\(\int \frac {a+b \log (c x^n)}{(d+e x)^4} \, dx\) [58]

Optimal result

Integrand size = 18, antiderivative size = 95 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\frac {b n}{6 d e (d+e x)^2}+\frac {b n}{3 d^2 e (d+e x)}+\frac {b n \log (x)}{3 d^3 e}-\frac {a+b \log \left (c x^n\right )}{3 e (d+e x)^3}-\frac {b n \log (d+e x)}{3 d^3 e} \]

[Out]

1/6*b*n/d/e/(e*x+d)^2+1/3*b*n/d^2/e/(e*x+d)+1/3*b*n*ln(x)/d^3/e+1/3*(-a-b*ln(c*x^n))/e/(e*x+d)^3-1/3*b*n*ln(e*
x+d)/d^3/e

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2356, 46} \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=-\frac {a+b \log \left (c x^n\right )}{3 e (d+e x)^3}+\frac {b n \log (x)}{3 d^3 e}-\frac {b n \log (d+e x)}{3 d^3 e}+\frac {b n}{3 d^2 e (d+e x)}+\frac {b n}{6 d e (d+e x)^2} \]

[In]

Int[(a + b*Log[c*x^n])/(d + e*x)^4,x]

[Out]

(b*n)/(6*d*e*(d + e*x)^2) + (b*n)/(3*d^2*e*(d + e*x)) + (b*n*Log[x])/(3*d^3*e) - (a + b*Log[c*x^n])/(3*e*(d +
e*x)^3) - (b*n*Log[d + e*x])/(3*d^3*e)

Rule 46

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

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \log \left (c x^n\right )}{3 e (d+e x)^3}+\frac {(b n) \int \frac {1}{x (d+e x)^3} \, dx}{3 e} \\ & = -\frac {a+b \log \left (c x^n\right )}{3 e (d+e x)^3}+\frac {(b n) \int \left (\frac {1}{d^3 x}-\frac {e}{d (d+e x)^3}-\frac {e}{d^2 (d+e x)^2}-\frac {e}{d^3 (d+e x)}\right ) \, dx}{3 e} \\ & = \frac {b n}{6 d e (d+e x)^2}+\frac {b n}{3 d^2 e (d+e x)}+\frac {b n \log (x)}{3 d^3 e}-\frac {a+b \log \left (c x^n\right )}{3 e (d+e x)^3}-\frac {b n \log (d+e x)}{3 d^3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\frac {-\frac {a+b \log \left (c x^n\right )}{(d+e x)^3}+\frac {b n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )}{2 d^3}}{3 e} \]

[In]

Integrate[(a + b*Log[c*x^n])/(d + e*x)^4,x]

[Out]

(-((a + b*Log[c*x^n])/(d + e*x)^3) + (b*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log[d + e*x]))/(2*d^3)
)/(3*e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(187\) vs. \(2(88)=176\).

Time = 0.49 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.98

method	result	size
parallelrisch	\(\frac {-5 x^{3} b \,e^{5} n -6 \ln \left (c \,x^{n}\right ) b \,d^{3} e^{2}+4 b \,d^{3} e^{2} n -6 a \,d^{3} e^{2}+18 \ln \left (x \right ) x^{2} b d \,e^{4} n -18 \ln \left (e x +d \right ) x^{2} b d \,e^{4} n +18 \ln \left (x \right ) x b \,d^{2} e^{3} n -18 \ln \left (e x +d \right ) x b \,d^{2} e^{3} n +6 \ln \left (x \right ) b \,d^{3} e^{2} n -6 \ln \left (e x +d \right ) b \,d^{3} e^{2} n -9 x^{2} b d \,e^{4} n +6 \ln \left (x \right ) x^{3} b \,e^{5} n -6 \ln \left (e x +d \right ) x^{3} b \,e^{5} n}{18 e^{3} d^{3} \left (e x +d \right )^{3}}\)	\(188\)
risch	\(-\frac {b \ln \left (x^{n}\right )}{3 e \left (e x +d \right )^{3}}-\frac {2 \ln \left (e x +d \right ) b \,e^{3} n \,x^{3}-2 \ln \left (-x \right ) b \,e^{3} n \,x^{3}-i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+6 \ln \left (e x +d \right ) b d \,e^{2} n \,x^{2}-6 \ln \left (-x \right ) b d \,e^{2} n \,x^{2}+6 \ln \left (e x +d \right ) b \,d^{2} e n x -6 \ln \left (-x \right ) b \,d^{2} e n x -2 b d \,e^{2} n \,x^{2}+2 \ln \left (e x +d \right ) b \,d^{3} n -2 \ln \left (-x \right ) b \,d^{3} n -5 b \,d^{2} e n x +2 d^{3} b \ln \left (c \right )-3 b \,d^{3} n +2 a \,d^{3}}{6 d^{3} e \left (e x +d \right )^{3}}\)	\(284\)

