3.248 \(\int \frac {a+b \log (c (d+e x)^n)}{x^3 (f+g x)} \, dx\)

Optimal. Leaf size=250 \[ \frac {g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac {b e^2 n \log (x)}{2 d^2 f}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}-\frac {b g^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^3}+\frac {b g^2 n \text {Li}_2\left (\frac {e x}{d}+1\right )}{f^3}-\frac {b e g n \log (x)}{d f^2}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {b e n}{2 d f x} \]

[Out]

-1/2*b*e*n/d/f/x-1/2*b*e^2*n*ln(x)/d^2/f-b*e*g*n*ln(x)/d/f^2+1/2*b*e^2*n*ln(e*x+d)/d^2/f+b*e*g*n*ln(e*x+d)/d/f
^2+1/2*(-a-b*ln(c*(e*x+d)^n))/f/x^2+g*(a+b*ln(c*(e*x+d)^n))/f^2/x+g^2*ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f^3-g^2
*(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/f^3-b*g^2*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/f^3+b*g^2*n*polyl
og(2,1+e*x/d)/f^3

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Rubi [A]  time = 0.25, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ -\frac {b g^2 n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^3}+\frac {b g^2 n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^3}+\frac {g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac {b e^2 n \log (x)}{2 d^2 f}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}-\frac {b e g n \log (x)}{d f^2}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {b e n}{2 d f x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(x^3*(f + g*x)),x]

[Out]

-(b*e*n)/(2*d*f*x) - (b*e^2*n*Log[x])/(2*d^2*f) - (b*e*g*n*Log[x])/(d*f^2) + (b*e^2*n*Log[d + e*x])/(2*d^2*f)
+ (b*e*g*n*Log[d + e*x])/(d*f^2) - (a + b*Log[c*(d + e*x)^n])/(2*f*x^2) + (g*(a + b*Log[c*(d + e*x)^n]))/(f^2*
x) + (g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^3 - (g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/
(e*f - d*g)])/f^3 - (b*g^2*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/f^3 + (b*g^2*n*PolyLog[2, 1 + (e*x)/d])
/f^3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)} \, dx &=\int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f x^3}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x^2}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac {g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3} \, dx}{f}-\frac {g \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2} \, dx}{f^2}+\frac {g^2 \int \frac {a+b \log \left (c (d+e x)^n\right )}{x} \, dx}{f^3}-\frac {g^3 \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{f^3}\\ &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^3}+\frac {(b e n) \int \frac {1}{x^2 (d+e x)} \, dx}{2 f}-\frac {(b e g n) \int \frac {1}{x (d+e x)} \, dx}{f^2}-\frac {\left (b e g^2 n\right ) \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx}{f^3}+\frac {\left (b e g^2 n\right ) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{f^3}\\ &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^3}+\frac {b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^3}+\frac {(b e n) \int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx}{2 f}-\frac {(b e g n) \int \frac {1}{x} \, dx}{d f^2}+\frac {\left (b e^2 g n\right ) \int \frac {1}{d+e x} \, dx}{d f^2}+\frac {\left (b g^2 n\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{f^3}\\ &=-\frac {b e n}{2 d f x}-\frac {b e^2 n \log (x)}{2 d^2 f}-\frac {b e g n \log (x)}{d f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^3}-\frac {b g^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^3}+\frac {b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 208, normalized size = 0.83 \[ -\frac {\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}+2 g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {2 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-2 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e f^2 n (-e x \log (d+e x)+d+e x \log (x))}{d^2 x}+2 b g^2 n \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )+\frac {2 b e f g n (\log (x)-\log (d+e x))}{d}-2 b g^2 n \text {Li}_2\left (\frac {e x}{d}+1\right )}{2 f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(x^3*(f + g*x)),x]

[Out]

-1/2*((2*b*e*f*g*n*(Log[x] - Log[d + e*x]))/d + (b*e*f^2*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x) + (f^2
*(a + b*Log[c*(d + e*x)^n]))/x^2 - (2*f*g*(a + b*Log[c*(d + e*x)^n]))/x - 2*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(
d + e*x)^n]) + 2*g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] + 2*b*g^2*n*PolyLog[2, (g*(d +
e*x))/(-(e*f) + d*g)] - 2*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^3

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g x^{4} + f x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f),x, algorithm="fricas")

[Out]

integral((b*log((e*x + d)^n*c) + a)/(g*x^4 + f*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/((g*x + f)*x^3), x)

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maple [C]  time = 0.25, size = 926, normalized size = 3.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(e*x+d)^n)+a)/x^3/(g*x+f),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {2 \, g^{2} \log \left (g x + f\right )}{f^{3}} - \frac {2 \, g^{2} \log \relax (x)}{f^{3}} - \frac {2 \, g x - f}{f^{2} x^{2}}\right )} + b \int \frac {\log \left ({\left (e x + d\right )}^{n}\right ) + \log \relax (c)}{g x^{4} + f x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f),x, algorithm="maxima")

[Out]

-1/2*a*(2*g^2*log(g*x + f)/f^3 - 2*g^2*log(x)/f^3 - (2*g*x - f)/(f^2*x^2)) + b*integrate((log((e*x + d)^n) + l
og(c))/(g*x^4 + f*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,\left (f+g\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x)^n))/(x^3*(f + g*x)),x)

[Out]

int((a + b*log(c*(d + e*x)^n))/(x^3*(f + g*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/x**3/(g*x+f),x)

[Out]

Timed out

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