3.209 \(\int (d+e x)^m \log (c (a+\frac{b}{x})^p) \, dx\)

Optimal. Leaf size=135 \[ \frac{(d+e x)^{m+1} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (m+1)}+\frac{a p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{a (d+e x)}{a d-b e}\right )}{e (m+1) (m+2) (a d-b e)}-\frac{p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )}{d e \left (m^2+3 m+2\right )} \]

[Out]

(a*p*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (a*(d + e*x))/(a*d - b*e)])/(e*(a*d - b*e)*(1 + m)*(
2 + m)) - (p*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, 1 + (e*x)/d])/(d*e*(2 + 3*m + m^2)) + ((d +
e*x)^(1 + m)*Log[c*(a + b/x)^p])/(e*(1 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.0926796, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2463, 514, 86, 65, 68} \[ \frac{(d+e x)^{m+1} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (m+1)}+\frac{a p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{a (d+e x)}{a d-b e}\right )}{e (m+1) (m+2) (a d-b e)}-\frac{p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )}{d e \left (m^2+3 m+2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*Log[c*(a + b/x)^p],x]

[Out]

(a*p*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (a*(d + e*x))/(a*d - b*e)])/(e*(a*d - b*e)*(1 + m)*(
2 + m)) - (p*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, 1 + (e*x)/d])/(d*e*(2 + 3*m + m^2)) + ((d +
e*x)^(1 + m)*Log[c*(a + b/x)^p])/(e*(1 + m))

Rule 2463

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

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 65

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int (d+e x)^m \log \left (c \left (a+\frac{b}{x}\right )^p\right ) \, dx &=\frac{(d+e x)^{1+m} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (1+m)}+\frac{(b p) \int \frac{(d+e x)^{1+m}}{\left (a+\frac{b}{x}\right ) x^2} \, dx}{e (1+m)}\\ &=\frac{(d+e x)^{1+m} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (1+m)}+\frac{(b p) \int \frac{(d+e x)^{1+m}}{x (b+a x)} \, dx}{e (1+m)}\\ &=\frac{(d+e x)^{1+m} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (1+m)}+\frac{p \int \frac{(d+e x)^{1+m}}{x} \, dx}{e (1+m)}-\frac{(a p) \int \frac{(d+e x)^{1+m}}{b+a x} \, dx}{e (1+m)}\\ &=\frac{a p (d+e x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac{a (d+e x)}{a d-b e}\right )}{e (a d-b e) (1+m) (2+m)}-\frac{p (d+e x)^{2+m} \, _2F_1\left (1,2+m;3+m;1+\frac{e x}{d}\right )}{d e \left (2+3 m+m^2\right )}+\frac{(d+e x)^{1+m} \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0826814, size = 123, normalized size = 0.91 \[ \frac{(d+e x)^{m+1} \left ((a d-b e) \left (p (d+e x) \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )-d (m+2) \log \left (c \left (a+\frac{b}{x}\right )^p\right )\right )-a d p (d+e x) \, _2F_1\left (1,m+2;m+3;\frac{a (d+e x)}{a d-b e}\right )\right )}{d e (m+1) (m+2) (b e-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(-(a*d*p*(d + e*x)*Hypergeometric2F1[1, 2 + m, 3 + m, (a*(d + e*x))/(a*d - b*e)]) + (a*d -
b*e)*(p*(d + e*x)*Hypergeometric2F1[1, 2 + m, 3 + m, 1 + (e*x)/d] - d*(2 + m)*Log[c*(a + b/x)^p])))/(d*e*(-(a*
d) + b*e)*(1 + m)*(2 + m))

________________________________________________________________________________________

Maple [F]  time = 2.635, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m}\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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