Integrand size = 31, antiderivative size = 86 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (1+p)}+\frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (3+2 p)} \]
-1/4*a*(b*x^2+a)^2*(b^2*x^4+2*a*b*x^2+a^2)^p/b^2/(p+1)+1/2*(b*x^2+a)^3*(b^ 2*x^4+2*a*b*x^2+a^2)^p/b^2/(3+2*p)
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.52 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (\left (a+b x^2\right )^2\right )^{1+p} \left (-a+2 b (1+p) x^2\right )}{4 b^2 (1+p) (3+2 p)} \]
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1383, 243, 53, 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.
\(\displaystyle \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\) |
\(\Big \downarrow \) 1383 |
\(\displaystyle \left (a+b x^2\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int x^3 \left (b x^2+a\right )^{2 p+1}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (a+b x^2\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int x^2 \left (b x^2+a\right )^{2 p+1}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \left (a+b x^2\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int \left (\frac {\left (b x^2+a\right )^{2 p+2}}{b}-\frac {a \left (b x^2+a\right )^{2 p+1}}{b}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (a+b x^2\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \left (\frac {\left (a+b x^2\right )^{2 p+3}}{b^2 (2 p+3)}-\frac {a \left (a+b x^2\right )^{2 (p+1)}}{2 b^2 (p+1)}\right )\) |
((a^2 + 2*a*b*x^2 + b^2*x^4)^p*(-1/2*(a*(a + b*x^2)^(2*(1 + p)))/(b^2*(1 + p)) + (a + b*x^2)^(3 + 2*p)/(b^2*(3 + 2*p))))/(2*(a + b*x^2)^(2*p))
3.1.93.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^p/(d + e*x^n)^(2 *p) Int[u*(d + e*x^n)^(q + 2*p), x], x] /; FreeQ[{a, b, c, d, e, n, p, q} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && !Int egerQ[p]
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72
method | result | size |
gosper | \(-\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (-2 x^{2} p b -2 b \,x^{2}+a \right ) \left (b \,x^{2}+a \right )^{2}}{4 b^{2} \left (2 p^{2}+5 p +3\right )}\) | \(62\) |
risch | \(-\frac {\left (-2 b^{3} p \,x^{6}-2 b^{3} x^{6}-4 a \,b^{2} p \,x^{4}-3 b^{2} x^{4} a -2 a^{2} b p \,x^{2}+a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{p}}{4 \left (1+p \right ) \left (3+2 p \right ) b^{2}}\) | \(79\) |
norman | \(-\frac {a^{3} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{2} \left (2 p^{2}+5 p +3\right )}+\frac {b \,x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{6+4 p}+\frac {a \left (4 p +3\right ) x^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{8 p^{2}+20 p +12}+\frac {p \,a^{2} x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b \left (2 p^{2}+5 p +3\right )}\) | \(170\) |
parallelrisch | \(\frac {2 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b^{3} p +2 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b^{3}+4 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2} p +3 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2}+2 x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b p -a^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 b^{2} \left (2 p^{2}+5 p +3\right )}\) | \(190\) |
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left (2 \, {\left (b^{3} p + b^{3}\right )} x^{6} + 2 \, a^{2} b p x^{2} + {\left (4 \, a b^{2} p + 3 \, a b^{2}\right )} x^{4} - a^{3}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]
1/4*(2*(b^3*p + b^3)*x^6 + 2*a^2*b*p*x^2 + (4*a*b^2*p + 3*a*b^2)*x^4 - a^3 )*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(2*b^2*p^2 + 5*b^2*p + 3*b^2)
\[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\begin {cases} \frac {a x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\\int \frac {x^{3} \left (a + b x^{2}\right )}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{3} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 a^{2} b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {4 a b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {3 a b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 b^{3} p x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 b^{3} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} & \text {otherwise} \end {cases} \]
Piecewise((a*x**4*(a**2)**p/4, Eq(b, 0)), (Integral(x**3*(a + b*x**2)/((a + b*x**2)**2)**(3/2), x), Eq(p, -3/2)), (-a*log(x - sqrt(-a/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)), (-a**3*(a**2 + 2 *a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 20*b**2*p + 12*b**2) + 2*a**2*b*p *x**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 20*b**2*p + 12*b** 2) + 4*a*b**2*p*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 20* b**2*p + 12*b**2) + 3*a*b**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b* *2*p**2 + 20*b**2*p + 12*b**2) + 2*b**3*p*x**6*(a**2 + 2*a*b*x**2 + b**2*x **4)**p/(8*b**2*p**2 + 20*b**2*p + 12*b**2) + 2*b**3*x**6*(a**2 + 2*a*b*x* *2 + b**2*x**4)**p/(8*b**2*p**2 + 20*b**2*p + 12*b**2), True))
Time = 0.20 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.57 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{2 \, p} a}{4 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{2 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \]
1/4*(b^2*(2*p + 1)*x^4 + 2*a*b*p*x^2 - a^2)*(b*x^2 + a)^(2*p)*a/((2*p^2 + 3*p + 1)*b^2) + 1/2*((2*p^2 + 3*p + 1)*b^3*x^6 + (2*p^2 + p)*a*b^2*x^4 - 2 *a^2*b*p*x^2 + a^3)*(b*x^2 + a)^(2*p)/((4*p^3 + 12*p^2 + 11*p + 3)*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (82) = 164\).
Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.28 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p x^{6} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} x^{6} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p x^{4} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} x^{4} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b p x^{2} - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3}}{4 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]
1/4*(2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^3*p*x^6 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^3*x^6 + 4*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^2*p*x^4 + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^2*x^4 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b*p*x ^2 - (b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^3)/(2*b^2*p^2 + 5*b^2*p + 3*b^2)
Time = 7.93 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.26 \[ \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx={\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {b\,x^6\,\left (p+1\right )}{2\,\left (2\,p^2+5\,p+3\right )}-\frac {a^3}{4\,b^2\,\left (2\,p^2+5\,p+3\right )}+\frac {a\,x^4\,\left (4\,p+3\right )}{4\,\left (2\,p^2+5\,p+3\right )}+\frac {a^2\,p\,x^2}{2\,b\,\left (2\,p^2+5\,p+3\right )}\right ) \]