3.102 \(\int \frac {1}{(a+b \log (c (d+e x)^n))^3} \, dx\)

Optimal. Leaf size=135 \[ \frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{2 b^3 e n^3}-\frac {d+e x}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \]

[Out]

1/2*(e*x+d)*Ei((a+b*ln(c*(e*x+d)^n))/b/n)/b^3/e/exp(a/b/n)/n^3/((c*(e*x+d)^n)^(1/n))+1/2*(-e*x-d)/b/e/n/(a+b*l
n(c*(e*x+d)^n))^2+1/2*(-e*x-d)/b^2/e/n^2/(a+b*ln(c*(e*x+d)^n))

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Rubi [A]  time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2389, 2297, 2300, 2178} \[ \frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{2 b^3 e n^3}-\frac {d+e x}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^(-3),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(2*b^3*e*E^(a/(b*n))*n^3*(c*(d + e*x)^n)^n^(-1)) -
(d + e*x)/(2*b*e*n*(a + b*Log[c*(d + e*x)^n])^2) - (d + e*x)/(2*b^2*e*n^2*(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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx,x,d+e x\right )}{e}\\ &=-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx,x,d+e x\right )}{2 b e n}\\ &=-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2}-\frac {d+e x}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{2 b^2 e n^2}\\ &=-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2}-\frac {d+e x}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac {\left ((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 )}{2 b^2 e n^3}\\ &=\frac {e^{-\frac {a}{b n}} (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 )}{2 b^3 e n^3}-\frac {d+e x}{2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2}-\frac {d+e x}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.87 \[ \frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{2 b^3 e n^3}-\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )+b n\right )}{2 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^(-3),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(2*b^3*e*E^(a/(b*n))*n^3*(c*(d + e*x)^n)^n^(-1)) -
((d + e*x)*(a + b*n + b*Log[c*(d + e*x)^n]))/(2*b^2*e*n^2*(a + b*Log[c*(d + e*x)^n])^2)

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fricas [B]  time = 0.49, size = 263, normalized size = 1.95 \[ -\frac {{\left ({\left (b^{2} d n^{2} + a b d n + {\left (b^{2} e n^{2} + a b e n\right )} x + {\left (b^{2} e n^{2} x + b^{2} d n^{2}\right )} \log \left (e x + d\right ) + {\left (b^{2} e n x + b^{2} d n\right )} \log \relax (c)\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )} - {\left (b^{2} n^{2} \log \left (e x + d\right )^{2} + b^{2} \log \relax (c)^{2} + 2 \, a b \log \relax (c) + a^{2} + 2 \, {\left (b^{2} n \log \relax (c) + a b n\right )} \log \left (e x + d\right )\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {b \log \relax (c) + a}{b n}\right )}}{2 \, {\left (b^{5} e n^{5} \log \left (e x + d\right )^{2} + b^{5} e n^{3} \log \relax (c)^{2} + 2 \, a b^{4} e n^{3} \log \relax (c) + a^{2} b^{3} e n^{3} + 2 \, {\left (b^{5} e n^{4} \log \relax (c) + a b^{4} e n^{4}\right )} \log \left (e x + d\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^3,x, algorithm="fricas")

[Out]

-1/2*((b^2*d*n^2 + a*b*d*n + (b^2*e*n^2 + a*b*e*n)*x + (b^2*e*n^2*x + b^2*d*n^2)*log(e*x + d) + (b^2*e*n*x + b
^2*d*n)*log(c))*e^((b*log(c) + a)/(b*n)) - (b^2*n^2*log(e*x + d)^2 + b^2*log(c)^2 + 2*a*b*log(c) + a^2 + 2*(b^
2*n*log(c) + a*b*n)*log(e*x + d))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n))))*e^(-(b*log(c) + a)/(b*n))/
(b^5*e*n^5*log(e*x + d)^2 + b^5*e*n^3*log(c)^2 + 2*a*b^4*e*n^3*log(c) + a^2*b^3*e*n^3 + 2*(b^5*e*n^4*log(c) +
a*b^4*e*n^4)*log(e*x + d))

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giac [B]  time = 0.26, size = 1322, normalized size = 9.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^3,x, algorithm="giac")

[Out]

