Integrand size = 18, antiderivative size = 145 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {a p^2 x^2}{b}+\frac {p^2 \left (a+b x^2\right )^2}{8 b^2}+\frac {a p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}-\frac {p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2} \] Output:
-a*p^2*x^2/b+1/8*p^2*(b*x^2+a)^2/b^2+a*p*(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^2-1 /4*p*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)/b^2-1/2*a*(b*x^2+a)*ln(c*(b*x^2+a)^p)^2 /b^2+1/4*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^2/b^2
Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {b p^2 x^2 \left (-6 a+b x^2\right )+2 a^2 p^2 \log \left (a+b x^2\right )+2 p \left (2 a^2+2 a b x^2-b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )-2 \left (a^2-b^2 x^4\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2} \] Input:
Integrate[x^3*Log[c*(a + b*x^2)^p]^2,x]
Output:
(b*p^2*x^2*(-6*a + b*x^2) + 2*a^2*p^2*Log[a + b*x^2] + 2*p*(2*a^2 + 2*a*b* x^2 - b^2*x^4)*Log[c*(a + b*x^2)^p] - 2*(a^2 - b^2*x^4)*Log[c*(a + b*x^2)^ p]^2)/(8*b^2)
Time = 0.59 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int x^2 \log ^2\left (c \left (b x^2+a\right )^p\right )dx^2\) |
\(\Big \downarrow \) 2848 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right ) \log ^2\left (c \left (b x^2+a\right )^p\right )}{b}-\frac {a \log ^2\left (c \left (b x^2+a\right )^p\right )}{b}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{b^2}-\frac {p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {2 a p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {p^2 \left (a+b x^2\right )^2}{4 b^2}-\frac {2 a p^2 x^2}{b}\right )\) |
Input:
Int[x^3*Log[c*(a + b*x^2)^p]^2,x]
Output:
((-2*a*p^2*x^2)/b + (p^2*(a + b*x^2)^2)/(4*b^2) + (2*a*p*(a + b*x^2)*Log[c *(a + b*x^2)^p])/b^2 - (p*(a + b*x^2)^2*Log[c*(a + b*x^2)^p])/(2*b^2) - (a *(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/b^2 + ((a + b*x^2)^2*Log[c*(a + b*x^2 )^p]^2)/(2*b^2))/2
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 = 1.75 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {2 x^{4} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} b^{2}-2 x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{2} p +b^{2} p^{2} x^{4}+4 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a b p -6 a b \,p^{2} x^{2}+10 \ln \left (b \,x^{2}+a \right ) a^{2} p^{2}-2 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{2}-4 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} p +6 a^{2} p^{2}}{8 b^{2}}\) | \(151\) |
risch | \(\text {Expression too large to display}\) | \(24313\) |
Input:
int(x^3*ln(c*(b*x^2+a)^p)^2,x,method=_RETURNVERBOSE)
Output:
1/8*(2*x^4*ln(c*(b*x^2+a)^p)^2*b^2-2*x^4*ln(c*(b*x^2+a)^p)*b^2*p+b^2*p^2*x ^4+4*x^2*ln(c*(b*x^2+a)^p)*a*b*p-6*a*b*p^2*x^2+10*ln(b*x^2+a)*a^2*p^2-2*ln (c*(b*x^2+a)^p)^2*a^2-4*ln(c*(b*x^2+a)^p)*a^2*p+6*a^2*p^2)/b^2
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.02 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {b^{2} p^{2} x^{4} + 2 \, b^{2} x^{4} \log \left (c\right )^{2} - 6 \, a b p^{2} x^{2} + 2 \, {\left (b^{2} p^{2} x^{4} - a^{2} p^{2}\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b^{2} p^{2} x^{4} - 2 \, a b p^{2} x^{2} - 3 \, a^{2} p^{2} - 2 \, {\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} \log \left (c\right )}{8 \, b^{2}} \] Input:
integrate(x^3*log(c*(b*x^2+a)^p)^2,x, algorithm="fricas")
Output:
1/8*(b^2*p^2*x^4 + 2*b^2*x^4*log(c)^2 - 6*a*b*p^2*x^2 + 2*(b^2*p^2*x^4 - a ^2*p^2)*log(b*x^2 + a)^2 - 2*(b^2*p^2*x^4 - 2*a*b*p^2*x^2 - 3*a^2*p^2 - 2* (b^2*p*x^4 - a^2*p)*log(c))*log(b*x^2 + a) - 2*(b^2*p*x^4 - 2*a*b*p*x^2)*l og(c))/b^2
