3.1.90 \(\int \frac {f+g x}{a+b \log (c (d+e x)^n)} \, dx\) [90]

Optimal. Leaf size=139 \[ \frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \]

[Out]

(-d*g+e*f)*(e*x+d)*Ei((a+b*ln(c*(e*x+d)^n))/b/n)/b/e^2/exp(a/b/n)/n/((c*(e*x+d)^n)^(1/n))+g*(e*x+d)^2*Ei(2*(a+
b*ln(c*(e*x+d)^n))/b/n)/b/e^2/exp(2*a/b/n)/n/((c*(e*x+d)^n)^(2/n))

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Rubi [A]
time = 0.12, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2446, 2436, 2337, 2209, 2437, 2347} \begin {gather*} \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2337

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2436

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 2437

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 2446

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]

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 126, normalized size = 0.91 \begin {gather*} \frac {e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )}{b e^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.82, size = 937, normalized size = 6.74

method result size
risch \(\text {Expression too large to display}\) \(937\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)/(a+b*ln(c*(e*x+d)^n)),x,method=_RETURNVERBOSE)

[Out]

-1/e^2*g/b/n*(e*x+d)^2*c^(-2/n)*((e*x+d)^n)^(-2/n)*exp(-(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^
n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+
d)^n)^3+2*a)/b/n)*Ei(1,-2*ln(e*x+d)-(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*
csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*
b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)-1/e*f/b/n*(e*x+d)*c^(-1/n)*((e*x+d)^n)^(-1/n)*exp(-1/2*(-I*b*Pi*csgn(I
*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn
(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*a)/b/n)*Ei(1,-ln(e*x+d)-1/2*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)
^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-
I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)+1/e^2*d*g/b/n*(e*x+d)*c^(-1/n
)*((e*x+d)^n)^(-1/n)*exp(-1/2*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I
*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*a)/b/n)*Ei(1,-ln
(e*x+d)-1/2*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*
b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*
x+d))+2*a)/b/n)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 108, normalized size = 0.78 \begin {gather*} -\frac {{\left ({\left (d g - f e\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (x e + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right ) - g \operatorname {log\_integral}\left ({\left (x^{2} e^{2} + 2 \, d x e + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n} - 2\right )}}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f + g x}{a + b \log {\left (c \left (d + e x\right )^{n} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(a+b*ln(c*(e*x+d)**n)),x)

[Out]

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

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Giac [A]
time = 5.46, size = 159, normalized size = 1.14 \begin {gather*} -\frac {d g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac {a}{b n} - 2\right )}}{b c^{\left (\frac {1}{n}\right )} n} + \frac {f {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac {a}{b n} - 1\right )}}{b c^{\left (\frac {1}{n}\right )} n} + \frac {g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (x e + d\right )\right ) e^{\left (-\frac {2 \, a}{b n} - 2\right )}}{b c^{\frac {2}{n}} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*g*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n) - 2)/(b*c^(1/n)*n) + f*Ei(log(c)/n + a/(b*n) + log(x*e
+ d))*e^(-a/(b*n) - 1)/(b*c^(1/n)*n) + g*Ei(2*log(c)/n + 2*a/(b*n) + 2*log(x*e + d))*e^(-2*a/(b*n) - 2)/(b*c^(
2/n)*n)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {f+g\,x}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)/(a + b*log(c*(d + e*x)^n)),x)

[Out]

int((f + g*x)/(a + b*log(c*(d + e*x)^n)), x)

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