3.192 \(\int \frac{(h+i x)^3}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx\)

Optimal. Leaf size=177 \[ \frac{3 i^2 e^{-\frac{2 a}{b}} (f h-e i) \text{Ei}\left (\frac{2 (a+b \log (c (e+f x)))}{b}\right )}{b c^2 d f^4}+\frac{i^3 e^{-\frac{3 a}{b}} \text{Ei}\left (\frac{3 (a+b \log (c (e+f x)))}{b}\right )}{b c^3 d f^4}+\frac{3 i e^{-\frac{a}{b}} (f h-e i)^2 \text{Ei}\left (\frac{a+b \log (c (e+f x))}{b}\right )}{b c d f^4}+\frac{(f h-e i)^3 \log (a+b \log (c (e+f x)))}{b d f^4} \]

[Out]

(3*i*(f*h - e*i)^2*ExpIntegralEi[(a + b*Log[c*(e + f*x)])/b])/(b*c*d*E^(a/b)*f^4) + (3*i^2*(f*h - e*i)*ExpInte
gralEi[(2*(a + b*Log[c*(e + f*x)]))/b])/(b*c^2*d*E^((2*a)/b)*f^4) + (i^3*ExpIntegralEi[(3*(a + b*Log[c*(e + f*
x)]))/b])/(b*c^3*d*E^((3*a)/b)*f^4) + ((f*h - e*i)^3*Log[a + b*Log[c*(e + f*x)]])/(b*d*f^4)

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Rubi [A]  time = 0.484022, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2411, 12, 2353, 2299, 2178, 2302, 29, 2309} \[ \frac{3 i^2 e^{-\frac{2 a}{b}} (f h-e i) \text{Ei}\left (\frac{2 (a+b \log (c (e+f x)))}{b}\right )}{b c^2 d f^4}+\frac{i^3 e^{-\frac{3 a}{b}} \text{Ei}\left (\frac{3 (a+b \log (c (e+f x)))}{b}\right )}{b c^3 d f^4}+\frac{3 i e^{-\frac{a}{b}} (f h-e i)^2 \text{Ei}\left (\frac{a+b \log (c (e+f x))}{b}\right )}{b c d f^4}+\frac{(f h-e i)^3 \log (a+b \log (c (e+f x)))}{b d f^4} \]

Antiderivative was successfully verified.

[In]

Int[(h + i*x)^3/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

(3*i*(f*h - e*i)^2*ExpIntegralEi[(a + b*Log[c*(e + f*x)])/b])/(b*c*d*E^(a/b)*f^4) + (3*i^2*(f*h - e*i)*ExpInte
gralEi[(2*(a + b*Log[c*(e + f*x)]))/b])/(b*c^2*d*E^((2*a)/b)*f^4) + (i^3*ExpIntegralEi[(3*(a + b*Log[c*(e + f*
x)]))/b])/(b*c^3*d*E^((3*a)/b)*f^4) + ((f*h - e*i)^3*Log[a + b*Log[c*(e + f*x)]])/(b*d*f^4)

Rule 2411

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]

Rule 12

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

Rule 2353

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 2299

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 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 2302

