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

Optimal. Leaf size=179 \[ \frac{(e+f x) e^{-\frac{a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f^2 p q}+\frac{h (e+f x)^2 e^{-\frac{2 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{2}{p q}} \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^2 p q} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.416135, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {2399, 2389, 2300, 2178, 2390, 2310, 2445} \[ \frac{(e+f x) e^{-\frac{a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f^2 p q}+\frac{h (e+f x)^2 e^{-\frac{2 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{2}{p q}} \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^2 p q} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((f*g - e*h)*(e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f^2*p*q*(c*(d
*(e + f*x)^p)^q)^(1/(p*q))) + (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^2*p*q*(c*(d*(e + f*x)^p)^q)^(2/(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}{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx &=\operatorname{Subst}\left (\int \frac{g+h x}{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}{f \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )}+\frac{h (e+f x)}{f \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 \int \frac{e+f x}{a+b \log \left (c d^q (e+f x)^{p q}\right )} \, dx}{f},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) \int \frac{1}{a+b \log \left (c d^q (e+f x)^{p q}\right )} \, dx}{f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\operatorname{Subst}\left (\frac{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^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) \operatorname{Subst}\left (\int \frac{1}{a+b \log \left (c d^q x^{p q}\right )} \, dx,x,e+f x\right )}{f^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\operatorname{Subst}\left (\frac{\left (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^2 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) (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^2 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) (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^2 p q}+\frac{e^{-\frac{2 a}{b p q}} 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^2 p q}\\ \end{align*}

Mathematica [A]  time = 0.260749, size = 164, normalized size = 0.92 \[ \frac{(e+f x) e^{-\frac{2 a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{2}{p q}} \left (e^{\frac{a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )+h (e+f x) \text{Ei}\left (\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )\right )}{b f^2 p q} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.266, size = 0, normalized size = 0. \begin{align*} \int{\frac{hx+g}{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)/(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.96442, size = 351, normalized size = 1.96 \begin{align*} \frac{{\left ({\left (f g - e h\right )} e^{\left (\frac{b q \log \left (d\right ) + b \log \left (c\right ) + a}{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 ) + h \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 )\right )} e^{\left (-\frac{2 \,{\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}}{b f^{2} p q} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{g + h x}{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)/(a+b*ln(c*(d*(f*x+e)**p)**q)),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.25138, size = 340, normalized size = 1.9 \begin{align*} \frac{g{\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{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{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

g*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) - h*Ei(l
og(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) + 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)