\(\int (a+b \log (c (d+\frac {e}{f+g x})^p))^2 \, dx\) [638]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=-\frac {2 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )}{d g}+\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}-\frac {2 b^2 e p^2 \operatorname {PolyLog}\left (2,1+\frac {e}{d (f+g x)}\right )}{d g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2533, 2499, 2504, 2441, 2352} \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=-\frac {2 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )}{d g}+\frac {(d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}-\frac {2 b^2 e p^2 \operatorname {PolyLog}\left (2,\frac {e}{d (f+g x)}+1\right )}{d g} \]

[In]

Int[(a + b*Log[c*(d + e/(f + g*x))^p])^2,x]

[Out]

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

Rule 2352

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

Rule 2441

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 2499

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

Rule 2504

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

Rule 2533

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.90 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=x \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2-\frac {b p \left (2 d f \log (f+g x) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )-2 (e+d f) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \log (e+d (f+g x))+b d f p \left (\log (f+g x) \left (\log (f+g x)-2 \log \left (\frac {e+d f+d g x}{e}\right )\right )-2 \operatorname {PolyLog}\left (2,-\frac {d (f+g x)}{e}\right )\right )-b (e+d f) p \left (\left (2 \log \left (-\frac {d (f+g x)}{e}\right )-\log (e+d f+d g x)\right ) \log (e+d (f+g x))+2 \operatorname {PolyLog}\left (2,\frac {e+d f+d g x}{e}\right )\right )\right )}{d g} \]

[In]

Integrate[(a + b*Log[c*(d + e/(f + g*x))^p])^2,x]

[Out]

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

Maple [F]

\[\int {\left (a +b \ln \left (c \left (d +\frac {e}{g x +f}\right )^{p}\right )\right )}^{2}d x\]

[In]

int((a+b*ln(c*(d+e/(g*x+f))^p))^2,x)

[Out]

int((a+b*ln(c*(d+e/(g*x+f))^p))^2,x)

Fricas [F]

\[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=\int { {\left (b \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((a+b*log(c*(d+e/(g*x+f))^p))^2,x, algorithm="fricas")

[Out]

integral(b^2*log(c*((d*g*x + d*f + e)/(g*x + f))^p)^2 + 2*a*b*log(c*((d*g*x + d*f + e)/(g*x + f))^p) + a^2, x)

Sympy [F]

\[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=\int \left (a + b \log {\left (c \left (d + \frac {e}{f + g x}\right )^{p} \right )}\right )^{2}\, dx \]

[In]

integrate((a+b*ln(c*(d+e/(g*x+f))**p))**2,x)

[Out]

Integral((a + b*log(c*(d + e/(f + g*x))**p))**2, x)

Maxima [F]

\[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=\int { {\left (b \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((a+b*log(c*(d+e/(g*x+f))^p))^2,x, algorithm="maxima")

[Out]

-2*a*b*e*g*p*(f*log(g*x + f)/(e*g^2) - (d*f + e)*log(d*g*x + d*f + e)/(d*e*g^2)) + 2*a*b*x*log(c*(d + e/(g*x +
f))^p) + a^2*x + b^2*((d*g*x*log((d*g*x + d*f + e)^p)^2 + d*g*x*log((g*x + f)^p)^2 - (d*f*p^2 + e*p^2)*log(d*
g*x + d*f + e)^2 + 2*(d*f*p^2 + e*p^2)*log(d*g*x + d*f + e)*log(g*x + f) - 2*(d*f*p*log(g*x + f) + d*g*x*log((
g*x + f)^p) - d*g*x*log(c) - (d*f*p + e*p)*log(d*g*x + d*f + e))*log((d*g*x + d*f + e)^p) + 2*(d*f*p*log(g*x +
f) - d*g*x*log(c) - (d*f*p + e*p)*log(d*g*x + d*f + e))*log((g*x + f)^p))/(d*g) - integrate(-(d*g^2*x^2*log(c
)^2 + (d*f^2 + e*f)*log(c)^2 + (2*e*g*p*log(c) + (2*d*f*g + e*g)*log(c)^2)*x - 2*(d*f^2*p^2 + 2*e*f*p^2 + (d*f
*g*p^2 + e*g*p^2)*x)*log(g*x + f))/(d*g^2*x^2 + d*f^2 + e*f + (2*d*f*g + e*g)*x), x))

Giac [F]

\[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=\int { {\left (b \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((a+b*log(c*(d+e/(g*x+f))^p))^2,x, algorithm="giac")

[Out]

integrate((b*log(c*(d + e/(g*x + f))^p) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2 \, dx=\int {\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{f+g\,x}\right )}^p\right )\right )}^2 \,d x \]

[In]

int((a + b*log(c*(d + e/(f + g*x))^p))^2,x)

[Out]

int((a + b*log(c*(d + e/(f + g*x))^p))^2, x)