Integrand size = 20, antiderivative size = 88 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=-\frac {a (b c-a d) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {(b c-2 a d) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {d \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)} \] Output:
-1/2*a*(-a*d+b*c)*(b*x^2+a)^(p+1)/b^3/(p+1)+1/2*(-2*a*d+b*c)*(b*x^2+a)^(2+ p)/b^3/(2+p)+1/2*d*(b*x^2+a)^(3+p)/b^3/(3+p)
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {1}{2} \left (-\frac {a (b c-a d) \left (a+b x^2\right )^{1+p}}{b^3 (1+p)}+\frac {(b c-2 a d) \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {d \left (a+b x^2\right )^{3+p}}{b^3 (3+p)}\right ) \] Input:
Integrate[x^3*(a + b*x^2)^p*(c + d*x^2),x]
Output:
(-((a*(b*c - a*d)*(a + b*x^2)^(1 + p))/(b^3*(1 + p))) + ((b*c - 2*a*d)*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (d*(a + b*x^2)^(3 + p))/(b^3*(3 + p)))/2
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {354, 86, 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.
\(\displaystyle \int x^3 \left (c+d x^2\right ) \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a (a d-b c) \left (b x^2+a\right )^p}{b^2}+\frac {(b c-2 a d) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {d \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a (b c-a d) \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}+\frac {(b c-2 a d) \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {d \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )\) |
Input:
Int[x^3*(a + b*x^2)^p*(c + d*x^2),x]
Output:
(-((a*(b*c - a*d)*(a + b*x^2)^(1 + p))/(b^3*(1 + p))) + ((b*c - 2*a*d)*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (d*(a + b*x^2)^(3 + p))/(b^3*(3 + p)))/2
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
Time = 0.46 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right )^{p +1} \left (b^{2} d \,p^{2} x^{4}+3 b^{2} d p \,x^{4}+b^{2} c \,p^{2} x^{2}+2 d \,b^{2} x^{4}-2 a b d p \,x^{2}+4 b^{2} c p \,x^{2}-2 a b d \,x^{2}+3 b^{2} c \,x^{2}-a b c p +2 d \,a^{2}-3 a b c \right )}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(127\) |
orering | \(\frac {\left (b \,x^{2}+a \right ) \left (b^{2} d \,p^{2} x^{4}+3 b^{2} d p \,x^{4}+b^{2} c \,p^{2} x^{2}+2 d \,b^{2} x^{4}-2 a b d p \,x^{2}+4 b^{2} c p \,x^{2}-2 a b d \,x^{2}+3 b^{2} c \,x^{2}-a b c p +2 d \,a^{2}-3 a b c \right ) \left (b \,x^{2}+a \right )^{p}}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(132\) |
norman | \(\frac {d \,x^{6} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 p +6}+\frac {a^{2} \left (-b c p +2 a d -3 b c \right ) {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}+\frac {\left (a d p +b c p +3 b c \right ) x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+5 p +6\right )}-\frac {p a \left (-b c p +2 a d -3 b c \right ) x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b^{2} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(164\) |
risch | \(\frac {\left (b^{3} d \,p^{2} x^{6}+3 b^{3} d p \,x^{6}+a \,b^{2} d \,p^{2} x^{4}+b^{3} c \,p^{2} x^{4}+2 d \,x^{6} b^{3}+a \,b^{2} d p \,x^{4}+4 b^{3} c p \,x^{4}+a \,b^{2} c \,p^{2} x^{2}+3 b^{3} c \,x^{4}-2 a^{2} b d p \,x^{2}+3 a \,b^{2} c p \,x^{2}-a^{2} b c p +2 a^{3} d -3 a^{2} b c \right ) \left (b \,x^{2}+a \right )^{p}}{2 \left (2+p \right ) \left (3+p \right ) \left (p +1\right ) b^{3}}\) | \(168\) |
