Integrand size = 18, antiderivative size = 56 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {(b c-a d) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {d \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)} \] Output:
1/2*(-a*d+b*c)*(b*x^2+a)^(p+1)/b^2/(p+1)+1/2*d*(b*x^2+a)^(2+p)/b^2/(2+p)
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {\left (a+b x^2\right )^{1+p} \left (-a d+b c (2+p)+b d (1+p) x^2\right )}{2 b^2 (1+p) (2+p)} \] Input:
Integrate[x*(a + b*x^2)^p*(c + d*x^2),x]
Output:
((a + b*x^2)^(1 + p)*(-(a*d) + b*c*(2 + p) + b*d*(1 + p)*x^2))/(2*b^2*(1 + p)*(2 + p))
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {353, 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.
\(\displaystyle \int x \left (c+d x^2\right ) \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \left (b x^2+a\right )^p \left (d x^2+c\right )dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(b c-a d) \left (b x^2+a\right )^p}{b}+\frac {d \left (b x^2+a\right )^{p+1}}{b}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {(b c-a d) \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac {d \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}\right )\) |
Input:
Int[x*(a + b*x^2)^p*(c + d*x^2),x]
Output:
(((b*c - a*d)*(a + b*x^2)^(1 + p))/(b^2*(1 + p)) + (d*(a + b*x^2)^(2 + p)) /(b^2*(2 + p)))/2
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_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Time = 0.52 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{p +1} \left (-b d p \,x^{2}-b d \,x^{2}-b c p +a d -2 b c \right )}{2 b^{2} \left (p^{2}+3 p +2\right )}\) | \(55\) |
orering | \(-\frac {\left (b \,x^{2}+a \right )^{p} \left (-b d p \,x^{2}-b d \,x^{2}-b c p +a d -2 b c \right ) \left (b \,x^{2}+a \right )}{2 b^{2} \left (p^{2}+3 p +2\right )}\) | \(60\) |
risch | \(-\frac {\left (-b^{2} d p \,x^{4}-d \,b^{2} x^{4}-a b d p \,x^{2}-b^{2} c p \,x^{2}-2 b^{2} c \,x^{2}-a b c p +d \,a^{2}-2 a b c \right ) \left (b \,x^{2}+a \right )^{p}}{2 b^{2} \left (2+p \right ) \left (p +1\right )}\) | \(89\) |
norman | \(\frac {d \,x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{4+2 p}-\frac {a \left (-b c p +a d -2 b c \right ) {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b^{2} \left (p^{2}+3 p +2\right )}+\frac {\left (a d p +b c p +2 b c \right ) x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+3 p +2\right )}\) | \(106\) |
parallelrisch | \(\frac {x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} d p +x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} d +x^{2} \left (b \,x^{2}+a \right )^{p} a^{2} b d p +x^{2} \left (b \,x^{2}+a \right )^{p} a \,b^{2} c p +2 x^{2} \left (b \,x^{2}+a \right )^{p} a \,b^{2} c +\left (b \,x^{2}+a \right )^{p} a^{2} b c p -\left (b \,x^{2}+a \right )^{p} a^{3} d +2 \left (b \,x^{2}+a \right )^{p} a^{2} b c}{2 \left (2+p \right ) \left (p +1\right ) a \,b^{2}}\) | \(161\) |
Input:
int(x*(b*x^2+a)^p*(d*x^2+c),x,method=_RETURNVERBOSE)
Output:
-1/2/b^2*(b*x^2+a)^(p+1)/(p^2+3*p+2)*(-b*d*p*x^2-b*d*x^2-b*c*p+a*d-2*b*c)
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.57 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left ({\left (b^{2} d p + b^{2} d\right )} x^{4} + a b c p + 2 \, a b c - a^{2} d + {\left (2 \, b^{2} c + {\left (b^{2} c + a b d\right )} p\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )}} \] Input:
integrate(x*(b*x^2+a)^p*(d*x^2+c),x, algorithm="fricas")
Output:
1/2*((b^2*d*p + b^2*d)*x^4 + a*b*c*p + 2*a*b*c - a^2*d + (2*b^2*c + (b^2*c + a*b*d)*p)*x^2)*(b*x^2 + a)^p/(b^2*p^2 + 3*b^2*p + 2*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (46) = 92\).
