Integrand size = 13, antiderivative size = 100 \[ \int x^7 \left (a+b x^2\right )^p \, dx=-\frac {a^3 \left (a+b x^2\right )^{1+p}}{2 b^4 (1+p)}+\frac {3 a^2 \left (a+b x^2\right )^{2+p}}{2 b^4 (2+p)}-\frac {3 a \left (a+b x^2\right )^{3+p}}{2 b^4 (3+p)}+\frac {\left (a+b x^2\right )^{4+p}}{2 b^4 (4+p)} \] Output:
-1/2*a^3*(b*x^2+a)^(p+1)/b^4/(p+1)+3/2*a^2*(b*x^2+a)^(2+p)/b^4/(2+p)-3/2*a *(b*x^2+a)^(3+p)/b^4/(3+p)+1/2*(b*x^2+a)^(4+p)/b^4/(4+p)
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\frac {1}{2} \left (-\frac {a^3 \left (a+b x^2\right )^{1+p}}{b^4 (1+p)}+\frac {3 a^2 \left (a+b x^2\right )^{2+p}}{b^4 (2+p)}-\frac {3 a \left (a+b x^2\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^2\right )^{4+p}}{b^4 (4+p)}\right ) \] Input:
Integrate[x^7*(a + b*x^2)^p,x]
Output:
(-((a^3*(a + b*x^2)^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*x^2)^(2 + p))/ (b^4*(2 + p)) - (3*a*(a + b*x^2)^(3 + p))/(b^4*(3 + p)) + (a + b*x^2)^(4 + p)/(b^4*(4 + p)))/2
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {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.
\(\displaystyle \int x^7 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^6 \left (b x^2+a\right )^pdx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {a^3 \left (b x^2+a\right )^p}{b^3}+\frac {3 a^2 \left (b x^2+a\right )^{p+1}}{b^3}-\frac {3 a \left (b x^2+a\right )^{p+2}}{b^3}+\frac {\left (b x^2+a\right )^{p+3}}{b^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^3 \left (a+b x^2\right )^{p+1}}{b^4 (p+1)}+\frac {3 a^2 \left (a+b x^2\right )^{p+2}}{b^4 (p+2)}-\frac {3 a \left (a+b x^2\right )^{p+3}}{b^4 (p+3)}+\frac {\left (a+b x^2\right )^{p+4}}{b^4 (p+4)}\right )\) |
Input:
Int[x^7*(a + b*x^2)^p,x]
Output:
(-((a^3*(a + b*x^2)^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*x^2)^(2 + p))/ (b^4*(2 + p)) - (3*a*(a + b*x^2)^(3 + p))/(b^4*(3 + p)) + (a + b*x^2)^(4 + p)/(b^4*(4 + 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_)^(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]
Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.32
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{p +1} \left (-b^{3} p^{3} x^{6}-6 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+3 a \,b^{2} p^{2} x^{4}-6 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+6 a \,b^{2} x^{4}-6 a^{2} b p \,x^{2}-6 a^{2} b \,x^{2}+6 a^{3}\right )}{2 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(132\) |
orering | \(-\frac {\left (b \,x^{2}+a \right )^{p} \left (-b^{3} p^{3} x^{6}-6 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+3 a \,b^{2} p^{2} x^{4}-6 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+6 a \,b^{2} x^{4}-6 a^{2} b p \,x^{2}-6 a^{2} b \,x^{2}+6 a^{3}\right ) \left (b \,x^{2}+a \right )}{2 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(137\) |
risch | \(-\frac {\left (-b^{4} p^{3} x^{8}-6 b^{4} p^{2} x^{8}-a \,b^{3} p^{3} x^{6}-11 b^{4} p \,x^{8}-3 a \,b^{3} p^{2} x^{6}-6 b^{4} x^{8}-2 a p \,x^{6} b^{3}+3 a^{2} b^{2} p^{2} x^{4}+3 a^{2} p \,x^{4} b^{2}-6 a^{3} p \,x^{2} b +6 a^{4}\right ) \left (b \,x^{2}+a \right )^{p}}{2 \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (p +1\right ) b^{4}}\) | \(150\) |
norman | \(\frac {x^{8} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 p +8}-\frac {3 a^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}+\frac {a p \,x^{6} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+7 p +12\right )}-\frac {3 a^{2} p \,x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b^{2} \left (p^{3}+9 p^{2}+26 p +24\right )}+\frac {3 p \,a^{3} x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{3} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(174\) |
