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

Optimal. Leaf size=115 \[ -\frac{2 b^2 e p^2 \text{PolyLog}\left (2,\frac{e}{d (f+g x)}+1\right )}{d g}-\frac{2 b e p \log \left (-\frac{e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )}{d g}+\frac{(d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )^2}{d g} \]

[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)

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Rubi [A]  time = 0.0900471, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {2483, 2449, 2454, 2394, 2315} \[ -\frac{2 b^2 e p^2 \text{PolyLog}\left (2,\frac{e}{d (f+g x)}+1\right )}{d g}-\frac{2 b e p \log \left (-\frac{e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )}{d g}+\frac{(d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )^2}{d g} \]

Antiderivative was successfully verified.

[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 2483

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])

Rule 2449

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 2454

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 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 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]

Rubi steps

\begin{align*} \int \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )^2 \, dx &=\frac{\operatorname{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) \operatorname{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) \operatorname{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 ) \operatorname{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]  time = 0.270018, size = 219, normalized size = 1.9 \[ x \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )^2-\frac{b p \left (b d f p \left (\log (f+g x) \left (\log (f+g x)-2 \log \left (\frac{d f+d g x+e}{e}\right )\right )-2 \text{PolyLog}\left (2,-\frac{d (f+g x)}{e}\right )\right )-b p (d f+e) \left (2 \text{PolyLog}\left (2,\frac{d f+d g x+e}{e}\right )+\left (2 \log \left (-\frac{d (f+g x)}{e}\right )-\log (d f+d g x+e)\right ) \log (d (f+g x)+e)\right )+2 d f \log (f+g x) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )-2 (d f+e) \log (d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac{e}{f+g x}\right )^p\right )\right )\right )}{d g} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.134, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\ln \left ( c \left ( d+{\frac{e}{gx+f}} \right ) ^{p} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -2 \, a b e g p{\left (\frac{f \log \left (g x + f\right )}{e g^{2}} - \frac{{\left (d f + e\right )} \log \left (d g x + d f + e\right )}{d e g^{2}}\right )} + 2 \, a b x \log \left (c{\left (d + \frac{e}{g x + f}\right )}^{p}\right ) + a^{2} x + b^{2}{\left (\frac{d g x \log \left ({\left (d g x + d f + e\right )}^{p}\right )^{2} + d g x \log \left ({\left (g x + f\right )}^{p}\right )^{2} -{\left (d f p^{2} + e p^{2}\right )} \log \left (d g x + d f + e\right )^{2} + 2 \,{\left (d f p^{2} + e p^{2}\right )} \log \left (d g x + d f + e\right ) \log \left (g x + f\right ) - 2 \,{\left (d f p \log \left (g x + f\right ) + d g x \log \left ({\left (g x + f\right )}^{p}\right ) - d g x \log \left (c\right ) -{\left (d f p + e p\right )} \log \left (d g x + d f + e\right )\right )} \log \left ({\left (d g x + d f + e\right )}^{p}\right ) + 2 \,{\left (d f p \log \left (g x + f\right ) - d g x \log \left (c\right ) -{\left (d f p + e p\right )} \log \left (d g x + d f + e\right )\right )} \log \left ({\left (g x + f\right )}^{p}\right )}{d g} - \int -\frac{d g^{2} x^{2} \log \left (c\right )^{2} +{\left (d f^{2} + e f\right )} \log \left (c\right )^{2} +{\left (2 \, e g p \log \left (c\right ) +{\left (2 \, d f g + e g\right )} \log \left (c\right )^{2}\right )} x - 2 \,{\left (d f^{2} p^{2} + 2 \, e f p^{2} +{\left (d f g p^{2} + e g p^{2}\right )} x\right )} \log \left (g x + f\right )}{d g^{2} x^{2} + d f^{2} + e f +{\left (2 \, d f g + e g\right )} x}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} \log \left (c \left (\frac{d g x + d f + e}{g x + f}\right )^{p}\right )^{2} + 2 \, a b \log \left (c \left (\frac{d g x + d f + e}{g x + f}\right )^{p}\right ) + a^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \log{\left (c \left (d + \frac{e}{f + g x}\right )^{p} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c{\left (d + \frac{e}{g x + f}\right )}^{p}\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)