\(\int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx\) [194]

Optimal result

Integrand size = 30, antiderivative size = 71 \[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {e^{-\frac {a}{b}} i \operatorname {ExpIntegralEi}\left (\frac {a+b \log (c (e+f x))}{b}\right )}{b c d f^2}+\frac {(f h-e i) \log (a+b \log (c (e+f x)))}{b d f^2} \]

[Out]

i*Ei((a+b*ln(c*(f*x+e)))/b)/b/c/d/exp(a/b)/f^2+(-e*i+f*h)*ln(a+b*ln(c*(f*x+e)))/b/d/f^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2458, 12, 2395, 2336, 2209, 2339, 29} \[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {i e^{-\frac {a}{b}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log (c (e+f x))}{b}\right )}{b c d f^2}+\frac {(f h-e i) \log (a+b \log (c (e+f x)))}{b d f^2} \]

[In]

Int[(h + i*x)/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2209

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

Rule 2336

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

Rule 2339

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

Rule 2395

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

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {e^{-\frac {a}{b}} \left (i \operatorname {ExpIntegralEi}\left (\frac {a}{b}+\log (c (e+f x))\right )+c e^{a/b} (f h-e i) \log (a+b \log (c (e+f x)))\right )}{b c d f^2} \]

[In]

Integrate[(h + i*x)/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

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

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(-\frac {\frac {i \,{\mathrm e}^{-\frac {a}{b}} \operatorname {Ei}_{1}\left (-\ln \left (c f x +c e \right )-\frac {a}{b}\right )}{b}-\frac {h c f \ln \left (a +b \ln \left (c f x +c e \right )\right )}{b}+\frac {c e i \ln \left (a +b \ln \left (c f x +c e \right )\right )}{b}}{c \,f^{2} d}\)	\(88\)
default	\(-\frac {\frac {i \,{\mathrm e}^{-\frac {a}{b}} \operatorname {Ei}_{1}\left (-\ln \left (c f x +c e \right )-\frac {a}{b}\right )}{b}-\frac {h c f \ln \left (a +b \ln \left (c f x +c e \right )\right )}{b}+\frac {c e i \ln \left (a +b \ln \left (c f x +c e \right )\right )}{b}}{c \,f^{2} d}\)	\(88\)
risch	\(-\frac {e i \ln \left (a +b \ln \left (c f x +c e \right )\right )}{f^{2} d b}+\frac {h \ln \left (a +b \ln \left (c f x +c e \right )\right )}{f d b}-\frac {i \,{\mathrm e}^{-\frac {a}{b}} \operatorname {Ei}_{1}\left (-\ln \left (c f x +c e \right )-\frac {a}{b}\right )}{c \,f^{2} d b}\)	\(96\)

[In]

int((i*x+h)/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x,method=_RETURNVERBOSE)

[Out]

-1/c/f^2/d*(i/b*exp(-a/b)*Ei(1,-ln(c*f*x+c*e)-a/b)-h*c*f*ln(a+b*ln(c*f*x+c*e))/b+c*e*i*ln(a+b*ln(c*f*x+c*e))/b
)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {{\left ({\left (c f h - c e i\right )} e^{\frac {a}{b}} \log \left (b \log \left (c f x + c e\right ) + a\right ) + i \operatorname {log\_integral}\left ({\left (c f x + c e\right )} e^{\frac {a}{b}}\right )\right )} e^{\left (-\frac {a}{b}\right )}}{b c d f^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\int \frac {h}{a e + a f x + b e \log {\left (c e + c f x \right )} + b f x \log {\left (c e + c f x \right )}}\, dx + \int \frac {i x}{a e + a f x + b e \log {\left (c e + c f x \right )} + b f x \log {\left (c e + c f x \right )}}\, dx}{d} \]

[In]

integrate((i*x+h)/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x)

[Out]

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

Maxima [F]

\[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\int { \frac {i x + h}{{\left (d f x + d e\right )} {\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}} \,d x } \]

[In]

integrate((i*x+h)/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="maxima")

[Out]

i*integrate(x/(b*d*e*log(c) + a*d*e + (b*d*f*log(c) + a*d*f)*x + (b*d*f*x + b*d*e)*log(f*x + e)), x) + h*log((
b*log(f*x + e) + b*log(c) + a)/b)/(b*d*f)

Giac [F]

\[ \int \frac {h+i x}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\int { \frac {i x + h}{{\left (d f x + d e\right )} {\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}} \,d x } \]

[In]

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

[Out]

integrate((i*x + h)/((d*f*x + d*e)*(b*log((f*x + e)*c) + a)), x)

Mupad [F(-1)]

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

[In]

int((h + i*x)/((d*e + d*f*x)*(a + b*log(c*(e + f*x)))),x)

[Out]

int((h + i*x)/((d*e + d*f*x)*(a + b*log(c*(e + f*x)))), x)