3.49 \(\int x (a+b \log (c \log ^p(d x^n))) \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{2} b p x^2 \left (d x^n\right )^{-2/n} \text {Ei}\left (\frac {2 \log \left (d x^n\right )}{n}\right ) \]

[Out]

-1/2*b*p*x^2*Ei(2*ln(d*x^n)/n)/((d*x^n)^(2/n))+1/2*x^2*(a+b*ln(c*ln(d*x^n)^p))

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2522, 2310, 2178} \[ \frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{2} b p x^2 \left (d x^n\right )^{-2/n} \text {Ei}\left (\frac {2 \log \left (d x^n\right )}{n}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Log[c*Log[d*x^n]^p]),x]

[Out]

-(b*p*x^2*ExpIntegralEi[(2*Log[d*x^n])/n])/(2*(d*x^n)^(2/n)) + (x^2*(a + b*Log[c*Log[d*x^n]^p]))/2

Rule 2178

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

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2522

Int[((a_.) + Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)]*(b_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[((e*x)^(m + 1
)*(a + b*Log[c*Log[d*x^n]^p]))/(e*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(e*x)^m/Log[d*x^n], x], x] /; FreeQ
[{a, b, c, d, e, m, n, p}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{2} (b n p) \int \frac {x}{\log \left (d x^n\right )} \, dx\\ &=\frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{2} \left (b p x^2 \left (d x^n\right )^{-2/n}\right ) \operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{n}}}{x} \, dx,x,\log \left (d x^n\right )\right )\\ &=-\frac {1}{2} b p x^2 \left (d x^n\right )^{-2/n} \text {Ei}\left (\frac {2 \log \left (d x^n\right )}{n}\right )+\frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 0.89 \[ \frac {1}{2} x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )-b p \left (d x^n\right )^{-2/n} \text {Ei}\left (\frac {2 \log \left (d x^n\right )}{n}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Log[c*Log[d*x^n]^p]),x]

[Out]

(x^2*(a - (b*p*ExpIntegralEi[(2*Log[d*x^n])/n])/(d*x^n)^(2/n) + b*Log[c*Log[d*x^n]^p]))/2

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fricas [A]  time = 0.46, size = 70, normalized size = 1.27 \[ \frac {b d^{\frac {2}{n}} p x^{2} \log \left (n \log \relax (x) + \log \relax (d)\right ) - b p \operatorname {log\_integral}\left (d^{\frac {2}{n}} x^{2}\right ) + {\left (b x^{2} \log \relax (c) + a x^{2}\right )} d^{\frac {2}{n}}}{2 \, d^{\frac {2}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*log(d*x^n)^p)),x, algorithm="fricas")

[Out]

1/2*(b*d^(2/n)*p*x^2*log(n*log(x) + log(d)) - b*p*log_integral(d^(2/n)*x^2) + (b*x^2*log(c) + a*x^2)*d^(2/n))/
d^(2/n)

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giac [A]  time = 0.20, size = 56, normalized size = 1.02 \[ \frac {1}{2} \, b p x^{2} \log \left (n \log \relax (x) + \log \relax (d)\right ) + \frac {1}{2} \, b x^{2} \log \relax (c) + \frac {1}{2} \, a x^{2} - \frac {b p {\rm Ei}\left (\frac {2 \, \log \relax (d)}{n} + 2 \, \log \relax (x)\right )}{2 \, d^{\frac {2}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*log(d*x^n)^p)),x, algorithm="giac")

[Out]

1/2*b*p*x^2*log(n*log(x) + log(d)) + 1/2*b*x^2*log(c) + 1/2*a*x^2 - 1/2*b*p*Ei(2*log(d)/n + 2*log(x))/d^(2/n)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \ln \left (d \,x^{n}\right )^{p}\right )+a \right ) x\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*ln(c*ln(d*x^n)^p)+a),x)

[Out]

int(x*(b*ln(c*ln(d*x^n)^p)+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a x^{2} - \frac {1}{2} \, {\left (2 \, n p \int \frac {x}{2 \, {\left (\log \relax (d) + \log \left (x^{n}\right )\right )}}\,{d x} - x^{2} \log \relax (c) - x^{2} \log \left ({\left (\log \relax (d) + \log \left (x^{n}\right )\right )}^{p}\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*log(d*x^n)^p)),x, algorithm="maxima")

[Out]

1/2*a*x^2 - 1/2*(2*n*p*integrate(1/2*x/(log(d) + log(x^n)), x) - x^2*log(c) - x^2*log((log(d) + log(x^n))^p))*
b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\left (a+b\,\ln \left (c\,{\ln \left (d\,x^n\right )}^p\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*log(c*log(d*x^n)^p)),x)

[Out]

int(x*(a + b*log(c*log(d*x^n)^p)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \log {\left (c \log {\left (d x^{n} \right )}^{p} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*ln(c*ln(d*x**n)**p)),x)

[Out]

Integral(x*(a + b*log(c*log(d*x**n)**p)), x)

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