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

Optimal. Leaf size=279 \[ \frac{2 h (e+f x)^2 e^{-\frac{2 a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{2}{p q}} \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^3 p q}+\frac{(e+f x) e^{-\frac{a}{b p q}} (f g-e h)^2 \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f^3 p q}+\frac{h^2 (e+f x)^3 e^{-\frac{3 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{3}{p q}} \text{Ei}\left (\frac{3 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^3 p q} \]

[Out]

((f*g - e*h)^2*(e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f^3*p*q*(c*
(d*(e + f*x)^p)^q)^(1/(p*q))) + (2*h*(f*g - e*h)*(e + f*x)^2*ExpIntegralEi[(2*(a + b*Log[c*(d*(e + f*x)^p)^q])
)/(b*p*q)])/(b*E^((2*a)/(b*p*q))*f^3*p*q*(c*(d*(e + f*x)^p)^q)^(2/(p*q))) + (h^2*(e + f*x)^3*ExpIntegralEi[(3*
(a + b*Log[c*(d*(e + f*x)^p)^q]))/(b*p*q)])/(b*E^((3*a)/(b*p*q))*f^3*p*q*(c*(d*(e + f*x)^p)^q)^(3/(p*q)))

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Rubi [A]  time = 0.778938, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2399, 2389, 2300, 2178, 2390, 2310, 2445} \[ \frac{2 h (e+f x)^2 e^{-\frac{2 a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{2}{p q}} \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^3 p q}+\frac{(e+f x) e^{-\frac{a}{b p q}} (f g-e h)^2 \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f^3 p q}+\frac{h^2 (e+f x)^3 e^{-\frac{3 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{3}{p q}} \text{Ei}\left (\frac{3 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^3 p q} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((f*g - e*h)^2*(e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f^3*p*q*(c*
(d*(e + f*x)^p)^q)^(1/(p*q))) + (2*h*(f*g - e*h)*(e + f*x)^2*ExpIntegralEi[(2*(a + b*Log[c*(d*(e + f*x)^p)^q])
)/(b*p*q)])/(b*E^((2*a)/(b*p*q))*f^3*p*q*(c*(d*(e + f*x)^p)^q)^(2/(p*q))) + (h^2*(e + f*x)^3*ExpIntegralEi[(3*
(a + b*Log[c*(d*(e + f*x)^p)^q]))/(b*p*q)])/(b*E^((3*a)/(b*p*q))*f^3*p*q*(c*(d*(e + f*x)^p)^q)^(3/(p*q)))

Rule 2399

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

Rule 2389

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

Rule 2300

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2390

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

Rule 2310

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

Rule 2445

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

Mathematica [A]  time = 0.815474, size = 252, normalized size = 0.9 \[ \frac{(e+f x) e^{-\frac{3 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{3}{p q}} \left (e^{\frac{2 a}{b p q}} (f g-e h)^2 \left (c \left (d (e+f x)^p\right )^q\right )^{\frac{2}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )-h (e+f x) \left (-2 e^{\frac{a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{\frac{1}{p q}} \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )-h (e+f x) \text{Ei}\left (\frac{3 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )\right )\right )}{b f^3 p q} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)*(E^((2*a)/(b*p*q))*(f*g - e*h)^2*(c*(d*(e + f*x)^p)^q)^(2/(p*q))*ExpIntegralEi[(a + b*Log[c*(d*(e +
f*x)^p)^q])/(b*p*q)] - h*(e + f*x)*(-2*E^(a/(b*p*q))*(f*g - e*h)*(c*(d*(e + f*x)^p)^q)^(1/(p*q))*ExpIntegralE
i[(2*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(b*p*q)] - h*(e + f*x)*ExpIntegralEi[(3*(a + b*Log[c*(d*(e + f*x)^p)^q]
))/(b*p*q)])))/(b*E^((3*a)/(b*p*q))*f^3*p*q*(c*(d*(e + f*x)^p)^q)^(3/(p*q)))