[In]

int((a+b*ln(c*x^n))/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

1/18*(-5*x^3*b*e^5*n-6*ln(c*x^n)*b*d^3*e^2+4*b*d^3*e^2*n-6*a*d^3*e^2+18*ln(x)*x^2*b*d*e^4*n-18*ln(e*x+d)*x^2*b
*d*e^4*n+18*ln(x)*x*b*d^2*e^3*n-18*ln(e*x+d)*x*b*d^2*e^3*n+6*ln(x)*b*d^3*e^2*n-6*ln(e*x+d)*b*d^3*e^2*n-9*x^2*b
*d*e^4*n+6*ln(x)*x^3*b*e^5*n-6*ln(e*x+d)*x^3*b*e^5*n)/e^3/d^3/(e*x+d)^3

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.68 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\frac {2 \, b d e^{2} n x^{2} + 5 \, b d^{2} e n x + 3 \, b d^{3} n - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} + 3 \, b d^{2} e n x + b d^{3} n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} + 3 \, b d^{2} e n x\right )} \log \left (x\right )}{6 \, {\left (d^{3} e^{4} x^{3} + 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x + d^{6} e\right )}} \]

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/6*(2*b*d*e^2*n*x^2 + 5*b*d^2*e*n*x + 3*b*d^3*n - 2*b*d^3*log(c) - 2*a*d^3 - 2*(b*e^3*n*x^3 + 3*b*d*e^2*n*x^2
+ 3*b*d^2*e*n*x + b*d^3*n)*log(e*x + d) + 2*(b*e^3*n*x^3 + 3*b*d*e^2*n*x^2 + 3*b*d^2*e*n*x)*log(x))/(d^3*e^4*
x^3 + 3*d^4*e^3*x^2 + 3*d^5*e^2*x + d^6*e)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (83) = 166\).

Time = 5.35 (sec) , antiderivative size = 700, normalized size of antiderivative = 7.37 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {a}{3 x^{3}} - \frac {b n}{9 x^{3}} - \frac {b \log {\left (c x^{n} \right )}}{3 x^{3}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {a x - b n x + b x \log {\left (c x^{n} \right )}}{d^{4}} & \text {for}\: e = 0 \\\frac {- \frac {a}{3 x^{3}} - \frac {b n}{9 x^{3}} - \frac {b \log {\left (c x^{n} \right )}}{3 x^{3}}}{e^{4}} & \text {for}\: d = 0 \\- \frac {2 a d^{3}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} - \frac {2 b d^{3} n \log {\left (\frac {d}{e} + x \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {3 b d^{3} n}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} - \frac {6 b d^{2} e n x \log {\left (\frac {d}{e} + x \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {5 b d^{2} e n x}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {6 b d^{2} e x \log {\left (c x^{n} \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} - \frac {6 b d e^{2} n x^{2} \log {\left (\frac {d}{e} + x \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {2 b d e^{2} n x^{2}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {6 b d e^{2} x^{2} \log {\left (c x^{n} \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} - \frac {2 b e^{3} n x^{3} \log {\left (\frac {d}{e} + x \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} + \frac {2 b e^{3} x^{3} \log {\left (c x^{n} \right )}}{6 d^{6} e + 18 d^{5} e^{2} x + 18 d^{4} e^{3} x^{2} + 6 d^{3} e^{4} x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

