Integrand size = 23, antiderivative size = 126 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=-\frac {b n}{2 d^2 x^2}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}-\frac {4 a-b n+4 b \log \left (c x^n\right )}{4 d^2 x^2}+\frac {e \log \left (1+\frac {d}{e x^2}\right ) \left (4 a-b n+4 b \log \left (c x^n\right )\right )}{4 d^3}-\frac {b e n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{2 d^3} \]
-1/2*b*n/d^2/x^2+1/2*(a+b*ln(c*x^n))/d/x^2/(e*x^2+d)+1/4*(-4*a+b*n-4*b*ln( c*x^n))/d^2/x^2+1/4*e*ln(1+d/e/x^2)*(4*a-b*n+4*b*ln(c*x^n))/d^3-1/2*b*e*n* polylog(2,-d/e/x^2)/d^3
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.65 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\frac {-\frac {2 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{x^2}-\frac {2 d e \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d+e x^2}-8 e \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+4 e \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )+b n \left (\frac {e^{3/2} x \log (x)}{-i \sqrt {d}+\sqrt {e} x}-4 e \log ^2(x)-\frac {d+2 d \log (x)}{x^2}-e \log \left (i \sqrt {d}-\sqrt {e} x\right )+\frac {-i e^{3/2} x \log (x)+e \left (-\sqrt {d}+i \sqrt {e} x\right ) \log \left (i \sqrt {d}+\sqrt {e} x\right )}{\sqrt {d}-i \sqrt {e} x}+4 e \left (\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )+4 e \left (\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )\right )}{4 d^3} \]
((-2*d*(a - b*n*Log[x] + b*Log[c*x^n]))/x^2 - (2*d*e*(a - b*n*Log[x] + b*L og[c*x^n]))/(d + e*x^2) - 8*e*Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]) + 4*e *(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] + b*n*((e^(3/2)*x*Log[x])/ ((-I)*Sqrt[d] + Sqrt[e]*x) - 4*e*Log[x]^2 - (d + 2*d*Log[x])/x^2 - e*Log[I *Sqrt[d] - Sqrt[e]*x] + ((-I)*e^(3/2)*x*Log[x] + e*(-Sqrt[d] + I*Sqrt[e]*x )*Log[I*Sqrt[d] + Sqrt[e]*x])/(Sqrt[d] - I*Sqrt[e]*x) + 4*e*(Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]]) + 4*e*(Lo g[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))) /(4*d^3)
Time = 0.58 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2785, 25, 2780, 2741, 2779, 2838}
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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2785 |
| \(\displaystyle \frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}-\frac {\int -\frac {4 a-b n+4 b \log \left (c x^n\right )}{x^3 \left (e x^2+d\right )}dx}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {4 a-b n+4 b \log \left (c x^n\right )}{x^3 \left (e x^2+d\right )}dx}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {\frac {\int \frac {4 a-b n+4 b \log \left (c x^n\right )}{x^3}dx}{d}-\frac {e \int \frac {4 a-b n+4 b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {\frac {-\frac {4 a+4 b \log \left (c x^n\right )-b n}{2 x^2}-\frac {b n}{x^2}}{d}-\frac {e \int \frac {4 a-b n+4 b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {\frac {-\frac {4 a+4 b \log \left (c x^n\right )-b n}{2 x^2}-\frac {b n}{x^2}}{d}-\frac {e \left (\frac {2 b n \int \frac {\log \left (\frac {d}{e x^2}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-b n\right )}{2 d}\right )}{d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {\frac {-\frac {4 a+4 b \log \left (c x^n\right )-b n}{2 x^2}-\frac {b n}{x^2}}{d}-\frac {e \left (\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-b n\right )}{2 d}\right )}{d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}\) |
(a + b*Log[c*x^n])/(2*d*x^2*(d + e*x^2)) + ((-((b*n)/x^2) - (4*a - b*n + 4 *b*Log[c*x^n])/(2*x^2))/d - (e*(-1/2*(Log[1 + d/(e*x^2)]*(4*a - b*n + 4*b* Log[c*x^n]))/d + (b*n*PolyLog[2, -(d/(e*x^2))])/d))/d)/(2*d)
3.3.25.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^2)^(q_.), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(d + e*x^2)^(q + 1)*((a + b*Log[c*x^n])/(2*d*f*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(f*x)^m*(d + e*x^2)^(q + 1)*(a*(m + 2*q + 3) + b*n + b*(m + 2*q + 3)*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && ILtQ[m, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.69 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.02
