3.3.47 \(\int \frac {(a+b \log (c (d x^m)^n))^p}{x^2} \, dx\) [247]

Optimal. Leaf size=107 \[ -\frac {e^{\frac {a}{b m n}} \left (c \left (d x^m\right )^n\right )^{\frac {1}{m n}} \Gamma \left (1+p,\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right ) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^p \left (\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right )^{-p}}{x} \]

[Out]

-exp(a/b/m/n)*(c*(d*x^m)^n)^(1/m/n)*GAMMA(1+p,(a+b*ln(c*(d*x^m)^n))/b/m/n)*(a+b*ln(c*(d*x^m)^n))^p/x/(((a+b*ln
(c*(d*x^m)^n))/b/m/n)^p)

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Rubi [A]
time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2347, 2212, 2495} \begin {gather*} -\frac {e^{\frac {a}{b m n}} \left (c \left (d x^m\right )^n\right )^{\frac {1}{m n}} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^p \left (\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right )^{-p} \text {Gamma}\left (p+1,\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*x^m)^n])^p/x^2,x]

[Out]

-((E^(a/(b*m*n))*(c*(d*x^m)^n)^(1/(m*n))*Gamma[1 + p, (a + b*Log[c*(d*x^m)^n])/(b*m*n)]*(a + b*Log[c*(d*x^m)^n
])^p)/(x*((a + b*Log[c*(d*x^m)^n])/(b*m*n))^p))

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

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 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 107, normalized size = 1.00 \begin {gather*} -\frac {e^{\frac {a}{b m n}} \left (c \left (d x^m\right )^n\right )^{\frac {1}{m n}} \Gamma \left (1+p,\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right ) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^p \left (\frac {a+b \log \left (c \left (d x^m\right )^n\right )}{b m n}\right )^{-p}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*x^m)^n])^p/x^2,x]

[Out]

-((E^(a/(b*m*n))*(c*(d*x^m)^n)^(1/(m*n))*Gamma[1 + p, (a + b*Log[c*(d*x^m)^n])/(b*m*n)]*(a + b*Log[c*(d*x^m)^n
])^p)/(x*((a + b*Log[c*(d*x^m)^n])/(b*m*n))^p))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \left (d \,x^{m}\right )^{n}\right )\right )^{p}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d*x^m)^n))^p/x^2,x)

[Out]

int((a+b*ln(c*(d*x^m)^n))^p/x^2,x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((d*x^m)^n*c) + a)^p/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c \left (d x^{m}\right )^{n} \right )}\right )^{p}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*x**m)**n))**p/x**2,x)

[Out]

Integral((a + b*log(c*(d*x**m)**n))**p/x**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log((d*x^m)^n*c) + a)^p/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\ln \left (c\,{\left (d\,x^m\right )}^n\right )\right )}^p}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d*x^m)^n))^p/x^2,x)

[Out]

int((a + b*log(c*(d*x^m)^n))^p/x^2, x)

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