Integrand size = 24, antiderivative size = 84 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=-\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (1+2 p)}+\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (1+p)} \]
-1/3*a*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^p/b^2/(1+2*p)+1/6*(b*x^3+a)^2*(b^ 2*x^6+2*a*b*x^3+a^2)^p/b^2/(p+1)
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (-a+b (1+2 p) x^3\right )}{6 b^2 (1+p) (1+2 p)} \]
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 = {1693, 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.
\(\displaystyle \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{3} \int x^3 \left (b^2 x^6+2 a b x^3+a^2\right )^pdx^3\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \int \left (b^2 x^6+2 a b x^3+a^2\right )^pdx^3}{b}\right )\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \left (a b+b^2 x^3\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \int \left (b^2 x^3+a b\right )^{2 p}dx^3}{b}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{p+1}}{2 b^2 (p+1)}-\frac {a \left (a b+b^2 x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^3 (2 p+1)}\right )\) |
(-((a*(a*b + b^2*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(b^3*(1 + 2*p))) + (a ^2 + 2*a*b*x^3 + b^2*x^6)^(1 + p)/(2*b^2*(1 + p)))/3
3.2.27.3.1 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_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ [Simplify[(m + 1)/n]]
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {\left (-2 b^{2} p \,x^{6}-b^{2} x^{6}-2 a b p \,x^{3}+a^{2}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{6 b^{2} \left (1+p \right ) \left (1+2 p \right )}\) | \(58\) |
gosper | \(-\frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} \left (-2 x^{3} p b -b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )}{6 b^{2} \left (2 p^{2}+3 p +1\right )}\) | \(60\) |
norman | \(\frac {x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 p +6}-\frac {a^{2} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 b^{2} \left (2 p^{2}+3 p +1\right )}+\frac {p a \,x^{3} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{3 b \left (2 p^{2}+3 p +1\right )}\) | \(120\) |
parallelrisch | \(\frac {2 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{2} p +x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{2}+2 x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a b p -\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2}}{6 b^{2} \left (2 p^{2}+3 p +1\right )}\) | \(128\) |
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, b^{2} p + b^{2}\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
1/6*((2*b^2*p + b^2)*x^6 + 2*a*b*p*x^3 - a^2)*(b^2*x^6 + 2*a*b*x^3 + a^2)^ p/(2*b^2*p^2 + 3*b^2*p + b^2)
\[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\begin {cases} \frac {x^{6} \left (a^{2}\right )^{p}}{6} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 a \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 b x^{3} \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} & \text {for}\: p = -1 \\\int \frac {x^{5}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 a b p x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 b^{2} p x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {b^{2} x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} & \text {otherwise} \end {cases} \]
Piecewise((x**6*(a**2)**p/6, Eq(b, 0)), (a*log(x - (-a/b)**(1/3))/(3*a*b** 2 + 3*b**3*x**3) + a*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(3* a*b**2 + 3*b**3*x**3) - 2*a*log(2)/(3*a*b**2 + 3*b**3*x**3) + a/(3*a*b**2 + 3*b**3*x**3) + b*x**3*log(x - (-a/b)**(1/3))/(3*a*b**2 + 3*b**3*x**3) + b*x**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(3*a*b**2 + 3*b** 3*x**3) - 2*b*x**3*log(2)/(3*a*b**2 + 3*b**3*x**3), Eq(p, -1)), (Integral( x**5/sqrt((a + b*x**3)**2), x), Eq(p, -1/2)), (-a**2*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2) + 2*a*b*p*x**3*(a**2 + 2 *a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2) + 2*b**2*p*x **6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2) + b**2*x**6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2), True))
Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{6 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]
Time = 0.32 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.57 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} p x^{6} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} x^{6} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b p x^{3} - {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
1/6*(2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^2*p*x^6 + (b^2*x^6 + 2*a*b*x^3 + a^ 2)^p*b^2*x^6 + 2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b*p*x^3 - (b^2*x^6 + 2*a* b*x^3 + a^2)^p*a^2)/(2*b^2*p^2 + 3*b^2*p + b^2)
Time = 8.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx={\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^6\,\left (2\,p+1\right )}{6\,\left (2\,p^2+3\,p+1\right )}-\frac {a^2}{6\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,p\,x^3}{3\,b\,\left (2\,p^2+3\,p+1\right )}\right ) \]