parallelrisch | \(\frac {x^{6} \left (b \,x^{2}+a \right )^{p} b^{3} d \,p^{2}+3 x^{6} \left (b \,x^{2}+a \right )^{p} b^{3} d p +2 x^{6} \left (b \,x^{2}+a \right )^{p} b^{3} d +x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} d \,p^{2}+x^{4} \left (b \,x^{2}+a \right )^{p} b^{3} c \,p^{2}+x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} d p +4 x^{4} \left (b \,x^{2}+a \right )^{p} b^{3} c p +3 x^{4} \left (b \,x^{2}+a \right )^{p} b^{3} c +x^{2} \left (b \,x^{2}+a \right )^{p} a \,b^{2} c \,p^{2}-2 x^{2} \left (b \,x^{2}+a \right )^{p} a^{2} b d p +3 x^{2} \left (b \,x^{2}+a \right )^{p} a \,b^{2} c p -\left (b \,x^{2}+a \right )^{p} a^{2} b c p +2 \left (b \,x^{2}+a \right )^{p} a^{3} d -3 \left (b \,x^{2}+a \right )^{p} a^{2} b c}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(285\) |
Input:
int(x^3*(b*x^2+a)^p*(d*x^2+c),x,method=_RETURNVERBOSE)
Output:
1/2/b^3*(b*x^2+a)^(p+1)/(p^3+6*p^2+11*p+6)*(b^2*d*p^2*x^4+3*b^2*d*p*x^4+b^ 2*c*p^2*x^2+2*b^2*d*x^4-2*a*b*d*p*x^2+4*b^2*c*p*x^2-2*a*b*d*x^2+3*b^2*c*x^ 2-a*b*c*p+2*a^2*d-3*a*b*c)
Time = 0.13 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.83 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left ({\left (b^{3} d p^{2} + 3 \, b^{3} d p + 2 \, b^{3} d\right )} x^{6} - a^{2} b c p + {\left (3 \, b^{3} c + {\left (b^{3} c + a b^{2} d\right )} p^{2} + {\left (4 \, b^{3} c + a b^{2} d\right )} p\right )} x^{4} - 3 \, a^{2} b c + 2 \, a^{3} d + {\left (a b^{2} c p^{2} + {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} p\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \] Input:
integrate(x^3*(b*x^2+a)^p*(d*x^2+c),x, algorithm="fricas")
Output:
1/2*((b^3*d*p^2 + 3*b^3*d*p + 2*b^3*d)*x^6 - a^2*b*c*p + (3*b^3*c + (b^3*c + a*b^2*d)*p^2 + (4*b^3*c + a*b^2*d)*p)*x^4 - 3*a^2*b*c + 2*a^3*d + (a*b^ 2*c*p^2 + (3*a*b^2*c - 2*a^2*b*d)*p)*x^2)*(b*x^2 + a)^p/(b^3*p^3 + 6*b^3*p ^2 + 11*b^3*p + 6*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 1586 vs. \(2 (71) = 142\).
Time = 10.44 (sec) , antiderivative size = 1586, normalized size of antiderivative = 18.02 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(x**3*(b*x**2+a)**p*(d*x**2+c),x)
Output:
Piecewise((a**p*(c*x**4/4 + d*x**6/6), Eq(b, 0)), (2*a**2*d*log(x - sqrt(- a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*d*log(x + sqrt( -a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 3*a**2*d/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) - a*b*c/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b **5*x**4) + 4*a*b*d*x**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*d*x**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4* x**2 + 4*b**5*x**4) + 4*a*b*d*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x **4) - 2*b**2*c*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2* d*x**4*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2 *b**2*d*x**4*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x** 4), Eq(p, -3)), (-2*a**2*d*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*d*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*d/(2*a*b**3 + 2*b**4*x**2) + a*b*c*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + a*b *c*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + a*b*c/(2*a*b**3 + 2*b**4 *x**2) - 2*a*b*d*x**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a*b *d*x**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*c*x**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*c*x**2*log(x + sqrt(-a/b))/(2 *a*b**3 + 2*b**4*x**2) + b**2*d*x**4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*d*log(x - sqrt(-a/b))/(2*b**3) + a**2*d*log(x + sqrt(-a/b))/(2*b**3 ) - a*c*log(x - sqrt(-a/b))/(2*b**2) - a*c*log(x + sqrt(-a/b))/(2*b**2)...
Time = 0.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left (b^{2} {\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{p} c}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} + \frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} + {\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{p} d}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \] Input:
integrate(x^3*(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*c/((p^2 + 3*p + 2)*b ^2) + 1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3)*(b*x^2 + a)^p*d/((p^3 + 6*p^2 + 11*p + 6)*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (82) = 164\).
Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.32 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} b c p + {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} d p - 2 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a d p + 3 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} b c + 2 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} d - 6 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a d}{2 \, {\left (b^{3} p^{2} + 5 \, b^{3} p + 6 \, b^{3}\right )}} - \frac {\frac {{\left (b x^{2} + a\right )}^{p + 1} a b c}{p + 1} - \frac {{\left (b x^{2} + a\right )}^{p + 1} a^{2} d}{p + 1}}{2 \, b^{3}} \] Input:
integrate(x^3*(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*b*c*p + (b*x^2 + a)^3*(b*x^2 + a)^p*d*p - 2*(b*x^2 + a)^2*(b*x^2 + a)^p*a*d*p + 3*(b*x^2 + a)^2*(b*x^2 + a)^p*b*c + 2*(b*x^2 + a)^3*(b*x^2 + a)^p*d - 6*(b*x^2 + a)^2*(b*x^2 + a)^p*a*d)/(b^3 *p^2 + 5*b^3*p + 6*b^3) - 1/2*((b*x^2 + a)^(p + 1)*a*b*c/(p + 1) - (b*x^2 + a)^(p + 1)*a^2*d/(p + 1))/b^3
Time = 0.54 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx={\left (b\,x^2+a\right )}^p\,\left (\frac {d\,x^6\,\left (p^2+3\,p+2\right )}{2\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {a^2\,\left (3\,b\,c-2\,a\,d+b\,c\,p\right )}{2\,b^3\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {x^4\,\left (p+1\right )\,\left (3\,b\,c+a\,d\,p+b\,c\,p\right )}{2\,b\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^2\,\left (3\,b\,c-2\,a\,d+b\,c\,p\right )}{2\,b^2\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:
int(x^3*(a + b*x^2)^p*(c + d*x^2),x)
Output:
(a + b*x^2)^p*((d*x^6*(3*p + p^2 + 2))/(2*(11*p + 6*p^2 + p^3 + 6)) - (a^2 *(3*b*c - 2*a*d + b*c*p))/(2*b^3*(11*p + 6*p^2 + p^3 + 6)) + (x^4*(p + 1)* (3*b*c + a*d*p + b*c*p))/(2*b*(11*p + 6*p^2 + p^3 + 6)) + (a*p*x^2*(3*b*c - 2*a*d + b*c*p))/(2*b^2*(11*p + 6*p^2 + p^3 + 6)))
Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.90 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {\left (b \,x^{2}+a \right )^{p} \left (b^{3} d \,p^{2} x^{6}+3 b^{3} d p \,x^{6}+a \,b^{2} d \,p^{2} x^{4}+b^{3} c \,p^{2} x^{4}+2 b^{3} d \,x^{6}+a \,b^{2} d p \,x^{4}+4 b^{3} c p \,x^{4}+a \,b^{2} c \,p^{2} x^{2}+3 b^{3} c \,x^{4}-2 a^{2} b d p \,x^{2}+3 a \,b^{2} c p \,x^{2}-a^{2} b c p +2 a^{3} d -3 a^{2} b c \right )}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^3*(b*x^2+a)^p*(d*x^2+c),x)
Output:
((a + b*x**2)**p*(2*a**3*d - a**2*b*c*p - 3*a**2*b*c - 2*a**2*b*d*p*x**2 + a*b**2*c*p**2*x**2 + 3*a*b**2*c*p*x**2 + a*b**2*d*p**2*x**4 + a*b**2*d*p* x**4 + b**3*c*p**2*x**4 + 4*b**3*c*p*x**4 + 3*b**3*c*x**4 + b**3*d*p**2*x* *6 + 3*b**3*d*p*x**6 + 2*b**3*d*x**6))/(2*b**3*(p**3 + 6*p**2 + 11*p + 6))