Integrand size = 23, antiderivative size = 121 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}+\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3} \]
1/4*b*d*n*x^2/e^2-1/16*b*n*x^4/e-1/2*d*x^2*(a+b*ln(c*x^n))/e^2+1/4*x^4*(a+ b*ln(c*x^n))/e+1/2*d^2*(a+b*ln(c*x^n))*ln(1+e*x^2/d)/e^3+1/4*b*d^2*n*polyl og(2,-e*x^2/d)/e^3
Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.44 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {4 b d e n x^2-b e^2 n x^4-8 d e x^2 \left (a+b \log \left (c x^n\right )\right )+4 e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+8 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+8 b d^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b d^2 n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{16 e^3} \]
(4*b*d*e*n*x^2 - b*e^2*n*x^4 - 8*d*e*x^2*(a + b*Log[c*x^n]) + 4*e^2*x^4*(a + b*Log[c*x^n]) + 8*d^2*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 8*d^2*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)] + 8*b*d^2*n*P olyLog[2, (Sqrt[e]*x)/Sqrt[-d]] + 8*b*d^2*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^ (3/2)])/(16*e^3)
Time = 0.37 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2793, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}-\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e}\) |
(b*d*n*x^2)/(4*e^2) - (b*n*x^4)/(16*e) - (d*x^2*(a + b*Log[c*x^n]))/(2*e^2 ) + (x^4*(a + b*Log[c*x^n]))/(4*e) + (d^2*(a + b*Log[c*x^n])*Log[1 + (e*x^ 2)/d])/(2*e^3) + (b*d^2*n*PolyLog[2, -((e*x^2)/d)])/(4*e^3)
3.3.10.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer Q[r]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.62 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.83
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x^{4}}{4 e}-\frac {b \ln \left (x^{n}\right ) d \,x^{2}}{2 e^{2}}+\frac {b \ln \left (x^{n}\right ) d^{2} \ln \left (e \,x^{2}+d \right )}{2 e^{3}}-\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}-\frac {b n \,x^{4}}{16 e}+\frac {b d n \,x^{2}}{4 e^{2}}-\frac {b n \,d^{2}}{4 e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} e \,x^{4}-d \,x^{2}}{2 e^{2}}+\frac {d^{2} \ln \left (e \,x^{2}+d \right )}{2 e^{3}}\right )\) | \(343\) |
1/4*b*ln(x^n)/e*x^4-1/2*b*ln(x^n)/e^2*d*x^2+1/2*b*ln(x^n)*d^2/e^3*ln(e*x^2 +d)-1/2*b*n*d^2/e^3*ln(x)*ln(e*x^2+d)+1/2*b*n*d^2/e^3*ln(x)*ln((-e*x+(-d*e )^(1/2))/(-d*e)^(1/2))+1/2*b*n*d^2/e^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^ (1/2))+1/2*b*n*d^2/e^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*d^2 /e^3*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/16*b*n*x^4/e+1/4*b*d*n*x^2/e ^2-1/4*b*n*d^2/e^3+(-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*(1/2*e*x^4-d*x^2)+1/2*d^2/e^3*l n(e*x^2+d))
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]
Time = 35.17 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.12 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {a d x^{2}}{2 e^{2}} + \frac {a x^{4}}{4 e} - \frac {b d^{2} n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e^{2}} + \frac {b d n x^{2}}{4 e^{2}} - \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} - \frac {b n x^{4}}{16 e} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4 e} \]
a*d**2*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2*e**2) - a*d*x**2/(2*e**2) + a*x**4/(4*e) - b*d**2*n*Piecewise((x**2/(2*d), Eq(e, 0)), (Piecewise((-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, A bs(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))/(2*e**2) + b*d**2*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/(2*e**2) + b*d*n*x**2/(4*e**2) - b*d*x**2 *log(c*x**n)/(2*e**2) - b*n*x**4/(16*e) + b*x**4*log(c*x**n)/(4*e)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]
1/4*a*(2*d^2*log(e*x^2 + d)/e^3 + (e*x^4 - 2*d*x^2)/e^2) + b*integrate((x^ 5*log(c) + x^5*log(x^n))/(e*x^2 + d), x)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \]