Integrand size = 23, antiderivative size = 129 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=-\frac {b n x^2}{4 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}-\frac {b d n \log \left (d+e x^2\right )}{4 e^3}-\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{e^3}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{2 e^3} \]
-1/4*b*n*x^2/e^2+1/2*x^2*(a+b*ln(c*x^n))/e^2+1/2*d*x^2*(a+b*ln(c*x^n))/e^2 /(e*x^2+d)-1/4*b*d*n*ln(e*x^2+d)/e^3-d*(a+b*ln(c*x^n))*ln(1+e*x^2/d)/e^3-1 /2*b*d*n*polylog(2,-e*x^2/d)/e^3
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.22 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {2 e x^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-\frac {2 d^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d+e x^2}-4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )+b n \left (\frac {d \sqrt {e} x \log (x)}{-i \sqrt {d}+\sqrt {e} x}+\frac {d \sqrt {e} x \log (x)}{i \sqrt {d}+\sqrt {e} x}+e x^2 (-1+2 \log (x))-d \log \left (i \sqrt {d}-\sqrt {e} x\right )-d \log \left (i \sqrt {d}+\sqrt {e} x\right )-4 d \left (\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-4 d \left (\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )\right )}{4 e^3} \]
(2*e*x^2*(a - b*n*Log[x] + b*Log[c*x^n]) - (2*d^2*(a - b*n*Log[x] + b*Log[ c*x^n]))/(d + e*x^2) - 4*d*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] + b*n*((d*Sqrt[e]*x*Log[x])/((-I)*Sqrt[d] + Sqrt[e]*x) + (d*Sqrt[e]*x*Log[ x])/(I*Sqrt[d] + Sqrt[e]*x) + e*x^2*(-1 + 2*Log[x]) - d*Log[I*Sqrt[d] - Sq rt[e]*x] - d*Log[I*Sqrt[d] + Sqrt[e]*x] - 4*d*(Log[x]*Log[1 + (I*Sqrt[e]*x )/Sqrt[d]] + PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]]) - 4*d*(Log[x]*Log[1 - ( I*Sqrt[e]*x)/Sqrt[d]] + PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])))/(4*e^3)
Time = 0.40 (sec) , antiderivative size = 129, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^2}-\frac {2 d x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{2 e^3}-\frac {b d n \log \left (d+e x^2\right )}{4 e^3}-\frac {b n x^2}{4 e^2}\) |
-1/4*(b*n*x^2)/e^2 + (x^2*(a + b*Log[c*x^n]))/(2*e^2) + (d*x^2*(a + b*Log[ c*x^n]))/(2*e^2*(d + e*x^2)) - (b*d*n*Log[d + e*x^2])/(4*e^3) - (d*(a + b* Log[c*x^n])*Log[1 + (e*x^2)/d])/e^3 - (b*d*n*PolyLog[2, -((e*x^2)/d)])/(2* e^3)
3.3.21.3.1 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 = 0.54 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.72
| method | result | size |
| risch | \(-\frac {b \ln \left (x^{n}\right ) d^{2}}{2 e^{3} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) d \ln \left (e \,x^{2}+d \right )}{e^{3}}+\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e^{2}}-\frac {b n \,x^{2}}{4 e^{2}}-\frac {b d n \ln \left (e \,x^{2}+d \right )}{4 e^{3}}+\frac {b n d \ln \left (x \right )}{2 e^{3}}+\frac {b n d \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{e^{3}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {d \left (\frac {d}{e \left (e \,x^{2}+d \right )}+\frac {2 \ln \left (e \,x^{2}+d \right )}{e}\right )}{2 e^{2}}+\frac {x^{2}}{2 e^{2}}\right )\) | \(351\) |
-1/2*b*ln(x^n)*d^2/e^3/(e*x^2+d)-b*ln(x^n)*d/e^3*ln(e*x^2+d)+1/2*b*ln(x^n) /e^2*x^2-1/4*b*n*x^2/e^2-1/4*b*d*n*ln(e*x^2+d)/e^3+1/2*b*n/e^3*d*ln(x)+b*n *d/e^3*ln(x)*ln(e*x^2+d)-b*n*d/e^3*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/ 2))-b*n*d/e^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-b*n*d/e^3*dilog((- e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-b*n*d/e^3*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*d/e^2*(d/e/(e*x^2+d)+2/e*ln(e*x^2+d))+1/2*x^2/ e^2)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Time = 44.05 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.45 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {a d^{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 )}{2 e^{2}} - \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 )}{e^{2}} + \frac {a x^{2}}{2 e^{2}} - \frac {b d^{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 e^{2}} + \frac {b d^{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 e^{2}} + \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 )}{e^{2}} - \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 )}}{e^{2}} - \frac {b n x^{2}}{4 e^{2}} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} \]
a*d**2*Piecewise((x**2/d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))/(2*e **2) - a*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/e**2 + a*x**2/(2*e**2) - b*d**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*e**2) + b*d**2*Piecewise((x**2/ d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))*log(c*x**n)/(2*e**2) + b*d* n*Piecewise((x**2/(2*d), Eq(e, 0)), (Piecewise((-polylog(2, e*x**2*exp_pol ar(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) - polylo g(2, e*x**2*exp_polar(I*pi)/d)/2, True))/e, True))/e**2 - b*d*Piecewise((x **2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/e**2 - b*n*x**2/( 4*e**2) + b*x**2*log(c*x**n)/(2*e**2)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
-1/2*a*(d^2/(e^4*x^2 + d*e^3) - x^2/e^2 + 2*d*log(e*x^2 + d)/e^3) + b*inte grate((x^5*log(c) + x^5*log(x^n))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \]