3.365 \(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))^2}{(d+e x^m)^3} \, dx\)

Optimal. Leaf size=214 \[ -\frac {b n x^{1-m} (f x)^{m-1} \log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e m^2}-\frac {b n x (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac {x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}+\frac {b^2 n^2 x^{1-m} (f x)^{m-1} \text {Li}_2\left (-\frac {d x^{-m}}{e}\right )}{d^2 e m^3}+\frac {b^2 n^2 x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{d^2 e m^3} \]

[Out]

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

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Rubi [A]  time = 0.51, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2339, 2338, 2349, 2345, 2391, 2335, 260} \[ \frac {b^2 n^2 x^{1-m} (f x)^{m-1} \text {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d^2 e m^3}-\frac {b n x^{1-m} (f x)^{m-1} \log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e m^2}-\frac {b n x (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac {x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}+\frac {b^2 n^2 x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{d^2 e m^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

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

Rule 2335

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

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :
> Simp[(f^m*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^p)/(e*r*(q + 1)), x] - Dist[(b*f^m*n*p)/(e*r*(q + 1)), Int[
((d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[
m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :>
Dist[(f*x)^m/x^m, Int[x^m*(d + e*x^r)^q*(a + b*Log[c*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r},
x] && EqQ[m, r - 1] && IGtQ[p, 0] &&  !(IntegerQ[m] || GtQ[f, 0])

Rule 2345

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> -Simp[(Log[1 +
d/(e*x^r)]*(a + b*Log[c*x^n])^p)/(d*r), x] + Dist[(b*n*p)/(d*r), Int[(Log[1 + d/(e*x^r)]*(a + b*Log[c*x^n])^(p
- 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2349

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_))/(x_), x_Symbol] :> Dist[1/d,
Int[((d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[x^(r - 1)*(d + e*x^r)^q*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0] && ILtQ[q, -1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 207, normalized size = 0.97 \[ \frac {x^{-m} (f x)^m \left (-\frac {m^2 \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2}+\frac {2 b m n \left (a+b \log \left (c x^n\right )\right )}{d \left (d+e x^m\right )}-\frac {2 a b m n \log \left (d-d x^m\right )}{d^2}+\frac {2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d^2}+\frac {2 b^2 n^2 \left (\text {Li}_2\left (\frac {e x^m}{d}+1\right )+\left (\log \left (-\frac {e x^m}{d}\right )-m \log (x)\right ) \log \left (d+e x^m\right )+\frac {1}{2} m^2 \log ^2(x)\right )}{d^2}+\frac {2 b^2 n^2 \log \left (d-d x^m\right )}{d^2}\right )}{2 e f m^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 535, normalized size = 2.50 \[ \frac {{\left (b^{2} e^{2} m^{2} n^{2} \log \relax (x)^{2} + 2 \, {\left (b^{2} e^{2} m^{2} n \log \relax (c) + a b e^{2} m^{2} n - b^{2} e^{2} m n^{2}\right )} \log \relax (x)\right )} f^{m - 1} x^{2 \, m} + 2 \, {\left (b^{2} d e m^{2} n^{2} \log \relax (x)^{2} + b^{2} d e m n \log \relax (c) + a b d e m n + {\left (2 \, b^{2} d e m^{2} n \log \relax (c) + 2 \, a b d e m^{2} n - b^{2} d e m n^{2}\right )} \log \relax (x)\right )} f^{m - 1} x^{m} - {\left (b^{2} d^{2} m^{2} \log \relax (c)^{2} + a^{2} d^{2} m^{2} - 2 \, a b d^{2} m n + 2 \, {\left (a b d^{2} m^{2} - b^{2} d^{2} m n\right )} \log \relax (c)\right )} f^{m - 1} - 2 \, {\left (b^{2} e^{2} f^{m - 1} n^{2} x^{2 \, m} + 2 \, b^{2} d e f^{m - 1} n^{2} x^{m} + b^{2} d^{2} f^{m - 1} n^{2}\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - 2 \, {\left ({\left (b^{2} e^{2} m n \log \relax (c) + a b e^{2} m n - b^{2} e^{2} n^{2}\right )} f^{m - 1} x^{2 \, m} + 2 \, {\left (b^{2} d e m n \log \relax (c) + a b d e m n - b^{2} d e n^{2}\right )} f^{m - 1} x^{m} + {\left (b^{2} d^{2} m n \log \relax (c) + a b d^{2} m n - b^{2} d^{2} n^{2}\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \, {\left (b^{2} e^{2} f^{m - 1} m n^{2} x^{2 \, m} \log \relax (x) + 2 \, b^{2} d e f^{m - 1} m n^{2} x^{m} \log \relax (x) + b^{2} d^{2} f^{m - 1} m n^{2} \log \relax (x)\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{2 \, {\left (d^{2} e^{3} m^{3} x^{2 \, m} + 2 \, d^{3} e^{2} m^{3} x^{m} + d^{4} e m^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2/(d+e*x^m)^3,x, algorithm="fricas")