Time = 4.41 (sec) , antiderivative size = 564, normalized size of antiderivative = 10.07 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\begin {cases} a^{p} \left (\frac {c x^{2}}{2} + \frac {d x^{4}}{4}\right ) & \text {for}\: b = 0 \\\frac {a d \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a d \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a d}{2 a b^{2} + 2 b^{3} x^{2}} - \frac {b c}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b d x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b d x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a d \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a d \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {c \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b} + \frac {c \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b} + \frac {d x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} d \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b c p \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {2 a b c \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b d p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} c p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {2 b^{2} c x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} d p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} d x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(b*x**2+a)**p*(d*x**2+c),x)
Output:
Piecewise((a**p*(c*x**2/2 + d*x**4/4), Eq(b, 0)), (a*d*log(x - sqrt(-a/b)) /(2*a*b**2 + 2*b**3*x**2) + a*d*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x** 2) + a*d/(2*a*b**2 + 2*b**3*x**2) - b*c/(2*a*b**2 + 2*b**3*x**2) + b*d*x** 2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*d*x**2*log(x + sqrt(-a/ b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*d*log(x - sqrt(-a/b))/(2*b** 2) - a*d*log(x + sqrt(-a/b))/(2*b**2) + c*log(x - sqrt(-a/b))/(2*b) + c*lo g(x + sqrt(-a/b))/(2*b) + d*x**2/(2*b), Eq(p, -1)), (-a**2*d*(a + b*x**2)* *p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*c*p*(a + b*x**2)**p/(2*b**2*p** 2 + 6*b**2*p + 4*b**2) + 2*a*b*c*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*d*p*x**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*c*p*x**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + 2*b**2 *c*x**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*d*p*x**4* (a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*d*x**4*(a + b*x** 2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2), True))
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.27 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left (b^{2} {\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{p} d}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} + \frac {{\left (b x^{2} + a\right )}^{p + 1} c}{2 \, b {\left (p + 1\right )}} \] Input:
integrate(x*(b*x^2+a)^p*(d*x^2+c),x, algorithm="maxima")
Output:
1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*d/((p^2 + 3*p + 2)*b ^2) + 1/2*(b*x^2 + a)^(p + 1)*c/(b*(p + 1))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} d}{2 \, b^{2} {\left (p + 2\right )}} + \frac {\frac {{\left (b x^{2} + a\right )}^{p + 1} b c}{p + 1} - \frac {{\left (b x^{2} + a\right )}^{p + 1} a d}{p + 1}}{2 \, b^{2}} \] Input:
integrate(x*(b*x^2+a)^p*(d*x^2+c),x, algorithm="giac")
Output:
1/2*(b*x^2 + a)^2*(b*x^2 + a)^p*d/(b^2*(p + 2)) + 1/2*((b*x^2 + a)^(p + 1) *b*c/(p + 1) - (b*x^2 + a)^(p + 1)*a*d/(p + 1))/b^2
Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.71 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx={\left (b\,x^2+a\right )}^p\,\left (\frac {a\,\left (2\,b\,c-a\,d+b\,c\,p\right )}{2\,b^2\,\left (p^2+3\,p+2\right )}+\frac {x^2\,\left (2\,b^2\,c+b^2\,c\,p+a\,b\,d\,p\right )}{2\,b^2\,\left (p^2+3\,p+2\right )}+\frac {d\,x^4\,\left (p+1\right )}{2\,\left (p^2+3\,p+2\right )}\right ) \] Input:
int(x*(a + b*x^2)^p*(c + d*x^2),x)
Output:
(a + b*x^2)^p*((a*(2*b*c - a*d + b*c*p))/(2*b^2*(3*p + p^2 + 2)) + (x^2*(2 *b^2*c + b^2*c*p + a*b*d*p))/(2*b^2*(3*p + p^2 + 2)) + (d*x^4*(p + 1))/(2* (3*p + p^2 + 2)))
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.50 \[ \int x \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {\left (b \,x^{2}+a \right )^{p} \left (b^{2} d p \,x^{4}+b^{2} d \,x^{4}+a b d p \,x^{2}+b^{2} c p \,x^{2}+2 b^{2} c \,x^{2}+a b c p -a^{2} d +2 a b c \right )}{2 b^{2} \left (p^{2}+3 p +2\right )} \] Input:
int(x*(b*x^2+a)^p*(d*x^2+c),x)
Output:
((a + b*x**2)**p*( - a**2*d + a*b*c*p + 2*a*b*c + a*b*d*p*x**2 + b**2*c*p* x**2 + 2*b**2*c*x**2 + b**2*d*p*x**4 + b**2*d*x**4))/(2*b**2*(p**2 + 3*p + 2))