3.58 \(\int (f x)^m \log (c (d+\frac{e}{x})^p) \, dx\)

Optimal. Leaf size=67 \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (m+1)}+\frac{e p (f x)^m \, _2F_1\left (1,-m;1-m;-\frac{e}{d x}\right )}{d m (m+1)} \]

[Out]

(e*p*(f*x)^m*Hypergeometric2F1[1, -m, 1 - m, -(e/(d*x))])/(d*m*(1 + m)) + ((f*x)^(1 + m)*Log[c*(d + e/x)^p])/(
f*(1 + m))

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Rubi [A]  time = 0.040617, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2455, 16, 339, 64} \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (m+1)}+\frac{e p (f x)^m \, _2F_1\left (1,-m;1-m;-\frac{e}{d x}\right )}{d m (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e/x)^p],x]

[Out]

(e*p*(f*x)^m*Hypergeometric2F1[1, -m, 1 - m, -(e/(d*x))])/(d*m*(1 + m)) + ((f*x)^(1 + m)*Log[c*(d + e/x)^p])/(
f*(1 + m))

Rule 2455

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int (f x)^m \log \left (c \left (d+\frac{e}{x}\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (1+m)}+\frac{(e p) \int \frac{(f x)^{1+m}}{\left (d+\frac{e}{x}\right ) x^2} \, dx}{f (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (1+m)}+\frac{(e f p) \int \frac{(f x)^{-1+m}}{d+\frac{e}{x}} \, dx}{1+m}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (1+m)}-\frac{\left (e p \left (\frac{1}{x}\right )^m (f x)^m\right ) \operatorname{Subst}\left (\int \frac{x^{-1-m}}{d+e x} \, dx,x,\frac{1}{x}\right )}{1+m}\\ &=\frac{e p (f x)^m \, _2F_1\left (1,-m;1-m;-\frac{e}{d x}\right )}{d m (1+m)}+\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{x}\right )^p\right )}{f (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0159608, size = 56, normalized size = 0.84 \[ \frac{(f x)^m \left (d m x \log \left (c \left (d+\frac{e}{x}\right )^p\right )+e p \, _2F_1\left (1,-m;1-m;-\frac{e}{d x}\right )\right )}{d m (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*Log[c*(d + e/x)^p],x]

[Out]

((f*x)^m*(e*p*Hypergeometric2F1[1, -m, 1 - m, -(e/(d*x))] + d*m*x*Log[c*(d + e/x)^p]))/(d*m*(1 + m))

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Maple [F]  time = 2.666, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m}\ln \left ( c \left ( d+{\frac{e}{x}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*ln(c*(d+e/x)^p),x)

[Out]

int((f*x)^m*ln(c*(d+e/x)^p),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (f x\right )^{m} \log \left (c \left (\frac{d x + e}{x}\right )^{p}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x)^p),x, algorithm="fricas")

[Out]

integral((f*x)^m*log(c*((d*x + e)/x)^p), x)

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Sympy [A]  time = 52.6949, size = 201, normalized size = 3. \begin{align*} e p \left (\begin{cases} \frac{0^{m} \log{\left (d x + e \right )}}{d} & \text{for}\: \left (f = 0 \wedge m \neq -1\right ) \vee f = 0 \\\frac{f^{m} m x^{m} \Phi \left (\frac{e e^{i \pi }}{d x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{d m \Gamma \left (1 - m\right ) + d \Gamma \left (1 - m\right )} & \text{for}\: m > -\infty \wedge m < \infty \wedge m \neq -1 \\\frac{\begin{cases} - \frac{1}{d x} & \text{for}\: e = 0 \\\frac{\begin{cases} \log{\left (d \right )} \log{\left (x \right )} + \operatorname{Li}_{2}\left (\frac{e e^{i \pi }}{d x}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (d \right )} \log{\left (\frac{1}{x} \right )} + \operatorname{Li}_{2}\left (\frac{e e^{i \pi }}{d x}\right ) & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (d \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (d \right )} + \operatorname{Li}_{2}\left (\frac{e e^{i \pi }}{d x}\right ) & \text{otherwise} \end{cases}}{e} & \text{otherwise} \end{cases}}{f} - \frac{\left (\begin{cases} \frac{1}{d x} & \text{for}\: e = 0 \\\frac{\log{\left (d + \frac{e}{x} \right )}}{e} & \text{otherwise} \end{cases}\right ) \log{\left (f x \right )}}{f} & \text{otherwise} \end{cases}\right ) + \left (\begin{cases} 0^{m} x & \text{for}\: f = 0 \\\frac{\begin{cases} \frac{\left (f x\right )^{m + 1}}{m + 1} & \text{for}\: m \neq -1 \\\log{\left (f x \right )} & \text{otherwise} \end{cases}}{f} & \text{otherwise} \end{cases}\right ) \log{\left (c \left (d + \frac{e}{x}\right )^{p} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*ln(c*(d+e/x)**p),x)

[Out]

e*p*Piecewise((0**m*log(d*x + e)/d, Eq(f, 0) | (Eq(f, 0) & Ne(m, -1))), (f**m*m*x**m*lerchphi(e*exp_polar(I*pi
)/(d*x), 1, m*exp_polar(I*pi))*gamma(-m)/(d*m*gamma(1 - m) + d*gamma(1 - m)), (m > -oo) & (m < oo) & Ne(m, -1)
), (Piecewise((-1/(d*x), Eq(e, 0)), (Piecewise((log(d)*log(x) + polylog(2, e*exp_polar(I*pi)/(d*x)), Abs(x) <
1), (-log(d)*log(1/x) + polylog(2, e*exp_polar(I*pi)/(d*x)), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), (
)), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) + polylog(2, e*exp_polar(I*pi)/(d*x)), True))/e,
True))/f - Piecewise((1/(d*x), Eq(e, 0)), (log(d + e/x)/e, True))*log(f*x)/f, True)) + Piecewise((0**m*x, Eq(
f, 0)), (Piecewise(((f*x)**(m + 1)/(m + 1), Ne(m, -1)), (log(f*x), True))/f, True))*log(c*(d + e/x)**p)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{m} \log \left (c{\left (d + \frac{e}{x}\right )}^{p}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x)^p),x, algorithm="giac")

[Out]

integrate((f*x)^m*log(c*(d + e/x)^p), x)