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*(e*x+d)^(2+m)*hypergeom([1, 2+m],[3+m],a*(e*x+d)/(a*d-b*e))/e/(a*d-b*e)/(1+m)/(2+m)-p*(e*x+d)^(2+m)*hyperg
eom([1, 2+m],[3+m],1+e*x/d)/d/e/(m^2+3*m+2)+(e*x+d)^(1+m)*ln(c*(a+b/x)^p)/e/(1+m)

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Rubi [A]  time = 0.09, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x + d\right )}^{m} \log \left (c \left (\frac {a x + b}{x}\right )^{p}\right ), x\right ) \]

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)

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

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)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )\, dx \]

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)

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

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]

(e*x + d)*(e*x + d)^m*log((a*x + b)^p)/(e*(m + 1)) - integrate(-(b*e*(m + 1)*log(c) - a*d*p + (e*(m + 1)*log(c
) - e*p)*a*x - (a*e*(m + 1)*x + b*e*(m + 1))*log(x^p))*(e*x + d)^m/(a*e*(m + 1)*x + b*e*(m + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x)^p)*(d + e*x)^m,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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