Integrand size = 18, antiderivative size = 150 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=-\frac {2 d p^2 x^3}{3 e}+\frac {p^2 \left (d+e x^3\right )^2}{12 e^2}+\frac {2 d p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e^2}-\frac {p \left (d+e x^3\right )^2 \log \left (c \left (d+e x^3\right )^p\right )}{6 e^2}-\frac {d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e^2}+\frac {\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{6 e^2} \]
-2/3*d*p^2*x^3/e+1/12*p^2*(e*x^3+d)^2/e^2+2/3*d*p*(e*x^3+d)*ln(c*(e*x^3+d) ^p)/e^2-1/6*p*(e*x^3+d)^2*ln(c*(e*x^3+d)^p)/e^2-1/3*d*(e*x^3+d)*ln(c*(e*x^ 3+d)^p)^2/e^2+1/6*(e*x^3+d)^2*ln(c*(e*x^3+d)^p)^2/e^2
Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.70 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {e p^2 x^3 \left (-6 d+e x^3\right )+2 d^2 p^2 \log \left (d+e x^3\right )+2 p \left (2 d^2+2 d e x^3-e^2 x^6\right ) \log \left (c \left (d+e x^3\right )^p\right )-2 \left (d^2-e^2 x^6\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{12 e^2} \]
(e*p^2*x^3*(-6*d + e*x^3) + 2*d^2*p^2*Log[d + e*x^3] + 2*p*(2*d^2 + 2*d*e* x^3 - e^2*x^6)*Log[c*(d + e*x^3)^p] - 2*(d^2 - e^2*x^6)*Log[c*(d + e*x^3)^ p]^2)/(12*e^2)
Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2904, 2848, 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.
\(\displaystyle \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{3} \int x^3 \log ^2\left (c \left (e x^3+d\right )^p\right )dx^3\) |
\(\Big \downarrow \) 2848 |
\(\displaystyle \frac {1}{3} \int \left (\frac {\left (e x^3+d\right ) \log ^2\left (c \left (e x^3+d\right )^p\right )}{e}-\frac {d \log ^2\left (c \left (e x^3+d\right )^p\right )}{e}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{2 e^2}-\frac {d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{e^2}-\frac {p \left (d+e x^3\right )^2 \log \left (c \left (d+e x^3\right )^p\right )}{2 e^2}+\frac {2 d p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{e^2}+\frac {p^2 \left (d+e x^3\right )^2}{4 e^2}-\frac {2 d p^2 x^3}{e}\right )\) |
((-2*d*p^2*x^3)/e + (p^2*(d + e*x^3)^2)/(4*e^2) + (2*d*p*(d + e*x^3)*Log[c *(d + e*x^3)^p])/e^2 - (p*(d + e*x^3)^2*Log[c*(d + e*x^3)^p])/(2*e^2) - (d *(d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/e^2 + ((d + e*x^3)^2*Log[c*(d + e*x^3 )^p]^2)/(2*e^2))/3
3.2.29.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[q, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
Time = 4.79 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01
method | result | size |
parallelrisch | \(\frac {2 x^{6} {\ln \left (c \left (e \,x^{3}+d \right )^{p}\right )}^{2} e^{2}-2 x^{6} \ln \left (c \left (e \,x^{3}+d \right )^{p}\right ) e^{2} p +e^{2} p^{2} x^{6}+4 x^{3} \ln \left (c \left (e \,x^{3}+d \right )^{p}\right ) d e p -6 d e \,p^{2} x^{3}+10 d^{2} p^{2} \ln \left (e \,x^{3}+d \right )-2 {\ln \left (c \left (e \,x^{3}+d \right )^{p}\right )}^{2} d^{2}-4 \ln \left (c \left (e \,x^{3}+d \right )^{p}\right ) d^{2} p +6 d^{2} p^{2}}{12 e^{2}}\) | \(151\) |
risch | \(\text {Expression too large to display}\) | \(24313\) |
1/12*(2*x^6*ln(c*(e*x^3+d)^p)^2*e^2-2*x^6*ln(c*(e*x^3+d)^p)*e^2*p+e^2*p^2* x^6+4*x^3*ln(c*(e*x^3+d)^p)*d*e*p-6*d*e*p^2*x^3+10*d^2*p^2*ln(e*x^3+d)-2*l n(c*(e*x^3+d)^p)^2*d^2-4*ln(c*(e*x^3+d)^p)*d^2*p+6*d^2*p^2)/e^2
Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {e^{2} p^{2} x^{6} + 2 \, e^{2} x^{6} \log \left (c\right )^{2} - 6 \, d e p^{2} x^{3} + 2 \, {\left (e^{2} p^{2} x^{6} - d^{2} p^{2}\right )} \log \left (e x^{3} + d\right )^{2} - 2 \, {\left (e^{2} p^{2} x^{6} - 2 \, d e p^{2} x^{3} - 3 \, d^{2} p^{2} - 2 \, {\left (e^{2} p x^{6} - d^{2} p\right )} \log \left (c\right )\right )} \log \left (e x^{3} + d\right ) - 2 \, {\left (e^{2} p x^{6} - 2 \, d e p x^{3}\right )} \log \left (c\right )}{12 \, e^{2}} \]
1/12*(e^2*p^2*x^6 + 2*e^2*x^6*log(c)^2 - 6*d*e*p^2*x^3 + 2*(e^2*p^2*x^6 - d^2*p^2)*log(e*x^3 + d)^2 - 2*(e^2*p^2*x^6 - 2*d*e*p^2*x^3 - 3*d^2*p^2 - 2 *(e^2*p*x^6 - d^2*p)*log(c))*log(e*x^3 + d) - 2*(e^2*p*x^6 - 2*d*e*p*x^3)* log(c))/e^2
Time = 3.67 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\begin {cases} \frac {d^{2} p \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{2 e^{2}} - \frac {d^{2} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{6 e^{2}} - \frac {d p^{2} x^{3}}{2 e} + \frac {d p x^{3} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{3 e} + \frac {p^{2} x^{6}}{12} - \frac {p x^{6} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{6} + \frac {x^{6} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{6} & \text {for}\: e \neq 0 \\\frac {x^{6} \log {\left (c d^{p} \right )}^{2}}{6} & \text {otherwise} \end {cases} \]
Piecewise((d**2*p*log(c*(d + e*x**3)**p)/(2*e**2) - d**2*log(c*(d + e*x**3 )**p)**2/(6*e**2) - d*p**2*x**3/(2*e) + d*p*x**3*log(c*(d + e*x**3)**p)/(3 *e) + p**2*x**6/12 - p*x**6*log(c*(d + e*x**3)**p)/6 + x**6*log(c*(d + e*x **3)**p)**2/6, Ne(e, 0)), (x**6*log(c*d**p)**2/6, True))
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.80 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {1}{6} \, x^{6} \log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2} - \frac {1}{6} \, e p {\left (\frac {2 \, d^{2} \log \left (e x^{3} + d\right )}{e^{3}} + \frac {e x^{6} - 2 \, d x^{3}}{e^{2}}\right )} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) + \frac {{\left (e^{2} x^{6} - 6 \, d e x^{3} + 2 \, d^{2} \log \left (e x^{3} + d\right )^{2} + 6 \, d^{2} \log \left (e x^{3} + d\right )\right )} p^{2}}{12 \, e^{2}} \]
1/6*x^6*log((e*x^3 + d)^p*c)^2 - 1/6*e*p*(2*d^2*log(e*x^3 + d)/e^3 + (e*x^ 6 - 2*d*x^3)/e^2)*log((e*x^3 + d)^p*c) + 1/12*(e^2*x^6 - 6*d*e*x^3 + 2*d^2 *log(e*x^3 + d)^2 + 6*d^2*log(e*x^3 + d))*p^2/e^2
Time = 0.31 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.44 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {2 \, {\left (e x^{3} + d\right )}^{2} p^{2} \log \left (e x^{3} + d\right )^{2} - 2 \, {\left (e x^{3} + d\right )}^{2} p^{2} \log \left (e x^{3} + d\right ) + 4 \, {\left (e x^{3} + d\right )}^{2} p \log \left (e x^{3} + d\right ) \log \left (c\right ) + {\left (e x^{3} + d\right )}^{2} p^{2} - 2 \, {\left (e x^{3} + d\right )}^{2} p \log \left (c\right ) + 2 \, {\left (e x^{3} + d\right )}^{2} \log \left (c\right )^{2}}{12 \, e^{2}} - \frac {{\left (2 \, e x^{3} + {\left (e x^{3} + d\right )} \log \left (e x^{3} + d\right )^{2} - 2 \, {\left (e x^{3} + d\right )} \log \left (e x^{3} + d\right ) + 2 \, d\right )} d p^{2} - 2 \, {\left (e x^{3} - {\left (e x^{3} + d\right )} \log \left (e x^{3} + d\right ) + d\right )} d p \log \left (c\right ) + {\left (e x^{3} + d\right )} d \log \left (c\right )^{2}}{3 \, e^{2}} \]
1/12*(2*(e*x^3 + d)^2*p^2*log(e*x^3 + d)^2 - 2*(e*x^3 + d)^2*p^2*log(e*x^3 + d) + 4*(e*x^3 + d)^2*p*log(e*x^3 + d)*log(c) + (e*x^3 + d)^2*p^2 - 2*(e *x^3 + d)^2*p*log(c) + 2*(e*x^3 + d)^2*log(c)^2)/e^2 - 1/3*((2*e*x^3 + (e* x^3 + d)*log(e*x^3 + d)^2 - 2*(e*x^3 + d)*log(e*x^3 + d) + 2*d)*d*p^2 - 2* (e*x^3 - (e*x^3 + d)*log(e*x^3 + d) + d)*d*p*log(c) + (e*x^3 + d)*d*log(c) ^2)/e^2
Time = 1.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.67 \[ \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {p^2\,x^6}{12}-\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )\,\left (\frac {p\,x^6}{6}-\frac {d\,p\,x^3}{3\,e}\right )+{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2\,\left (\frac {x^6}{6}-\frac {d^2}{6\,e^2}\right )-\frac {d\,p^2\,x^3}{2\,e}+\frac {d^2\,p^2\,\ln \left (e\,x^3+d\right )}{2\,e^2} \]