Time = 1.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \frac {3 a^{2} p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b^{2}} - \frac {a^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{4 b^{2}} - \frac {3 a p^{2} x^{2}}{4 b} + \frac {a p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 b} + \frac {p^{2} x^{4}}{8} - \frac {p x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4} + \frac {x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{4} & \text {for}\: b \neq 0 \\\frac {x^{4} \log {\left (a^{p} c \right )}^{2}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*ln(c*(b*x**2+a)**p)**2,x)
Output:
Piecewise((3*a**2*p*log(c*(a + b*x**2)**p)/(4*b**2) - a**2*log(c*(a + b*x* *2)**p)**2/(4*b**2) - 3*a*p**2*x**2/(4*b) + a*p*x**2*log(c*(a + b*x**2)**p )/(2*b) + p**2*x**4/8 - p*x**4*log(c*(a + b*x**2)**p)/4 + x**4*log(c*(a + b*x**2)**p)**2/4, Ne(b, 0)), (x**4*log(a**p*c)**2/4, True))
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} - \frac {1}{4} \, b p {\left (\frac {2 \, a^{2} \log \left (b x^{2} + a\right )}{b^{3}} + \frac {b x^{4} - 2 \, a x^{2}}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {{\left (b^{2} x^{4} - 6 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 6 \, a^{2} \log \left (b x^{2} + a\right )\right )} p^{2}}{8 \, b^{2}} \] Input:
integrate(x^3*log(c*(b*x^2+a)^p)^2,x, algorithm="maxima")
Output:
1/4*x^4*log((b*x^2 + a)^p*c)^2 - 1/4*b*p*(2*a^2*log(b*x^2 + a)/b^3 + (b*x^ 4 - 2*a*x^2)/b^2)*log((b*x^2 + a)^p*c) + 1/8*(b^2*x^4 - 6*a*b*x^2 + 2*a^2* log(b*x^2 + a)^2 + 6*a^2*log(b*x^2 + a))*p^2/b^2
Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.49 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right ) + 4 \, {\left (b x^{2} + a\right )}^{2} p \log \left (b x^{2} + a\right ) \log \left (c\right ) + {\left (b x^{2} + a\right )}^{2} p^{2} - 2 \, {\left (b x^{2} + a\right )}^{2} p \log \left (c\right ) + 2 \, {\left (b x^{2} + a\right )}^{2} \log \left (c\right )^{2}}{8 \, b^{2}} - \frac {{\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} a p^{2} - 2 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a p \log \left (c\right ) + {\left (b x^{2} + a\right )} a \log \left (c\right )^{2}}{2 \, b^{2}} \] Input:
integrate(x^3*log(c*(b*x^2+a)^p)^2,x, algorithm="giac")
Output:
1/8*(2*(b*x^2 + a)^2*p^2*log(b*x^2 + a)^2 - 2*(b*x^2 + a)^2*p^2*log(b*x^2 + a) + 4*(b*x^2 + a)^2*p*log(b*x^2 + a)*log(c) + (b*x^2 + a)^2*p^2 - 2*(b* x^2 + a)^2*p*log(c) + 2*(b*x^2 + a)^2*log(c)^2)/b^2 - 1/2*((2*b*x^2 + (b*x ^2 + a)*log(b*x^2 + a)^2 - 2*(b*x^2 + a)*log(b*x^2 + a) + 2*a)*a*p^2 - 2*( b*x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*a*p*log(c) + (b*x^2 + a)*a*log(c)^ 2)/b^2
Time = 14.78 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.69 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {p^2\,x^4}{8}-\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {p\,x^4}{4}-\frac {a\,p\,x^2}{2\,b}\right )+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {x^4}{4}-\frac {a^2}{4\,b^2}\right )-\frac {3\,a\,p^2\,x^2}{4\,b}+\frac {3\,a^2\,p^2\,\ln \left (b\,x^2+a\right )}{4\,b^2} \] Input:
int(x^3*log(c*(a + b*x^2)^p)^2,x)
Output:
(p^2*x^4)/8 - log(c*(a + b*x^2)^p)*((p*x^4)/4 - (a*p*x^2)/(2*b)) + log(c*( a + b*x^2)^p)^2*(x^4/4 - a^2/(4*b^2)) - (3*a*p^2*x^2)/(4*b) + (3*a^2*p^2*l og(a + b*x^2))/(4*b^2)
Time = 0.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.87 \[ \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-2 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a^{2}+2 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} b^{2} x^{4}+6 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} p +4 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b p \,x^{2}-2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} p \,x^{4}-6 a b \,p^{2} x^{2}+b^{2} p^{2} x^{4}}{8 b^{2}} \] Input:
int(x^3*log(c*(b*x^2+a)^p)^2,x)
Output:
( - 2*log((a + b*x**2)**p*c)**2*a**2 + 2*log((a + b*x**2)**p*c)**2*b**2*x* *4 + 6*log((a + b*x**2)**p*c)*a**2*p + 4*log((a + b*x**2)**p*c)*a*b*p*x**2 - 2*log((a + b*x**2)**p*c)*b**2*p*x**4 - 6*a*b*p**2*x**2 + b**2*p**2*x**4 )/(8*b**2)