3.96 \(\int \frac {f+g x}{(a+b \log (c (d+e x)^n))^2} \, dx\)

Optimal. Leaf size=177 \[ \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e^2 n^2}+\frac {2 g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b^2 e^2 n^2}-\frac {(d+e x) (f+g x)}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2400, 2399, 2389, 2300, 2178, 2390, 2310} \[ \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e^2 n^2}+\frac {2 g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b^2 e^2 n^2}-\frac {(d+e x) (f+g x)}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*f - d*g)*(d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b^2*e^2*E^(a/(b*n))*n^2*(c*(d + e*x)^
n)^n^(-1)) + (2*g*(d + e*x)^2*ExpIntegralEi[(2*(a + b*Log[c*(d + e*x)^n]))/(b*n)])/(b^2*e^2*E^((2*a)/(b*n))*n^
2*(c*(d + e*x)^n)^(2/n)) - ((d + e*x)*(f + g*x))/(b*e*n*(a + b*Log[c*(d + e*x)^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 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 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 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 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 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 2400

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 208, normalized size = 1.18 \[ -\frac {e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (-e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )-2 g (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )+b e n e^{\frac {2 a}{b n}} (f+g x) \left (c (d+e x)^n\right )^{2/n}\right )}{b^2 e^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 239, normalized size = 1.35 \[ \frac {{\left ({\left (a e f - a d g + {\left (b e f - b d g\right )} n \log \left (e x + d\right ) + {\left (b e f - b d g\right )} \log \relax (c)\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )}\right ) - {\left (b e^{2} g n x^{2} + b d e f n + {\left (b e^{2} f + b d e g\right )} n x\right )} e^{\left (\frac {2 \, {\left (b \log \relax (c) + a\right )}}{b n}\right )} + 2 \, {\left (b g n \log \left (e x + d\right ) + b g \log \relax (c) + a g\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \relax (c) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \relax (c) + a\right )}}{b n}\right )}}{b^{3} e^{2} n^{3} \log \left (e x + d\right ) + b^{3} e^{2} n^{2} \log \relax (c) + a b^{2} e^{2} n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*e*f - a*d*g + (b*e*f - b*d*g)*n*log(e*x + d) + (b*e*f - b*d*g)*log(c))*e^((b*log(c) + a)/(b*n))*log_integr
al((e*x + d)*e^((b*log(c) + a)/(b*n))) - (b*e^2*g*n*x^2 + b*d*e*f*n + (b*e^2*f + b*d*e*g)*n*x)*e^(2*(b*log(c)
+ a)/(b*n)) + 2*(b*g*n*log(e*x + d) + b*g*log(c) + a*g)*log_integral((e^2*x^2 + 2*d*e*x + d^2)*e^(2*(b*log(c)
+ a)/(b*n))))*e^(-2*(b*log(c) + a)/(b*n))/(b^3*e^2*n^3*log(e*x + d) + b^3*e^2*n^2*log(c) + a*b^2*e^2*n^2)

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giac [B]  time = 0.37, size = 984, normalized size = 5.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*e + d)^2*b*g*n*e/(b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3) + (x*e + d)*b*d*g*n*e/(b^
3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3) - b*d*g*n*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e
^(-a/(b*n) + 1)*log(x*e + d)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(1/n)) - (x*e
+ d)*b*f*n*e^2/(b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3) + b*f*n*Ei(log(c)/n + a/(b*n) +
log(x*e + d))*e^(-a/(b*n) + 2)*log(x*e + d)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*
c^(1/n)) + 2*b*g*n*Ei(2*log(c)/n + 2*a/(b*n) + 2*log(x*e + d))*e^(-2*a/(b*n) + 1)*log(x*e + d)/((b^3*n^3*e^3*l
og(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(2/n)) - b*d*g*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a
/(b*n) + 1)*log(c)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(1/n)) - a*d*g*Ei(log(c)
/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n) + 1)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)
*c^(1/n)) + b*f*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n) + 2)*log(c)/((b^3*n^3*e^3*log(x*e + d) + b^3
*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(1/n)) + 2*b*g*Ei(2*log(c)/n + 2*a/(b*n) + 2*log(x*e + d))*e^(-2*a/(b*n) +
1)*log(c)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(2/n)) + a*f*Ei(log(c)/n + a/(b*n
) + log(x*e + d))*e^(-a/(b*n) + 2)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3*log(c) + a*b^2*n^2*e^3)*c^(1/n)) +
2*a*g*Ei(2*log(c)/n + 2*a/(b*n) + 2*log(x*e + d))*e^(-2*a/(b*n) + 1)/((b^3*n^3*e^3*log(x*e + d) + b^3*n^2*e^3
*log(c) + a*b^2*n^2*e^3)*c^(2/n))

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maple [F]  time = 4.83, size = 0, normalized size = 0.00 \[ \int \frac {g x +f}{\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {e g x^{2} + d f + {\left (e f + d g\right )} x}{b^{2} e n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} e n \log \relax (c) + a b e n} + \int \frac {2 \, e g x + e f + d g}{b^{2} e n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} e n \log \relax (c) + a b e n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {f+g\,x}{{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f + g x}{\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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