3.155 \(\int \frac {(d x)^m}{(a+b \log (c x^n))^3} \, dx\)

Optimal. Leaf size=142 \[ \frac {(m+1)^2 (d x)^{m+1} e^{-\frac {a (m+1)}{b n}} \left (c x^n\right )^{-\frac {m+1}{n}} \text {Ei}\left (\frac {(m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 b^3 d n^3}-\frac {(m+1) (d x)^{m+1}}{2 b^2 d n^2 \left (a+b \log \left (c x^n\right )\right )}-\frac {(d x)^{m+1}}{2 b d n \left (a+b \log \left (c x^n\right )\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2306, 2310, 2178} \[ \frac {(m+1)^2 (d x)^{m+1} e^{-\frac {a (m+1)}{b n}} \left (c x^n\right )^{-\frac {m+1}{n}} \text {Ei}\left (\frac {(m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 b^3 d n^3}-\frac {(m+1) (d x)^{m+1}}{2 b^2 d n^2 \left (a+b \log \left (c x^n\right )\right )}-\frac {(d x)^{m+1}}{2 b d n \left (a+b \log \left (c x^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(a + b*Log[c*x^n])^3,x]

[Out]

((1 + m)^2*(d*x)^(1 + m)*ExpIntegralEi[((1 + m)*(a + b*Log[c*x^n]))/(b*n)])/(2*b^3*d*E^((a*(1 + m))/(b*n))*n^3
*(c*x^n)^((1 + m)/n)) - (d*x)^(1 + m)/(2*b*d*n*(a + b*Log[c*x^n])^2) - ((1 + m)*(d*x)^(1 + m))/(2*b^2*d*n^2*(a
+ b*Log[c*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 2306

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.37, size = 113, normalized size = 0.80 \[ \frac {(d x)^m \left ((m+1)^2 x^{-m} \exp \left (-\frac {(m+1) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{b n}\right ) \text {Ei}\left (\frac {(m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )-\frac {b n x \left (a m+a+b (m+1) \log \left (c x^n\right )+b n\right )}{\left (a+b \log \left (c x^n\right )\right )^2}\right )}{2 b^3 n^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 322, normalized size = 2.27 \[ \frac {{\left ({\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} n^{2} \log \relax (x)^{2} + a^{2} m^{2} + 2 \, a^{2} m + {\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} \log \relax (c)^{2} + a^{2} + 2 \, {\left (a b m^{2} + 2 \, a b m + a b\right )} \log \relax (c) + 2 \, {\left ({\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} n \log \relax (c) + {\left (a b m^{2} + 2 \, a b m + a b\right )} n\right )} \log \relax (x)\right )} {\rm Ei}\left (\frac {{\left (b m + b\right )} n \log \relax (x) + a m + {\left (b m + b\right )} \log \relax (c) + a}{b n}\right ) e^{\left (\frac {b m n \log \relax (d) - a m - {\left (b m + b\right )} \log \relax (c) - a}{b n}\right )} - {\left ({\left (b^{2} m + b^{2}\right )} n^{2} x \log \relax (x) + {\left (b^{2} m + b^{2}\right )} n x \log \relax (c) + {\left (b^{2} n^{2} + {\left (a b m + a b\right )} n\right )} x\right )} e^{\left (m \log \relax (d) + m \log \relax (x)\right )}}{2 \, {\left (b^{5} n^{5} \log \relax (x)^{2} + b^{5} n^{3} \log \relax (c)^{2} + 2 \, a b^{4} n^{3} \log \relax (c) + a^{2} b^{3} n^{3} + 2 \, {\left (b^{5} n^{4} \log \relax (c) + a b^{4} n^{4}\right )} \log \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m/(b*log(c*x^n) + a)^3, x)

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maple [F]  time = 1.84, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\left (b \ln \left (c \,x^{n}\right )+a \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a + b*log(c*x^n))^3,x)

[Out]

int((d*x)^m/(a + b*log(c*x^n))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m/(a + b*log(c*x**n))**3, x)

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