[Out]

1/2*((b^2*e^2*m^2*n^2*log(x)^2 + 2*(b^2*e^2*m^2*n*log(c) + a*b*e^2*m^2*n - b^2*e^2*m*n^2)*log(x))*f^(m - 1)*x^
(2*m) + 2*(b^2*d*e*m^2*n^2*log(x)^2 + b^2*d*e*m*n*log(c) + a*b*d*e*m*n + (2*b^2*d*e*m^2*n*log(c) + 2*a*b*d*e*m
^2*n - b^2*d*e*m*n^2)*log(x))*f^(m - 1)*x^m - (b^2*d^2*m^2*log(c)^2 + a^2*d^2*m^2 - 2*a*b*d^2*m*n + 2*(a*b*d^2
*m^2 - b^2*d^2*m*n)*log(c))*f^(m - 1) - 2*(b^2*e^2*f^(m - 1)*n^2*x^(2*m) + 2*b^2*d*e*f^(m - 1)*n^2*x^m + b^2*d
^2*f^(m - 1)*n^2)*dilog(-(e*x^m + d)/d + 1) - 2*((b^2*e^2*m*n*log(c) + a*b*e^2*m*n - b^2*e^2*n^2)*f^(m - 1)*x^
(2*m) + 2*(b^2*d*e*m*n*log(c) + a*b*d*e*m*n - b^2*d*e*n^2)*f^(m - 1)*x^m + (b^2*d^2*m*n*log(c) + a*b*d^2*m*n -
b^2*d^2*n^2)*f^(m - 1))*log(e*x^m + d) - 2*(b^2*e^2*f^(m - 1)*m*n^2*x^(2*m)*log(x) + 2*b^2*d*e*f^(m - 1)*m*n^
2*x^m*log(x) + b^2*d^2*f^(m - 1)*m*n^2*log(x))*log((e*x^m + d)/d))/(d^2*e^3*m^3*x^(2*m) + 2*d^3*e^2*m^3*x^m +
d^4*e*m^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2/(d+e*x^m)^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.95, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{2} \left (f x \right )^{m -1}}{\left (e \,x^{m}+d \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(m-1)*(b*ln(c*x^n)+a)^2/(e*x^m+d)^3,x)

[Out]

int((f*x)^(m-1)*(b*ln(c*x^n)+a)^2/(e*x^m+d)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2/(d+e*x^m)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (f\,x\right )}^{m-1}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x^m\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f*x)^(m - 1)*(a + b*log(c*x^n))^2)/(d + e*x^m)^3,x)

[Out]

int(((f*x)^(m - 1)*(a + b*log(c*x^n))^2)/(d + e*x^m)^3, 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((f*x)**(-1+m)*(a+b*ln(c*x**n))**2/(d+e*x**m)**3,x)

[Out]

Timed out

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