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

Optimal. Leaf size=78 \[ -\frac {p (a d-b e)^2 \log (a x+b)}{2 a^2 e}+\frac {(d+e x)^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}+\frac {b e p x}{2 a}+\frac {d^2 p \log (x)}{2 e} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2463, 514, 72} \[ -\frac {p (a d-b e)^2 \log (a x+b)}{2 a^2 e}+\frac {(d+e x)^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}+\frac {b e p x}{2 a}+\frac {d^2 p \log (x)}{2 e} \]

Antiderivative was successfully verified.

[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 72

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 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) \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.03, size = 85, normalized size = 1.09 \[ \frac {1}{2} b e p \left (\frac {x}{a}-\frac {b \log (a x+b)}{a^2}\right )+d x \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{2} e x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b d p \log \left (a+\frac {b}{x}\right )}{a}+\frac {b d p \log (x)}{a} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.42, size = 80, normalized size = 1.03 \[ \frac {a b e p x + {\left (2 \, a b d - b^{2} e\right )} p \log \left (a x + b\right ) + {\left (a^{2} e x^{2} + 2 \, a^{2} d x\right )} \log \relax (c) + {\left (a^{2} e p x^{2} + 2 \, a^{2} d p x\right )} \log \left (\frac {a x + b}{x}\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.19, size = 394, normalized size = 5.05 \[ -\frac {2 \, a^{3} b^{2} d p \log \left (-a + \frac {a x + b}{x}\right ) - a^{2} b^{3} p e \log \left (-a + \frac {a x + b}{x}\right ) + a^{2} b^{3} p e - \frac {4 \, {\left (a x + b\right )} a^{2} b^{2} d p \log \left (-a + \frac {a x + b}{x}\right )}{x} + \frac {2 \, {\left (a x + b\right )} a b^{3} p e \log \left (-a + \frac {a x + b}{x}\right )}{x} + 2 \, a^{3} b^{2} d \log \relax (c) - a^{2} b^{3} e \log \relax (c) + \frac {2 \, {\left (a x + b\right )} a^{2} b^{2} d p \log \left (\frac {a x + b}{x}\right )}{x} - \frac {2 \, {\left (a x + b\right )} a b^{3} p e \log \left (\frac {a x + b}{x}\right )}{x} - \frac {{\left (a x + b\right )} a b^{3} p e}{x} + \frac {2 \, {\left (a x + b\right )}^{2} a b^{2} d p \log \left (-a + \frac {a x + b}{x}\right )}{x^{2}} - \frac {{\left (a x + b\right )}^{2} b^{3} p e \log \left (-a + \frac {a x + b}{x}\right )}{x^{2}} - \frac {2 \, {\left (a x + b\right )} a^{2} b^{2} d \log \relax (c)}{x} - \frac {2 \, {\left (a x + b\right )}^{2} a b^{2} d p \log \left (\frac {a x + b}{x}\right )}{x^{2}} + \frac {{\left (a x + b\right )}^{2} b^{3} p e \log \left (\frac {a x + b}{x}\right )}{x^{2}}}{2 \, {\left (a^{4} - \frac {2 \, {\left (a x + b\right )} a^{3}}{x} + \frac {{\left (a x + b\right )}^{2} a^{2}}{x^{2}}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3*b^2*d*p*log(-a + (a*x + b)/x) - a^2*b^3*p*e*log(-a + (a*x + b)/x) + a^2*b^3*p*e - 4*(a*x + b)*a^2*
b^2*d*p*log(-a + (a*x + b)/x)/x + 2*(a*x + b)*a*b^3*p*e*log(-a + (a*x + b)/x)/x + 2*a^3*b^2*d*log(c) - a^2*b^3
*e*log(c) + 2*(a*x + b)*a^2*b^2*d*p*log((a*x + b)/x)/x - 2*(a*x + b)*a*b^3*p*e*log((a*x + b)/x)/x - (a*x + b)*
a*b^3*p*e/x + 2*(a*x + b)^2*a*b^2*d*p*log(-a + (a*x + b)/x)/x^2 - (a*x + b)^2*b^3*p*e*log(-a + (a*x + b)/x)/x^
2 - 2*(a*x + b)*a^2*b^2*d*log(c)/x - 2*(a*x + b)^2*a*b^2*d*p*log((a*x + b)/x)/x^2 + (a*x + b)^2*b^3*p*e*log((a
*x + b)/x)/x^2)/((a^4 - 2*(a*x + b)*a^3/x + (a*x + b)^2*a^2/x^2)*b)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right ) \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 55, normalized size = 0.71 \[ \frac {1}{2} \, b p {\left (\frac {e x}{a} + \frac {{\left (2 \, a d - b e\right )} \log \left (a x + b\right )}{a^{2}}\right )} + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.30, size = 57, normalized size = 0.73 \[ \ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )\,\left (\frac {e\,x^2}{2}+d\,x\right )-\frac {\ln \left (b+a\,x\right )\,\left (b^2\,e\,p-2\,a\,b\,d\,p\right )}{2\,a^2}+\frac {b\,e\,p\,x}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 2.35, size = 156, normalized size = 2.00 \[ \begin {cases} d p x \log {\left (a + \frac {b}{x} \right )} + d x \log {\relax (c )} + \frac {e p x^{2} \log {\left (a + \frac {b}{x} \right )}}{2} + \frac {e x^{2} \log {\relax (c )}}{2} + \frac {b d p \log {\left (x + \frac {b}{a} \right )}}{a} + \frac {b e p x}{2 a} - \frac {b^{2} e p \log {\left (x + \frac {b}{a} \right )}}{2 a^{2}} & \text {for}\: a \neq 0 \\d p x \log {\relax (b )} - d p x \log {\relax (x )} + d p x + d x \log {\relax (c )} + \frac {e p x^{2} \log {\relax (b )}}{2} - \frac {e p x^{2} \log {\relax (x )}}{2} + \frac {e p x^{2}}{4} + \frac {e x^{2} \log {\relax (c )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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