Integrand size = 23, antiderivative size = 152 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b d n}{8 e^3 \left (d+e x^2\right )}+\frac {b n \log (x)}{4 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {3 b n \log \left (d+e x^2\right )}{8 e^3}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3} \]
1/8*b*d*n/e^3/(e*x^2+d)+1/4*b*n*ln(x)/e^3-1/4*d^2*(a+b*ln(c*x^n))/e^3/(e*x ^2+d)^2-x^2*(a+b*ln(c*x^n))/e^2/(e*x^2+d)+3/8*b*n*ln(e*x^2+d)/e^3+1/2*(a+b *ln(c*x^n))*ln(1+e*x^2/d)/e^3+1/4*b*n*polylog(2,-e*x^2/d)/e^3
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 498, normalized size of antiderivative = 3.28 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {-2 d^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+8 d \left (d+e x^2\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+4 \left (d+e x^2\right )^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )+b n \left (d^2+d e x^2-4 d e x^2 \log (x)-6 e^2 x^4 \log (x)+3 d^2 \log \left (i \sqrt {d}-\sqrt {e} x\right )+6 d e x^2 \log \left (i \sqrt {d}-\sqrt {e} x\right )+3 e^2 x^4 \log \left (i \sqrt {d}-\sqrt {e} x\right )+3 d^2 \log \left (i \sqrt {d}+\sqrt {e} x\right )+6 d e x^2 \log \left (i \sqrt {d}+\sqrt {e} x\right )+3 e^2 x^4 \log \left (i \sqrt {d}+\sqrt {e} x\right )+4 d^2 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+8 d e x^2 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 e^2 x^4 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 d^2 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+8 d e x^2 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 e^2 x^4 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 \left (d+e x^2\right )^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 \left (d+e x^2\right )^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )}{8 e^3 \left (d+e x^2\right )^2} \]
(-2*d^2*(a - b*n*Log[x] + b*Log[c*x^n]) + 8*d*(d + e*x^2)*(a - b*n*Log[x] + b*Log[c*x^n]) + 4*(d + e*x^2)^2*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] + b*n*(d^2 + d*e*x^2 - 4*d*e*x^2*Log[x] - 6*e^2*x^4*Log[x] + 3*d^2* Log[I*Sqrt[d] - Sqrt[e]*x] + 6*d*e*x^2*Log[I*Sqrt[d] - Sqrt[e]*x] + 3*e^2* x^4*Log[I*Sqrt[d] - Sqrt[e]*x] + 3*d^2*Log[I*Sqrt[d] + Sqrt[e]*x] + 6*d*e* x^2*Log[I*Sqrt[d] + Sqrt[e]*x] + 3*e^2*x^4*Log[I*Sqrt[d] + Sqrt[e]*x] + 4* d^2*Log[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 8*d*e*x^2*Log[x]*Log[1 - (I*Sq rt[e]*x)/Sqrt[d]] + 4*e^2*x^4*Log[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 4*d^ 2*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 8*d*e*x^2*Log[x]*Log[1 + (I*Sqrt [e]*x)/Sqrt[d]] + 4*e^2*x^4*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 4*(d + e*x^2)^2*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]] + 4*(d + e*x^2)^2*PolyLog[2 , (I*Sqrt[e]*x)/Sqrt[d]]))/(8*e^3*(d + e*x^2)^2)
Time = 0.48 (sec) , antiderivative size = 152, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^3}-\frac {2 d x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3}+\frac {b d n}{8 e^3 \left (d+e x^2\right )}+\frac {3 b n \log \left (d+e x^2\right )}{8 e^3}+\frac {b n \log (x)}{4 e^3}\) |
(b*d*n)/(8*e^3*(d + e*x^2)) + (b*n*Log[x])/(4*e^3) - (d^2*(a + b*Log[c*x^n ]))/(4*e^3*(d + e*x^2)^2) - (x^2*(a + b*Log[c*x^n]))/(e^2*(d + e*x^2)) + ( 3*b*n*Log[d + e*x^2])/(8*e^3) + ((a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2 *e^3) + (b*n*PolyLog[2, -((e*x^2)/d)])/(4*e^3)
3.3.31.3.1 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 = 0.70 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.36
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {b \ln \left (x^{n}\right ) d}{e^{3} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) d^{2}}{4 e^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {3 b n \ln \left (e \,x^{2}+d \right )}{8 e^{3}}+\frac {b d n}{8 e^{3} \left (e \,x^{2}+d \right )}-\frac {3 b n \ln \left (x \right )}{4 e^{3}}-\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {d}{e^{3} \left (e \,x^{2}+d \right )}-\frac {d^{2}}{4 e^{3} \left (e \,x^{2}+d \right )^{2}}\right )\) | \(359\) |
1/2*b*ln(x^n)/e^3*ln(e*x^2+d)+b*ln(x^n)*d/e^3/(e*x^2+d)-1/4*b*ln(x^n)*d^2/ e^3/(e*x^2+d)^2+3/8*b*n*ln(e*x^2+d)/e^3+1/8*b*d*n/e^3/(e*x^2+d)-3/4*b*n*ln (x)/e^3-1/2*b*n/e^3*ln(x)*ln(e*x^2+d)+1/2*b*n/e^3*ln(x)*ln((-e*x+(-d*e)^(1 /2))/(-d*e)^(1/2))+1/2*b*n/e^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1 /2*b*n/e^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n/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/e^3*ln(e*x^2+d)+d/e^3/(e*x ^2+d)-1/4*d^2/e^3/(e*x^2+d)^2)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Time = 63.33 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.65 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x^{2}}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x^{2}\right )^{2}} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {a d \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 )}{e^{2}} + \frac {a \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 {b d^{2} n \left (\begin {cases} \frac {x^{2}}{2 d^{3}} & \text {for}\: e = 0 \\- \frac {1}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {\log {\left (x \right )}}{2 d^{2} e} + \frac {\log {\left (\frac {d}{e} + x^{2} \right )}}{4 d^{2} e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x^{2}}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x^{2}\right )^{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^{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 )}{e^{2}} - \frac {b d \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 )}}{e^{2}} - \frac {b 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 \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}} \]
a*d**2*Piecewise((x**2/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x**2)**2), True))/ (2*e**2) - a*d*Piecewise((x**2/d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), Tru e))/e**2 + a*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2*e **2) - b*d**2*n*Piecewise((x**2/(2*d**3), Eq(e, 0)), (-1/(4*d**2*e + 4*d*e **2*x**2) - log(x)/(2*d**2*e) + log(d/e + x**2)/(4*d**2*e), True))/(2*e**2 ) + b*d**2*Piecewise((x**2/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x**2)**2), Tru e))*log(c*x**n)/(2*e**2) + b*d*n*Piecewise((x**2/(2*d**2), Eq(e, 0)), (-lo g(x)/(d*e) + log(d/e + x**2)/(2*d*e), True))/e**2 - b*d*Piecewise((x**2/d* *2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))*log(c*x**n)/e**2 - b*n*Piecew ise((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*e xp_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*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/(2*e**2)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 + d)/e^3) + b*integrate((x^5*log(c) + x^5*log(x^n))/(e^3*x^6 + 3*d*e^2*x^ 4 + 3*d^2*e*x^2 + d^3), x)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]