Integrand size = 18, antiderivative size = 76 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\frac {b n}{2 d e (d+e x)}+\frac {b n \log (x)}{2 d^2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}-\frac {b n \log (d+e x)}{2 d^2 e} \]
1/2*b*n/d/e/(e*x+d)+1/2*b*n*ln(x)/d^2/e+1/2*(-a-b*ln(c*x^n))/e/(e*x+d)^2-1 /2*b*n*ln(e*x+d)/d^2/e
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.70 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\frac {-\frac {a+b \log \left (c x^n\right )}{(d+e x)^2}+\frac {b n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )}{d^2}}{2 e} \]
Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2756, 54, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 2756 |
\(\displaystyle \frac {b n \int \frac {1}{x (d+e x)^2}dx}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {b n \int \left (-\frac {e}{d^2 (d+e x)}-\frac {e}{d (d+e x)^2}+\frac {1}{d^2 x}\right )dx}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b n \left (-\frac {\log (d+e x)}{d^2}+\frac {\log (x)}{d^2}+\frac {1}{d (d+e x)}\right )}{2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}\) |
-1/2*(a + b*Log[c*x^n])/(e*(d + e*x)^2) + (b*n*(1/(d*(d + e*x)) + Log[x]/d ^2 - Log[d + e*x]/d^2))/(2*e)
3.1.49.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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] - Simp[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] || (IntegersQ[2*p, 2*q] && !IGtQ[q, 0]) || (EqQ[p, 2] & & NeQ[q, 1]))
Time = 0.62 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(\frac {2 \ln \left (x \right ) x^{2} b \,e^{3} n -2 \ln \left (e x +d \right ) x^{2} b \,e^{3} n +4 \ln \left (x \right ) x b d \,e^{2} n -4 \ln \left (e x +d \right ) x b d \,e^{2} n -b \,e^{3} n \,x^{2}+2 \ln \left (x \right ) b \,d^{2} e n -2 \ln \left (e x +d \right ) b \,d^{2} e n -2 b \ln \left (c \,x^{n}\right ) d^{2} e +e \,d^{2} b n -2 e \,d^{2} a}{4 e^{2} d^{2} \left (e x +d \right )^{2}}\) | \(132\) |
risch | \(-\frac {b \ln \left (x^{n}\right )}{2 e \left (e x +d \right )^{2}}-\frac {-i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-2 \ln \left (-x \right ) b \,e^{2} n \,x^{2}+2 \ln \left (e x +d \right ) b \,e^{2} n \,x^{2}-4 \ln \left (-x \right ) b d e n x +4 \ln \left (e x +d \right ) b d e n x -2 \ln \left (-x \right ) b \,d^{2} n +2 \ln \left (e x +d \right ) b \,d^{2} n -2 b d e n x +2 d^{2} b \ln \left (c \right )-2 b \,d^{2} n +2 a \,d^{2}}{4 d^{2} e \left (e x +d \right )^{2}}\) | \(235\) |
1/4*(2*ln(x)*x^2*b*e^3*n-2*ln(e*x+d)*x^2*b*e^3*n+4*ln(x)*x*b*d*e^2*n-4*ln( e*x+d)*x*b*d*e^2*n-b*e^3*n*x^2+2*ln(x)*b*d^2*e*n-2*ln(e*x+d)*b*d^2*e*n-2*b *ln(c*x^n)*d^2*e+e*d^2*b*n-2*e*d^2*a)/e^2/d^2/(e*x+d)^2
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\frac {b d e n x + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2} - {\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (e x + d\right ) + {\left (b e^{2} n x^{2} + 2 \, b d e n x\right )} \log \left (x\right )}{2 \, {\left (d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e\right )}} \]
1/2*(b*d*e*n*x + b*d^2*n - b*d^2*log(c) - a*d^2 - (b*e^2*n*x^2 + 2*b*d*e*n *x + b*d^2*n)*log(e*x + d) + (b*e^2*n*x^2 + 2*b*d*e*n*x)*log(x))/(d^2*e^3* x^2 + 2*d^3*e^2*x + d^4*e)
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (66) = 132\).
Time = 1.88 (sec) , antiderivative size = 415, normalized size of antiderivative = 5.46 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {a}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 x^{2}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {a x - b n x + b x \log {\left (c x^{n} \right )}}{d^{3}} & \text {for}\: e = 0 \\\frac {- \frac {a}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 x^{2}}}{e^{3}} & \text {for}\: d = 0 \\- \frac {a d^{2}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {b d^{2} n \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b d^{2} n}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {2 b d e n x \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b d e n x}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {2 b d e x \log {\left (c x^{n} \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {b e^{2} n x^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b e^{2} x^{2} \log {\left (c x^{n} \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(-a/(2*x**2) - b*n/(4*x**2) - b*log(c*x**n)/(2*x**2)), Eq(d , 0) & Eq(e, 0)), ((a*x - b*n*x + b*x*log(c*x**n))/d**3, Eq(e, 0)), ((-a/( 2*x**2) - b*n/(4*x**2) - b*log(c*x**n)/(2*x**2))/e**3, Eq(d, 0)), (-a*d**2 /(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) - b*d**2*n*log(d/e + x)/(2* d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) + b*d**2*n/(2*d**4*e + 4*d**3*e **2*x + 2*d**2*e**3*x**2) - 2*b*d*e*n*x*log(d/e + x)/(2*d**4*e + 4*d**3*e* *2*x + 2*d**2*e**3*x**2) + b*d*e*n*x/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e* *3*x**2) + 2*b*d*e*x*log(c*x**n)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x **2) - b*e**2*n*x**2*log(d/e + x)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3* x**2) + b*e**2*x**2*log(c*x**n)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x* *2), True))
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\frac {1}{2} \, b n {\left (\frac {1}{d e^{2} x + d^{2} e} - \frac {\log \left (e x + d\right )}{d^{2} e} + \frac {\log \left (x\right )}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {a}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]
1/2*b*n*(1/(d*e^2*x + d^2*e) - log(e*x + d)/(d^2*e) + log(x)/(d^2*e)) - 1/ 2*b*log(c*x^n)/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 1/2*a/(e^3*x^2 + 2*d*e^2*x + d^2*e)
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.37 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=-\frac {b n \log \left (x\right )}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} + \frac {b e n x + b d n - b d \log \left (c\right ) - a d}{2 \, {\left (d e^{3} x^{2} + 2 \, d^{2} e^{2} x + d^{3} e\right )}} - \frac {b n \log \left (e x + d\right )}{2 \, d^{2} e} + \frac {b n \log \left (x\right )}{2 \, d^{2} e} \]
-1/2*b*n*log(x)/(e^3*x^2 + 2*d*e^2*x + d^2*e) + 1/2*(b*e*n*x + b*d*n - b*d *log(c) - a*d)/(d*e^3*x^2 + 2*d^2*e^2*x + d^3*e) - 1/2*b*n*log(e*x + d)/(d ^2*e) + 1/2*b*n*log(x)/(d^2*e)
Time = 0.77 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx=\frac {b\,n-a+\frac {b\,e\,n\,x}{d}}{2\,d^2\,e+4\,d\,e^2\,x+2\,e^3\,x^2}-\frac {b\,\ln \left (c\,x^n\right )}{2\,e\,\left (d^2+2\,d\,e\,x+e^2\,x^2\right )}-\frac {b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{d^2\,e} \]