1/2*b^2*n^2*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(x*e + d)^2/((b^5*n^5*e*log(x*e + d)^2 + 2*b
^5*n^4*e*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^
3*n^3*e)*c^(1/n)) - 1/2*(x*e + d)*b^2*n^2*log(x*e + d)/(b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x*e + d)*lo
g(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e) + b^2*n*Ei(log(
c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(x*e + d)*log(c)/((b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x
*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)*c^(1/
n)) - 1/2*(x*e + d)*b^2*n^2/(b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*
e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e) + a*b*n*Ei(log(c)/n + a/(b*n) + log(x*e +
d))*e^(-a/(b*n))*log(x*e + d)/((b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log
(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)*c^(1/n)) - 1/2*(x*e + d)*b^2*n*log(c)/(
b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 +
2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e) + 1/2*b^2*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(c)^2/(
(b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2
+ 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)*c^(1/n)) - 1/2*(x*e + d)*a*b*n/(b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e
*log(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)
+ a*b*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(c)/((b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log(
x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)*c^(1
/n)) + 1/2*a^2*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))/((b^5*n^5*e*log(x*e + d)^2 + 2*b^5*n^4*e*log
(x*e + d)*log(c) + 2*a*b^4*n^4*e*log(x*e + d) + b^5*n^3*e*log(c)^2 + 2*a*b^4*n^3*e*log(c) + a^2*b^3*n^3*e)*c^(
1/n))

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maple [C]  time = 0.05, size = 734, normalized size = 5.44 \[ -\frac {\left (e x +d \right ) c^{-\frac {1}{n}} \left (\left (e x +d \right )^{n}\right )^{-\frac {1}{n}} \Ei \left (1, -\ln \left (e x +d \right )-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 b \ln \relax (c )+2 a +2 \left (-n \ln \left (e x +d \right )+\ln \left (\left (e x +d \right )^{n}\right )\right ) b}{2 b n}\right ) {\mathrm e}^{-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 a}{2 b n}}}{2 b^{3} e \,n^{3}}-\frac {-i \pi b e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b e x \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b e x \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}-i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b d \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b d \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 b e n x +2 b e x \ln \relax (c )+2 b e x \ln \left (\left (e x +d \right )^{n}\right )+2 a e x +2 b d n +2 b d \ln \relax (c )+2 b d \ln \left (\left (e x +d \right )^{n}\right )+2 a d}{\left (-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 b \ln \relax (c )+2 b \ln \left (\left (e x +d \right )^{n}\right )+2 a \right )^{2} b^{2} e \,n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*ln(c*(e*x+d)^n)+a)^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (d n + d \log \relax (c)\right )} b + a d + {\left ({\left (e n + e \log \relax (c)\right )} b + a e\right )} x + {\left (b e x + b d\right )} \log \left ({\left (e x + d\right )}^{n}\right )}{2 \, {\left (b^{4} e n^{2} \log \left ({\left (e x + d\right )}^{n}\right )^{2} + b^{4} e n^{2} \log \relax (c)^{2} + 2 \, a b^{3} e n^{2} \log \relax (c) + a^{2} b^{2} e n^{2} + 2 \, {\left (b^{4} e n^{2} \log \relax (c) + a b^{3} e n^{2}\right )} \log \left ({\left (e x + d\right )}^{n}\right )\right )}} + \int \frac {1}{2 \, {\left (b^{3} n^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^3,x, algorithm="maxima")

[Out]

-1/2*((d*n + d*log(c))*b + a*d + ((e*n + e*log(c))*b + a*e)*x + (b*e*x + b*d)*log((e*x + d)^n))/(b^4*e*n^2*log
((e*x + d)^n)^2 + b^4*e*n^2*log(c)^2 + 2*a*b^3*e*n^2*log(c) + a^2*b^2*e*n^2 + 2*(b^4*e*n^2*log(c) + a*b^3*e*n^
2)*log((e*x + d)^n)) + integrate(1/2/(b^3*n^2*log((e*x + d)^n) + b^3*n^2*log(c) + a*b^2*n^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*(d + e*x)^n))^3,x)

[Out]

int(1/(a + b*log(c*(d + e*x)^n))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(e*x+d)**n))**3,x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**(-3), x)

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