\(\int (d+e x) \log (c (a+\frac {b}{x})^p) \, dx\) [200]

Optimal result

Integrand size = 18, antiderivative size = 78 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b e p x}{2 a}+\frac {(d+e x)^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}+\frac {d^2 p \log (x)}{2 e}-\frac {(a d-b e)^2 p \log (b+a x)}{2 a^2 e} \]

[Out]

1/2*b*e*p*x/a+1/2*(e*x+d)^2*ln(c*(a+b/x)^p)/e+1/2*d^2*p*ln(x)/e-1/2*(a*d-b*e)^2*p*ln(a*x+b)/a^2/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2513, 528, 84} \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {p (a d-b e)^2 \log (a x+b)}{2 a^2 e}+\frac {(d+e x)^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}+\frac {b e p x}{2 a}+\frac {d^2 p \log (x)}{2 e} \]

[In]

Int[(d + e*x)*Log[c*(a + b/x)^p],x]

[Out]

(b*e*p*x)/(2*a) + ((d + e*x)^2*Log[c*(a + b/x)^p])/(2*e) + (d^2*p*Log[x])/(2*e) - ((a*d - b*e)^2*p*Log[b + a*x
])/(2*a^2*e)

Rule 84

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

Rule 528

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 2513

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.09 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b d p \log \left (a+\frac {b}{x}\right )}{a}+d x \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{2} e x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b d p \log (x)}{a}+\frac {1}{2} b e p \left (\frac {x}{a}-\frac {b \log (b+a x)}{a^2}\right ) \]

[In]

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

[Out]

(b*d*p*Log[a + b/x])/a + d*x*Log[c*(a + b/x)^p] + (e*x^2*Log[c*(a + b/x)^p])/2 + (b*d*p*Log[x])/a + (b*e*p*(x/
a - (b*Log[b + a*x])/a^2))/2

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83

method	result	size
parts	\(\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e \,x^{2}}{2}+\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) d x +\frac {p b \left (\frac {e x}{a}+\frac {\left (2 a d -b e \right ) \ln \left (a x +b \right )}{a^{2}}\right )}{2}\)	\(65\)
parallelrisch	\(-\frac {-x^{2} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{2} e +2 \ln \left (x \right ) a b d p -4 \ln \left (a x +b \right ) a b d p +\ln \left (a x +b \right ) b^{2} e p -2 x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{2} d -a b e p x +2 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a b d +b^{2} e p}{2 a^{2}}\)	\(115\)

[In]

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

[Out]

1/2*ln(c*(a+b/x)^p)*e*x^2+ln(c*(a+b/x)^p)*d*x+1/2*p*b*(e/a*x+(2*a*d-b*e)/a^2*ln(a*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b e p x + {\left (2 \, a b d - b^{2} e\right )} p \log \left (a x + b\right ) + {\left (a^{2} e x^{2} + 2 \, a^{2} d x\right )} \log \left (c\right ) + {\left (a^{2} e p x^{2} + 2 \, a^{2} d p x\right )} \log \left (\frac {a x + b}{x}\right )}{2 \, a^{2}} \]

[In]

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

[Out]

1/2*(a*b*e*p*x + (2*a*b*d - b^2*e)*p*log(a*x + b) + (a^2*e*x^2 + 2*a^2*d*x)*log(c) + (a^2*e*p*x^2 + 2*a^2*d*p*
x)*log((a*x + b)/x))/a^2

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} d x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {e x^{2} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{2} + \frac {b d p \log {\left (x + \frac {b}{a} \right )}}{a} + \frac {b e p x}{2 a} - \frac {b^{2} e p \log {\left (x + \frac {b}{a} \right )}}{2 a^{2}} & \text {for}\: a \neq 0 \\d p x + d x \log {\left (c \left (\frac {b}{x}\right )^{p} \right )} + \frac {e p x^{2}}{4} + \frac {e x^{2} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d*x*log(c*(a + b/x)**p) + e*x**2*log(c*(a + b/x)**p)/2 + b*d*p*log(x + b/a)/a + b*e*p*x/(2*a) - b**
2*e*p*log(x + b/a)/(2*a**2), Ne(a, 0)), (d*p*x + d*x*log(c*(b/x)**p) + e*p*x**2/4 + e*x**2*log(c*(b/x)**p)/2,
True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{2} \, b p {\left (\frac {e x}{a} + \frac {{\left (2 \, a d - b e\right )} \log \left (a x + b\right )}{a^{2}}\right )} + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

[In]

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

[Out]

1/2*b*p*(e*x/a + (2*a*d - b*e)*log(a*x + b)/a^2) + 1/2*(e*x^2 + 2*d*x)*log((a + b/x)^p*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (70) = 140\).

Time = 0.45 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.99 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {{\left (2 \, a b^{2} d p - b^{3} e p - \frac {2 \, {\left (a x + b\right )} b^{2} d p}{x}\right )} \log \left (\frac {a x + b}{x}\right )}{a^{2} - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}}} + \frac {a b^{3} e p + 2 \, a^{2} b^{2} d \log \left (c\right ) - a b^{3} e \log \left (c\right ) - \frac {{\left (a x + b\right )} b^{3} e p}{x} - \frac {2 \, {\left (a x + b\right )} a b^{2} d \log \left (c\right )}{x}}{a^{3} - \frac {2 \, {\left (a x + b\right )} a^{2}}{x} + \frac {{\left (a x + b\right )}^{2} a}{x^{2}}} + \frac {{\left (2 \, a b^{2} d p - b^{3} e p\right )} \log \left (-a + \frac {a x + b}{x}\right )}{a^{2}} - \frac {{\left (2 \, a b^{2} d p - b^{3} e p\right )} \log \left (\frac {a x + b}{x}\right )}{a^{2}}}{2 \, b} \]

[In]

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

[Out]

-1/2*((2*a*b^2*d*p - b^3*e*p - 2*(a*x + b)*b^2*d*p/x)*log((a*x + b)/x)/(a^2 - 2*(a*x + b)*a/x + (a*x + b)^2/x^
2) + (a*b^3*e*p + 2*a^2*b^2*d*log(c) - a*b^3*e*log(c) - (a*x + b)*b^3*e*p/x - 2*(a*x + b)*a*b^2*d*log(c)/x)/(a
^3 - 2*(a*x + b)*a^2/x + (a*x + b)^2*a/x^2) + (2*a*b^2*d*p - b^3*e*p)*log(-a + (a*x + b)/x)/a^2 - (2*a*b^2*d*p
- b^3*e*p)*log((a*x + b)/x)/a^2)/b

Mupad [B] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.73 \[ \int (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )\,\left (\frac {e\,x^2}{2}+d\,x\right )-\frac {\ln \left (b+a\,x\right )\,\left (b^2\,e\,p-2\,a\,b\,d\,p\right )}{2\,a^2}+\frac {b\,e\,p\,x}{2\,a} \]

[In]

int(log(c*(a + b/x)^p)*(d + e*x),x)

[Out]

log(c*(a + b/x)^p)*(d*x + (e*x^2)/2) - (log(b + a*x)*(b^2*e*p - 2*a*b*d*p))/(2*a^2) + (b*e*p*x)/(2*a)