| method | result | size |
| risch | \(-\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} \left (e \,x^{2}+d \right )}+\frac {b \ln \left (x^{n}\right ) e \ln \left (e \,x^{2}+d \right )}{d^{3}}-\frac {b \ln \left (x^{n}\right )}{2 d^{2} x^{2}}-\frac {2 b \ln \left (x^{n}\right ) e \ln \left (x \right )}{d^{3}}+\frac {b n e \ln \left (x \right )^{2}}{d^{3}}-\frac {b n e \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{d^{3}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3}}+\frac {b n e \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3}}+\frac {b n e \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3}}-\frac {b n e \ln \left (e \,x^{2}+d \right )}{4 d^{3}}-\frac {b n}{4 d^{2} x^{2}}+\frac {b n e \ln \left (x \right )}{2 d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {e^{2} \left (-\frac {d}{e \left (e \,x^{2}+d \right )}+\frac {2 \ln \left (e \,x^{2}+d \right )}{e}\right )}{2 d^{3}}-\frac {1}{2 d^{2} x^{2}}-\frac {2 e \ln \left (x \right )}{d^{3}}\right )\) | \(380\) |
-1/2*b*ln(x^n)*e/d^2/(e*x^2+d)+b*ln(x^n)*e/d^3*ln(e*x^2+d)-1/2*b*ln(x^n)/d ^2/x^2-2*b*ln(x^n)/d^3*e*ln(x)+b*n/d^3*e*ln(x)^2-b*n/d^3*e*ln(x)*ln(e*x^2+ d)+b*n/d^3*e*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+b*n/d^3*e*ln(x)*ln ((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+b*n/d^3*e*dilog((-e*x+(-d*e)^(1/2))/(-d* e)^(1/2))+b*n/d^3*e*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/4*b*n*e/d^3*l n(e*x^2+d)-1/4*b*n/d^2/x^2+1/2*b*n/d^3*e*ln(x)+(-1/2*I*b*Pi*csgn(I*c)*csgn (I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn (I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)*(1/2*e^2/d^3 *(-d/e/(e*x^2+d)+2/e*ln(e*x^2+d))-1/2/d^2/x^2-2/d^3*e*ln(x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
Time = 157.73 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.89 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\frac {a e^{2} \left (\begin {cases} \frac {x^{2}}{2 d^{2}} & \text {for}\: e = 0 \\- \frac {1}{2 d e + 2 e^{2} x^{2}} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a}{2 d^{2} x^{2}} - \frac {2 a e \log {\left (x \right )}}{d^{3}} + \frac {a e \log {\left (d + e x^{2} \right )}}{d^{3}} - \frac {b e^{2} n \left (\begin {cases} \frac {x^{2}}{2 d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x^{2} \right )}}{2 d e} & \text {otherwise} \end {cases}\right )}{2 d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x^{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 d^{2}} - \frac {b n}{4 d^{2} x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 d^{2} x^{2}} - \frac {b e^{2} n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {b e^{2} \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} + \frac {b e n \log {\left (x^{2} \right )}^{2}}{4 d^{3}} - \frac {b e \log {\left (x^{2} \right )} \log {\left (c x^{n} \right )}}{d^{3}} \]
a*e**2*Piecewise((x**2/(2*d**2), Eq(e, 0)), (-1/(2*d*e + 2*e**2*x**2), Tru e))/d**2 - a/(2*d**2*x**2) - 2*a*e*log(x)/d**3 + a*e*log(d + e*x**2)/d**3 - b*e**2*n*Piecewise((x**2/(2*d**2), Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x**2)/(2*d*e), True))/(2*d**2) + b*e**2*Piecewise((x**2/d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))*log(c*x**n)/(2*d**2) - b*n/(4*d**2*x**2) - b *log(c*x**n)/(2*d**2*x**2) - b*e**2*n*Piecewise((x**2/(2*d), Eq(e, 0)), (P iecewise((-polylog(2, e*x**2*exp_polar(I*pi)/d)/2, (Abs(x) < 1) & (1/Abs(x ) < 1)), (log(d)*log(x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ( )), ((), (0, 0)), x)*log(d) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, True ))/e, True))/d**3 + b*e**2*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/ e, True))*log(c*x**n)/d**3 + b*e*n*log(x**2)**2/(4*d**3) - b*e*log(x**2)*l og(c*x**n)/d**3
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
-1/2*a*((2*e*x^2 + d)/(d^2*e*x^4 + d^3*x^2) - 2*e*log(e*x^2 + d)/d^3 + 4*e *log(x)/d^3) + b*integrate((log(c) + log(x^n))/(e^2*x^7 + 2*d*e*x^5 + d^2* x^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (e\,x^2+d\right )}^2} \,d x \]