Integrand size = 24, antiderivative size = 84 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (1+2 p)}+\frac {\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (1+p)} \]
-1/2*a*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p/b^2/(1+2*p)+1/4*(b*x^2+a)^2*(b^ 2*x^4+2*a*b*x^2+a^2)^p/b^2/(p+1)
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (-a+b (1+2 p) x^2\right )}{4 b^2 (1+p) (1+2 p)} \]
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1434, 1100, 1079, 17}
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.
\(\displaystyle \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b^2 x^4+2 a b x^2+a^2\right )^pdx^2\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \int \left (b^2 x^4+2 a b x^2+a^2\right )^pdx^2}{b}\right )\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \left (a b+b^2 x^2\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int \left (b^2 x^2+a b\right )^{2 p}dx^2}{b}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \left (a b+b^2 x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (2 p+1)}\right )\) |
(-((a*(a*b + b^2*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p)/(b^3*(1 + 2*p))) + (a ^2 + 2*a*b*x^2 + b^2*x^4)^(1 + p)/(2*b^2*(1 + p)))/2
3.8.100.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c *x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(b/2 + c *x)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b* e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {\left (-2 b^{2} p \,x^{4}-b^{2} x^{4}-2 a b p \,x^{2}+a^{2}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{p}}{4 b^{2} \left (1+p \right ) \left (1+2 p \right )}\) | \(58\) |
gosper | \(-\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (-2 x^{2} p b -b \,x^{2}+a \right ) \left (b \,x^{2}+a \right )}{4 b^{2} \left (2 p^{2}+3 p +1\right )}\) | \(60\) |
norman | \(\frac {x^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 p +4}-\frac {a^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{2} \left (2 p^{2}+3 p +1\right )}+\frac {p a \,x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b \left (2 p^{2}+3 p +1\right )}\) | \(120\) |
parallelrisch | \(\frac {2 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2} p +x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2}+2 x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b p -\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3}}{4 b^{2} \left (1+p \right ) \left (1+2 p \right ) a}\) | \(135\) |
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left (2 \, a b p x^{2} + {\left (2 \, b^{2} p + b^{2}\right )} x^{4} - a^{2}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
1/4*(2*a*b*p*x^2 + (2*b^2*p + b^2)*x^4 - a^2)*(b^2*x^4 + 2*a*b*x^2 + a^2)^ p/(2*b^2*p^2 + 3*b^2*p + b^2)
\[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\begin {cases} \frac {x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -1 \\\int \frac {x^{3}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {2 a b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {2 b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases} \]
Piecewise((x**4*(a**2)**p/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b* *2 + 2*b**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -1)), (Integral (x**3/sqrt((a + b*x**2)**2), x), Eq(p, -1/2)), (-a**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2) + 2*a*b*p*x**2*(a**2 + 2 *a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2) + 2*b**2*p*x* *4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2) + b**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4 *b**2), True))
Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.57 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} p x^{4} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} x^{4} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b p x^{2} - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2}}{4 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
1/4*(2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^2*p*x^4 + (b^2*x^4 + 2*a*b*x^2 + a^ 2)^p*b^2*x^4 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b*p*x^2 - (b^2*x^4 + 2*a* b*x^2 + a^2)^p*a^2)/(2*b^2*p^2 + 3*b^2*p + b^2)
Time = 12.91 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx={\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^4\,\left (2\,p+1\right )}{4\,\left (2\,p^2+3\,p+1\right )}-\frac {a^2}{4\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,p\,x^2}{2\,b\,\left (2\,p^2+3\,p+1\right )}\right ) \]