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 29

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

Rule 2309

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

Rubi steps

\begin{align*} \int \frac{(h+192 x)^3}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{-192 e+f h}{f}+\frac{192 x}{f}\right )^3}{d x (a+b \log (c x))} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{-192 e+f h}{f}+\frac{192 x}{f}\right )^3}{x (a+b \log (c x))} \, dx,x,e+f x\right )}{d f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{576 (192 e-f h)^2}{f^3 (a+b \log (c x))}-\frac{(192 e-f h)^3}{f^3 x (a+b \log (c x))}-\frac{110592 (192 e-f h) x}{f^3 (a+b \log (c x))}+\frac{7077888 x^2}{f^3 (a+b \log (c x))}\right ) \, dx,x,e+f x\right )}{d f}\\ &=\frac{7077888 \operatorname{Subst}\left (\int \frac{x^2}{a+b \log (c x)} \, dx,x,e+f x\right )}{d f^4}-\frac{(110592 (192 e-f h)) \operatorname{Subst}\left (\int \frac{x}{a+b \log (c x)} \, dx,x,e+f x\right )}{d f^4}+\frac{\left (576 (192 e-f h)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b \log (c x)} \, dx,x,e+f x\right )}{d f^4}-\frac{(192 e-f h)^3 \operatorname{Subst}\left (\int \frac{1}{x (a+b \log (c x))} \, dx,x,e+f x\right )}{d f^4}\\ &=\frac{7077888 \operatorname{Subst}\left (\int \frac{e^{3 x}}{a+b x} \, dx,x,\log (c (e+f x))\right )}{c^3 d f^4}-\frac{(110592 (192 e-f h)) \operatorname{Subst}\left (\int \frac{e^{2 x}}{a+b x} \, dx,x,\log (c (e+f x))\right )}{c^2 d f^4}+\frac{\left (576 (192 e-f h)^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{a+b x} \, dx,x,\log (c (e+f x))\right )}{c d f^4}-\frac{(192 e-f h)^3 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,a+b \log (c (e+f x))\right )}{b d f^4}\\ &=\frac{576 e^{-\frac{a}{b}} (192 e-f h)^2 \text{Ei}\left (\frac{a+b \log (c (e+f x))}{b}\right )}{b c d f^4}-\frac{110592 e^{-\frac{2 a}{b}} (192 e-f h) \text{Ei}\left (\frac{2 (a+b \log (c (e+f x)))}{b}\right )}{b c^2 d f^4}+\frac{7077888 e^{-\frac{3 a}{b}} \text{Ei}\left (\frac{3 (a+b \log (c (e+f x)))}{b}\right )}{b c^3 d f^4}-\frac{(192 e-f h)^3 \log (a+b \log (c (e+f x)))}{b d f^4}\\ \end{align*}

Mathematica [A]  time = 0.482417, size = 279, normalized size = 1.58 \[ \frac{e^{-\frac{3 a}{b}} \left (3 c^3 e^2 f h i^2 e^{\frac{3 a}{b}} \log (a+b \log (c (e+f x)))+c^3 \left (-e^3\right ) i^3 e^{\frac{3 a}{b}} \log (a+b \log (c (e+f x)))+3 c^2 i e^{\frac{2 a}{b}} (f h-e i)^2 \text{Ei}\left (\frac{a}{b}+\log (c (e+f x))\right )-3 c^3 e f^2 h^2 i e^{\frac{3 a}{b}} \log (a+b \log (c (e+f x)))+c^3 f^3 h^3 e^{\frac{3 a}{b}} \log (f (a+b \log (c (e+f x))))+3 c f h i^2 e^{a/b} \text{Ei}\left (\frac{2 (a+b \log (c (e+f x)))}{b}\right )-3 c e i^3 e^{a/b} \text{Ei}\left (2 \left (\frac{a}{b}+\log (c (e+f x))\right )\right )+i^3 \text{Ei}\left (3 \left (\frac{a}{b}+\log (c (e+f x))\right )\right )\right )}{b c^3 d f^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(h + i*x)^3/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

(3*c^2*E^((2*a)/b)*i*(f*h - e*i)^2*ExpIntegralEi[a/b + Log[c*(e + f*x)]] - 3*c*e*E^(a/b)*i^3*ExpIntegralEi[2*(
a/b + Log[c*(e + f*x)])] + i^3*ExpIntegralEi[3*(a/b + Log[c*(e + f*x)])] + 3*c*E^(a/b)*f*h*i^2*ExpIntegralEi[(
2*(a + b*Log[c*(e + f*x)]))/b] - 3*c^3*e*E^((3*a)/b)*f^2*h^2*i*Log[a + b*Log[c*(e + f*x)]] + 3*c^3*e^2*E^((3*a
)/b)*f*h*i^2*Log[a + b*Log[c*(e + f*x)]] - c^3*e^3*E^((3*a)/b)*i^3*Log[a + b*Log[c*(e + f*x)]] + c^3*E^((3*a)/
b)*f^3*h^3*Log[f*(a + b*Log[c*(e + f*x)])])/(b*c^3*d*E^((3*a)/b)*f^4)

