.37 \begin {gather*} -{\left (\log \left (a\right ) + \log \left (b\right )\right )}^{2} \Gamma \left (-2, -x {\left (\log \left (a\right ) + \log \left (b\right )\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x/x^3,x, algorithm="maxima")

[Out]

-(log(a) + log(b))^2*gamma(-2, -x*(log(a) + log(b)))

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 61, normalized size = 1.20 \begin {gather*} -\frac {{\left (x \log \left (a\right ) + x \log \left (b\right ) + 1\right )} a^{x} b^{x} - {\left (x^{2} \log \left (a\right )^{2} + 2 \, x^{2} \log \left (a\right ) \log \left (b\right ) + x^{2} \log \left (b\right )^{2}\right )} {\rm Ei}\left (x \log \left (a\right ) + x \log \left (b\right )\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x/x^3,x, algorithm="fricas")

[Out]

-1/2*((x*log(a) + x*log(b) + 1)*a^x*b^x - (x^2*log(a)^2 + 2*x^2*log(a)*log(b) + x^2*log(b)^2)*Ei(x*log(a) + x*
log(b)))/x^2

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{x} b^{x}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a**x*b**x/x**3,x)

[Out]

Integral(a**x*b**x/x**3, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x/x^3,x, algorithm="giac")

[Out]

integrate(a^x*b^x/x^3, x)

________________________________________________________________________________________

Mupad [B]
time = 0.05, size = 59, normalized size = 1.16 \begin {gather*} -\frac {\mathrm {expint}\left (-x\,\left (\ln \left (a\right )+\ln \left (b\right )\right )\right )\,{\left (\ln \left (a\right )+\ln \left (b\right )\right )}^2}{2}-a^x\,b^x\,\left (\frac {1}{2\,x\,\left (\ln \left (a\right )+\ln \left (b\right )\right )}+\frac {1}{2\,x^2\,{\left (\ln \left (a\right )+\ln \left (b\right )\right )}^2}\right )\,{\left (\ln \left (a\right )+\ln \left (b\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^x*b^x)/x^3,x)

[Out]

- (expint(-x*(log(a) + log(b)))*(log(a) + log(b))^2)/2 - a^x*b^x*(1/(2*x*(log(a) + log(b))) + 1/(2*x^2*(log(a)
+ log(b))^2))*(log(a) + log(b))^2

________________________________________________________________________________________