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Maple [F]  time = 0.5, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( hx+g \right ) ^{2}}{a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03072, size = 597, normalized size = 2.14 \begin{align*} \frac{{\left (h^{2} \logintegral \left ({\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )} e^{\left (\frac{3 \,{\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}\right ) + 2 \,{\left (f g h - e h^{2}\right )} e^{\left (\frac{b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )} \logintegral \left ({\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )} e^{\left (\frac{2 \,{\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}\right ) +{\left (f^{2} g^{2} - 2 \, e f g h + e^{2} h^{2}\right )} e^{\left (\frac{2 \,{\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )} \logintegral \left ({\left (f x + e\right )} e^{\left (\frac{b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )}\right )\right )} e^{\left (-\frac{3 \,{\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}}{b f^{3} p q} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(h^2*log_integral((f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*e^(3*(b*q*log(d) + b*log(c) + a)/(b*p*q))) + 2*(f*
g*h - e*h^2)*e^((b*q*log(d) + b*log(c) + a)/(b*p*q))*log_integral((f^2*x^2 + 2*e*f*x + e^2)*e^(2*(b*q*log(d) +
b*log(c) + a)/(b*p*q))) + (f^2*g^2 - 2*e*f*g*h + e^2*h^2)*e^(2*(b*q*log(d) + b*log(c) + a)/(b*p*q))*log_integ
ral((f*x + e)*e^((b*q*log(d) + b*log(c) + a)/(b*p*q))))*e^(-3*(b*q*log(d) + b*log(c) + a)/(b*p*q))/(b*f^3*p*q)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39634, size = 707, normalized size = 2.53 \begin{align*} \frac{g^{2}{\rm Ei}\left (\frac{\log \left (d\right )}{p} + \frac{\log \left (c\right )}{p q} + \frac{a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac{a}{b p q}\right )}}{b c^{\frac{1}{p q}} d^{\left (\frac{1}{p}\right )} f p q} - \frac{2 \, g h{\rm Ei}\left (\frac{\log \left (d\right )}{p} + \frac{\log \left (c\right )}{p q} + \frac{a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac{a}{b p q} + 1\right )}}{b c^{\frac{1}{p q}} d^{\left (\frac{1}{p}\right )} f^{2} p q} + \frac{2 \, g h{\rm Ei}\left (\frac{2 \, \log \left (d\right )}{p} + \frac{2 \, \log \left (c\right )}{p q} + \frac{2 \, a}{b p q} + 2 \, \log \left (f x + e\right )\right ) e^{\left (-\frac{2 \, a}{b p q}\right )}}{b c^{\frac{2}{p q}} d^{\frac{2}{p}} f^{2} p q} + \frac{h^{2}{\rm Ei}\left (\frac{\log \left (d\right )}{p} + \frac{\log \left (c\right )}{p q} + \frac{a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac{a}{b p q} + 2\right )}}{b c^{\frac{1}{p q}} d^{\left (\frac{1}{p}\right )} f^{3} p q} - \frac{2 \, h^{2}{\rm Ei}\left (\frac{2 \, \log \left (d\right )}{p} + \frac{2 \, \log \left (c\right )}{p q} + \frac{2 \, a}{b p q} + 2 \, \log \left (f x + e\right )\right ) e^{\left (-\frac{2 \, a}{b p q} + 1\right )}}{b c^{\frac{2}{p q}} d^{\frac{2}{p}} f^{3} p q} + \frac{h^{2}{\rm Ei}\left (\frac{3 \, \log \left (d\right )}{p} + \frac{3 \, \log \left (c\right )}{p q} + \frac{3 \, a}{b p q} + 3 \, \log \left (f x + e\right )\right ) e^{\left (-\frac{3 \, a}{b p q}\right )}}{b c^{\frac{3}{p q}} d^{\frac{3}{p}} f^{3} p q} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

g^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))/(b*c^(1/(p*q))*d^(1/p)*f*p*q) - 2*g*
h*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q) + 1)/(b*c^(1/(p*q))*d^(1/p)*f^2*p*q) +
2*g*h*Ei(2*log(d)/p + 2*log(c)/(p*q) + 2*a/(b*p*q) + 2*log(f*x + e))*e^(-2*a/(b*p*q))/(b*c^(2/(p*q))*d^(2/p)*f
^2*p*q) + h^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q) + 2)/(b*c^(1/(p*q))*d^(1/p)
*f^3*p*q) - 2*h^2*Ei(2*log(d)/p + 2*log(c)/(p*q) + 2*a/(b*p*q) + 2*log(f*x + e))*e^(-2*a/(b*p*q) + 1)/(b*c^(2/
(p*q))*d^(2/p)*f^3*p*q) + h^2*Ei(3*log(d)/p + 3*log(c)/(p*q) + 3*a/(b*p*q) + 3*log(f*x + e))*e^(-3*a/(b*p*q))/
(b*c^(3/(p*q))*d^(3/p)*f^3*p*q)