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Maple [F]  time = 0.721, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ix+h \right ) ^{3}}{ \left ( dfx+de \right ) \left ( a+b\ln \left ( c \left ( fx+e \right ) \right ) \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x+h)^3/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x)

[Out]

int((i*x+h)^3/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{h^{3} \log \left (\frac{b \log \left (f x + e\right ) + b \log \left (c\right ) + a}{b}\right )}{b d f} + \int \frac{i^{3} x^{3} + 3 \, h i^{2} x^{2} + 3 \, h^{2} i x}{b d e \log \left (c\right ) + a d e +{\left (b d f \log \left (c\right ) + a d f\right )} x +{\left (b d f x + b d e\right )} \log \left (f x + e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^3/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="maxima")

[Out]

h^3*log((b*log(f*x + e) + b*log(c) + a)/b)/(b*d*f) + integrate((i^3*x^3 + 3*h*i^2*x^2 + 3*h^2*i*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)

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Fricas [A]  time = 1.65964, size = 562, normalized size = 3.18 \begin{align*} \frac{{\left (i^{3} \logintegral \left ({\left (c^{3} f^{3} x^{3} + 3 \, c^{3} e f^{2} x^{2} + 3 \, c^{3} e^{2} f x + c^{3} e^{3}\right )} e^{\left (\frac{3 \, a}{b}\right )}\right ) +{\left (c^{3} f^{3} h^{3} - 3 \, c^{3} e f^{2} h^{2} i + 3 \, c^{3} e^{2} f h i^{2} - c^{3} e^{3} i^{3}\right )} e^{\left (\frac{3 \, a}{b}\right )} \log \left (b \log \left (c f x + c e\right ) + a\right ) + 3 \,{\left (c f h i^{2} - c e i^{3}\right )} e^{\frac{a}{b}} \logintegral \left ({\left (c^{2} f^{2} x^{2} + 2 \, c^{2} e f x + c^{2} e^{2}\right )} e^{\left (\frac{2 \, a}{b}\right )}\right ) + 3 \,{\left (c^{2} f^{2} h^{2} i - 2 \, c^{2} e f h i^{2} + c^{2} e^{2} i^{3}\right )} e^{\left (\frac{2 \, a}{b}\right )} \logintegral \left ({\left (c f x + c e\right )} e^{\frac{a}{b}}\right )\right )} e^{\left (-\frac{3 \, a}{b}\right )}}{b c^{3} d f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^3/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="fricas")

[Out]

(i^3*log_integral((c^3*f^3*x^3 + 3*c^3*e*f^2*x^2 + 3*c^3*e^2*f*x + c^3*e^3)*e^(3*a/b)) + (c^3*f^3*h^3 - 3*c^3*
e*f^2*h^2*i + 3*c^3*e^2*f*h*i^2 - c^3*e^3*i^3)*e^(3*a/b)*log(b*log(c*f*x + c*e) + a) + 3*(c*f*h*i^2 - c*e*i^3)
*e^(a/b)*log_integral((c^2*f^2*x^2 + 2*c^2*e*f*x + c^2*e^2)*e^(2*a/b)) + 3*(c^2*f^2*h^2*i - 2*c^2*e*f*h*i^2 +
c^2*e^2*i^3)*e^(2*a/b)*log_integral((c*f*x + c*e)*e^(a/b)))*e^(-3*a/b)/(b*c^3*d*f^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{h^{3}}{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^{3} x^{3}}{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{3 h i^{2} x^{2}}{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{3 h^{2} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)**3/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x)

[Out]

(Integral(h**3/(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**3*x**3/(a*e + a
*f*x + b*e*log(c*e + c*f*x) + b*f*x*log(c*e + c*f*x)), x) + Integral(3*h*i**2*x**2/(a*e + a*f*x + b*e*log(c*e
+ c*f*x) + b*f*x*log(c*e + c*f*x)), x) + Integral(3*h**2*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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i x + h\right )}^{3}}{{\left (d f x + d e\right )}{\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^3/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="giac")

[Out]

integrate((i*x + h)^3/((d*f*x + d*e)*(b*log((f*x + e)*c) + a)), x)