Piecewise((zoo*(-a/(3*x**3) - b*n/(9*x**3) - b*log(c*x**n)/(3*x**3)), Eq(d, 0) & Eq(e, 0)), ((a*x - b*n*x + b*
x*log(c*x**n))/d**4, Eq(e, 0)), ((-a/(3*x**3) - b*n/(9*x**3) - b*log(c*x**n)/(3*x**3))/e**4, Eq(d, 0)), (-2*a*
d**3/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) - 2*b*d**3*n*log(d/e + x)/(6*d**6*e +
18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) + 3*b*d**3*n/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*
x**2 + 6*d**3*e**4*x**3) - 6*b*d**2*e*n*x*log(d/e + x)/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3
*e**4*x**3) + 5*b*d**2*e*n*x/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) + 6*b*d**2*e*x
*log(c*x**n)/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) - 6*b*d*e**2*n*x**2*log(d/e +
x)/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) + 2*b*d*e**2*n*x**2/(6*d**6*e + 18*d**5*
e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4*x**3) + 6*b*d*e**2*x**2*log(c*x**n)/(6*d**6*e + 18*d**5*e**2*x + 18*d
**4*e**3*x**2 + 6*d**3*e**4*x**3) - 2*b*e**3*n*x**3*log(d/e + x)/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**
2 + 6*d**3*e**4*x**3) + 2*b*e**3*x**3*log(c*x**n)/(6*d**6*e + 18*d**5*e**2*x + 18*d**4*e**3*x**2 + 6*d**3*e**4
*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.52 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\frac {1}{6} \, b n {\left (\frac {2 \, e x + 3 \, d}{d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e} - \frac {2 \, \log \left (e x + d\right )}{d^{3} e} + \frac {2 \, \log \left (x\right )}{d^{3} e}\right )} - \frac {b \log \left (c x^{n}\right )}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} - \frac {a}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \]

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*b*n*((2*e*x + 3*d)/(d^2*e^3*x^2 + 2*d^3*e^2*x + d^4*e) - 2*log(e*x + d)/(d^3*e) + 2*log(x)/(d^3*e)) - 1/3*
b*log(c*x^n)/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) - 1/3*a/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*
e)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=-\frac {b n \log \left (x\right )}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {2 \, b e^{2} n x^{2} + 5 \, b d e n x + 3 \, b d^{2} n - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2}}{6 \, {\left (d^{2} e^{4} x^{3} + 3 \, d^{3} e^{3} x^{2} + 3 \, d^{4} e^{2} x + d^{5} e\right )}} - \frac {b n \log \left (e x + d\right )}{3 \, d^{3} e} + \frac {b n \log \left (x\right )}{3 \, d^{3} e} \]

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="giac")

[Out]

-1/3*b*n*log(x)/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) + 1/6*(2*b*e^2*n*x^2 + 5*b*d*e*n*x + 3*b*d^2*n -
2*b*d^2*log(c) - 2*a*d^2)/(d^2*e^4*x^3 + 3*d^3*e^3*x^2 + 3*d^4*e^2*x + d^5*e) - 1/3*b*n*log(e*x + d)/(d^3*e)
+ 1/3*b*n*log(x)/(d^3*e)

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.34 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx=\frac {\frac {3\,b\,n}{2}-a+\frac {b\,e^2\,n\,x^2}{d^2}+\frac {5\,b\,e\,n\,x}{2\,d}}{3\,d^3\,e+9\,d^2\,e^2\,x+9\,d\,e^3\,x^2+3\,e^4\,x^3}-\frac {b\,\ln \left (c\,x^n\right )}{3\,e\,\left (d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3\right )}-\frac {2\,b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{3\,d^3\,e} \]

[In]

int((a + b*log(c*x^n))/(d + e*x)^4,x)

[Out]

((3*b*n)/2 - a + (b*e^2*n*x^2)/d^2 + (5*b*e*n*x)/(2*d))/(3*d^3*e + 3*e^4*x^3 + 9*d^2*e^2*x + 9*d*e^3*x^2) - (b
*log(c*x^n))/(3*e*(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)) - (2*b*n*atanh((2*e*x)/d + 1))/(3*d^3*e)