Integrand size = 23, antiderivative size = 83 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=-\frac {b n x^2}{4 e}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^2}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^2} \]
-1/4*b*n*x^2/e+1/2*x^2*(a+b*ln(c*x^n))/e-1/2*d*(a+b*ln(c*x^n))*ln(1+e*x^2/ d)/e^2-1/4*b*d*n*polylog(2,-e*x^2/d)/e^2
Time = 0.05 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.63 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=-\frac {b e n x^2-2 e x^2 \left (a+b \log \left (c x^n\right )\right )+2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+2 b d n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 b d n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{4 e^2} \]
-1/4*(b*e*n*x^2 - 2*e*x^2*(a + b*Log[c*x^n]) + 2*d*(a + b*Log[c*x^n])*Log[ 1 + (Sqrt[e]*x)/Sqrt[-d]] + 2*d*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/( -d)^(3/2)] + 2*b*d*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]] + 2*b*d*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/e^2
Time = 0.32 (sec) , antiderivative size = 83, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {x \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e \left (d+e x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^2}-\frac {b n x^2}{4 e}\) |
-1/4*(b*n*x^2)/e + (x^2*(a + b*Log[c*x^n]))/(2*e) - (d*(a + b*Log[c*x^n])* Log[1 + (e*x^2)/d])/(2*e^2) - (b*d*n*PolyLog[2, -((e*x^2)/d)])/(4*e^2)
3.3.11.3.1 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 = 0.44 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.42
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e}-\frac {b \ln \left (x^{n}\right ) d \ln \left (e \,x^{2}+d \right )}{2 e^{2}}-\frac {b n \,x^{2}}{4 e}+\frac {b n d \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{2}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}-\frac {b n d \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}-\frac {b n d \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x^{2}}{2 e}-\frac {d \ln \left (e \,x^{2}+d \right )}{2 e^{2}}\right )\) | \(284\) |
1/2*b*ln(x^n)/e*x^2-1/2*b*ln(x^n)*d/e^2*ln(e*x^2+d)-1/4*b*n*x^2/e+1/2*b*n* d/e^2*ln(x)*ln(e*x^2+d)-1/2*b*n*d/e^2*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^ (1/2))-1/2*b*n*d/e^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*d/e ^2*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*d/e^2*dilog((e*x+(-d*e) ^(1/2))/(-d*e)^(1/2))+(-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*x^2/e-1/2*d/e^2*ln(e*x^2+d))
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{e x^{2} + d} \,d x } \]
Time = 17.42 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.43 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=- \frac {a d \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e} + \frac {a x^{2}}{2 e} + \frac {b d n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{2 e} - \frac {b d \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e} - \frac {b n x^{2}}{4 e} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e} \]
-a*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2*e) + a*x* *2/(2*e) + b*d*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, 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))/(2*e) - b*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/( 2*e) - b*n*x**2/(4*e) + b*x**2*log(c*x**n)/(2*e)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{e x^{2} + d} \,d x } \]
1/2*a*(x^2/e - d*log(e*x^2 + d)/e^2) + b*integrate((x^3*log(c) + x^3*log(x ^n))/(e*x^2 + d), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \]