3.159 \(\int \frac {(a+b \log (c x^n)) (d+e \log (f x^r))}{x} \, dx\)

Optimal. Leaf size=57 \[ \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}-\frac {e r \left (a+b \log \left (c x^n\right )\right )^3}{6 b^2 n^2} \]

[Out]

-1/6*e*r*(a+b*ln(c*x^n))^3/b^2/n^2+1/2*(a+b*ln(c*x^n))^2*(d+e*ln(f*x^r))/b/n

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Rubi [A]  time = 0.07, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2301, 2366, 12, 2302, 30} \[ \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}-\frac {e r \left (a+b \log \left (c x^n\right )\right )^3}{6 b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*(d + e*Log[f*x^r]))/x,x]

[Out]

-(e*r*(a + b*Log[c*x^n])^3)/(6*b^2*n^2) + ((a + b*Log[c*x^n])^2*(d + e*Log[f*x^r]))/(2*b*n)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2302

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

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{x} \, dx &=\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}-(e r) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b n x} \, dx\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}-\frac {(e r) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx}{2 b n}\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}-\frac {(e r) \operatorname {Subst}\left (\int x^2 \, dx,x,a+b \log \left (c x^n\right )\right )}{2 b^2 n^2}\\ &=-\frac {e r \left (a+b \log \left (c x^n\right )\right )^3}{6 b^2 n^2}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 b n}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 72, normalized size = 1.26 \[ \frac {1}{6} \log (x) \left (-3 \log (x) \left (a e r+b e r \log \left (c x^n\right )+b d n+b e n \log \left (f x^r\right )\right )+6 \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )+2 b e n r \log ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*(d + e*Log[f*x^r]))/x,x]

[Out]

(Log[x]*(2*b*e*n*r*Log[x]^2 + 6*(a + b*Log[c*x^n])*(d + e*Log[f*x^r]) - 3*Log[x]*(b*d*n + a*e*r + b*e*r*Log[c*
x^n] + b*e*n*Log[f*x^r])))/6

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fricas [A]  time = 0.68, size = 62, normalized size = 1.09 \[ \frac {1}{3} \, b e n r \log \relax (x)^{3} + \frac {1}{2} \, {\left (b e r \log \relax (c) + b e n \log \relax (f) + b d n + a e r\right )} \log \relax (x)^{2} + {\left (b d \log \relax (c) + a d + {\left (b e \log \relax (c) + a e\right )} \log \relax (f)\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(d+e*log(f*x^r))/x,x, algorithm="fricas")

[Out]

1/3*b*e*n*r*log(x)^3 + 1/2*(b*e*r*log(c) + b*e*n*log(f) + b*d*n + a*e*r)*log(x)^2 + (b*d*log(c) + a*d + (b*e*l
og(c) + a*e)*log(f))*log(x)

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giac [A]  time = 0.29, size = 85, normalized size = 1.49 \[ \frac {1}{3} \, b n r e \log \relax (x)^{3} + \frac {1}{2} \, b r e \log \relax (c) \log \relax (x)^{2} + \frac {1}{2} \, b n e \log \relax (f) \log \relax (x)^{2} + b e \log \relax (c) \log \relax (f) \log \relax (x) + \frac {1}{2} \, b d n \log \relax (x)^{2} + \frac {1}{2} \, a r e \log \relax (x)^{2} + b d \log \relax (c) \log \relax (x) + a e \log \relax (f) \log \relax (x) + a d \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(d+e*log(f*x^r))/x,x, algorithm="giac")

[Out]

1/3*b*n*r*e*log(x)^3 + 1/2*b*r*e*log(c)*log(x)^2 + 1/2*b*n*e*log(f)*log(x)^2 + b*e*log(c)*log(f)*log(x) + 1/2*
b*d*n*log(x)^2 + 1/2*a*r*e*log(x)^2 + b*d*log(c)*log(x) + a*e*log(f)*log(x) + a*d*log(x)

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maple [C]  time = 0.62, size = 1597, normalized size = 28.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)*(d+e*ln(f*x^r))/x,x)

[Out]

-1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)*csgn(I*c)+(b*e*ln(x)*ln(x^n)
-1/2*b*e*n*ln(x)^2+1/2*I*Pi*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)
*csgn(I*c)-1/2*I*Pi*ln(x)*b*e*csgn(I*c*x^n)^3+1/2*I*Pi*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*c)+ln(c)*ln(x)*b*e+ln(
x)*a*e)*ln(x^r)+ln(f)*ln(x)*a*e-1/2*ln(x)^2*a*e*r+a*d*ln(x)+b*d*ln(c)*ln(x)-1/2*b*d*n*ln(x)^2+1/2*I*ln(x)*Pi*b
*d*csgn(I*x^n)*csgn(I*c*x^n)^2+ln(x^n)*b*d*ln(x)+ln(x^n)*ln(f)*b*e*ln(x)-1/2*ln(x^n)*r*b*e*ln(x)^2+ln(c)*ln(f)
*ln(x)*b*e+1/3*b*e*n*r*ln(x)^3-1/2*ln(x)^2*ln(c)*b*e*r-1/2*ln(x)^2*ln(f)*b*e*n-1/2*I*ln(x)*Pi*b*d*csgn(I*x^n)*
csgn(I*c*x^n)*csgn(I*c)-1/2*I*ln(c)*Pi*ln(x)*b*e*csgn(I*f*x^r)^3-1/2*I*Pi*ln(f)*ln(x)*b*e*csgn(I*c*x^n)^3+1/2*
I*Pi*ln(x)*a*e*csgn(I*f)*csgn(I*f*x^r)^2+1/2*I*Pi*ln(x)*a*e*csgn(I*x^r)*csgn(I*f*x^r)^2-1/2*I*ln(c)*Pi*ln(x)*b
*e*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)+1/4*I*ln(x)^2*Pi*b*e*r*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*Pi^2*ln(
x)*b*e*csgn(I*c*x^n)^3*csgn(I*f)*csgn(I*f*x^r)^2+1/4*Pi^2*ln(x)*b*e*csgn(I*c*x^n)^3*csgn(I*x^r)*csgn(I*f*x^r)^
2+1/4*Pi^2*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*f*x^r)^3*csgn(I*c)-1/2*I*Pi*ln(x)*a*e*csgn(I*f*x^r)^3-1/4*Pi^2*ln(
x)*b*e*csgn(I*c*x^n)^3*csgn(I*f*x^r)^3-1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*csgn(I*f)*csgn(I*f*x^r)^
2-1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*csgn(I*x^r)*csgn(I*f*x^r)^2-1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*cs
gn(I*c*x^n)*csgn(I*f*x^r)^3*csgn(I*c)+1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*csgn(I*f*x^r)^3-1/2*I*ln(
x^n)*Pi*b*e*csgn(I*f*x^r)^3*ln(x)+1/4*I*ln(x)^2*Pi*b*e*n*csgn(I*f*x^r)^3-1/2*I*Pi*ln(f)*ln(x)*b*e*csgn(I*x^n)*
csgn(I*c*x^n)*csgn(I*c)+1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)+1/4
*Pi^2*ln(x)*b*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*f)*csgn(I*f*x^r)^2*csgn(I*c)+1/4*Pi^2*ln(x)*b*e*csgn(I*x^n)*c
sgn(I*c*x^n)*csgn(I*x^r)*csgn(I*f*x^r)^2*csgn(I*c)+1/2*I*ln(x)*Pi*b*d*csgn(I*c*x^n)^2*csgn(I*c)-1/2*I*ln(x)*Pi
*b*d*csgn(I*c*x^n)^3+1/2*I*ln(c)*Pi*ln(x)*b*e*csgn(I*x^r)*csgn(I*f*x^r)^2+1/2*I*Pi*ln(f)*ln(x)*b*e*csgn(I*x^n)
*csgn(I*c*x^n)^2+1/2*I*Pi*ln(f)*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*c)-1/2*I*Pi*ln(x)*a*e*csgn(I*f)*csgn(I*x^r)*c
sgn(I*f*x^r)+1/4*Pi^2*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)*csgn(I*c)+1/2*I*ln(x^n)*Pi
*b*e*csgn(I*f)*csgn(I*f*x^r)^2*ln(x)+1/2*I*ln(x^n)*Pi*b*e*csgn(I*x^r)*csgn(I*f*x^r)^2*ln(x)-1/4*I*ln(x)^2*Pi*b
*e*n*csgn(I*f)*csgn(I*f*x^r)^2+1/4*I*ln(x)^2*Pi*b*e*r*csgn(I*c*x^n)^3+1/4*I*ln(x)^2*Pi*b*e*n*csgn(I*f)*csgn(I*
x^r)*csgn(I*f*x^r)-1/2*I*ln(x^n)*Pi*b*e*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)*ln(x)-1/4*Pi^2*ln(x)*b*e*csgn(I*c*
x^n)^3*csgn(I*f)*csgn(I*x^r)*csgn(I*f*x^r)-1/4*Pi^2*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*f)*csgn(I*f*x^r)^2*csgn(I
*c)-1/4*Pi^2*ln(x)*b*e*csgn(I*c*x^n)^2*csgn(I*x^r)*csgn(I*f*x^r)^2*csgn(I*c)-1/4*I*ln(x)^2*Pi*b*e*r*csgn(I*c*x
^n)^2*csgn(I*c)+1/2*I*ln(c)*Pi*ln(x)*b*e*csgn(I*f)*csgn(I*f*x^r)^2-1/4*I*ln(x)^2*Pi*b*e*n*csgn(I*x^r)*csgn(I*f
*x^r)^2-1/4*I*ln(x)^2*Pi*b*e*r*csgn(I*x^n)*csgn(I*c*x^n)^2

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maxima [A]  time = 0.62, size = 73, normalized size = 1.28 \[ \frac {b e \log \left (c x^{n}\right ) \log \left (f x^{r}\right )^{2}}{2 \, r} - \frac {b e n \log \left (f x^{r}\right )^{3}}{6 \, r^{2}} + \frac {b d \log \left (c x^{n}\right )^{2}}{2 \, n} + \frac {a e \log \left (f x^{r}\right )^{2}}{2 \, r} + a d \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*(d+e*log(f*x^r))/x,x, algorithm="maxima")

[Out]

1/2*b*e*log(c*x^n)*log(f*x^r)^2/r - 1/6*b*e*n*log(f*x^r)^3/r^2 + 1/2*b*d*log(c*x^n)^2/n + 1/2*a*e*log(f*x^r)^2
/r + a*d*log(x)

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mupad [B]  time = 3.87, size = 73, normalized size = 1.28 \[ a\,d\,\ln \relax (x)+\frac {b\,d\,{\ln \left (c\,x^n\right )}^2}{2\,n}+\frac {a\,e\,{\ln \left (f\,x^r\right )}^2}{2\,r}-\frac {b\,e\,r\,{\ln \left (c\,x^n\right )}^3}{6\,n^2}+\frac {b\,e\,{\ln \left (c\,x^n\right )}^2\,\ln \left (f\,x^r\right )}{2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*log(f*x^r))*(a + b*log(c*x^n)))/x,x)

[Out]

a*d*log(x) + (b*d*log(c*x^n)^2)/(2*n) + (a*e*log(f*x^r)^2)/(2*r) - (b*e*r*log(c*x^n)^3)/(6*n^2) + (b*e*log(c*x
^n)^2*log(f*x^r))/(2*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*(d+e*ln(f*x**r))/x,x)

[Out]

Integral((a + b*log(c*x**n))*(d + e*log(f*x**r))/x, x)

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