parallelrisch | \(\frac {x^{8} \left (b \,x^{2}+a \right )^{p} a \,b^{4} p^{3}+6 x^{8} \left (b \,x^{2}+a \right )^{p} a \,b^{4} p^{2}+11 x^{8} \left (b \,x^{2}+a \right )^{p} a \,b^{4} p +x^{6} \left (b \,x^{2}+a \right )^{p} a^{2} b^{3} p^{3}+6 x^{8} \left (b \,x^{2}+a \right )^{p} a \,b^{4}+3 x^{6} \left (b \,x^{2}+a \right )^{p} a^{2} b^{3} p^{2}+2 x^{6} \left (b \,x^{2}+a \right )^{p} a^{2} b^{3} p -3 x^{4} \left (b \,x^{2}+a \right )^{p} a^{3} b^{2} p^{2}-3 x^{4} \left (b \,x^{2}+a \right )^{p} a^{3} b^{2} p +6 x^{2} \left (b \,x^{2}+a \right )^{p} a^{4} b p -6 a^{5} \left (b \,x^{2}+a \right )^{p}}{2 \left (4+p \right ) \left (p^{2}+5 p +6\right ) a \left (p +1\right ) b^{4}}\) | \(251\) |
Input:
int(x^7*(b*x^2+a)^p,x,method=_RETURNVERBOSE)
Output:
-1/2/b^4*(b*x^2+a)^(p+1)/(p^4+10*p^3+35*p^2+50*p+24)*(-b^3*p^3*x^6-6*b^3*p ^2*x^6-11*b^3*p*x^6+3*a*b^2*p^2*x^4-6*b^3*x^6+9*a*b^2*p*x^4+6*a*b^2*x^4-6* a^2*b*p*x^2-6*a^2*b*x^2+6*a^3)
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.48 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\frac {{\left ({\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{8} + 6 \, a^{3} b p x^{2} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x^{6} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} - 6 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )}} \] Input:
integrate(x^7*(b*x^2+a)^p,x, algorithm="fricas")
Output:
1/2*((b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^8 + 6*a^3*b*p*x^2 + (a*b^3 *p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*x^6 - 3*(a^2*b^2*p^2 + a^2*b^2*p)*x^4 - 6* a^4)*(b*x^2 + a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 1923 vs. \(2 (85) = 170\).
Time = 2.76 (sec) , antiderivative size = 1923, normalized size of antiderivative = 19.23 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\text {Too large to display} \] Input:
integrate(x**7*(b*x**2+a)**p,x)
Output:
Piecewise((a**p*x**8/8, Eq(b, 0)), (6*a**3*log(x - sqrt(-a/b))/(12*a**3*b* *4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*a**3*log(x + s qrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x* *6) + 11*a**3/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7 *x**6) + 18*a**2*b*x**2*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x **2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a**2*b*x**2*log(x + sqrt(-a/b))/ (12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 27*a* *2*b*x**2/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x** 6) + 18*a*b**2*x**4*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2*x**4*log(x + sqrt(-a/b))/(12* a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2 *x**4/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*b**3*x**6*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a* b**6*x**4 + 12*b**7*x**6) + 6*b**3*x**6*log(x + sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6), Eq(p, -4)), (-6*a**3 *log(x - sqrt(-a/b))/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 6*a**3* log(x + sqrt(-a/b))/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 9*a**3/( 4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x**2*log(x - sqrt(- a/b))/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x**2*log(x + sqrt(-a/b))/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x*...
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.06 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{8} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{6} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 6 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \] Input:
integrate(x^7*(b*x^2+a)^p,x, algorithm="maxima")
Output:
1/2*((p^3 + 6*p^2 + 11*p + 6)*b^4*x^8 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^6 - 3* (p^2 + p)*a^2*b^2*x^4 + 6*a^3*b*p*x^2 - 6*a^4)*(b*x^2 + a)^p/((p^4 + 10*p^ 3 + 35*p^2 + 50*p + 24)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (92) = 184\).
Time = 0.13 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.60 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\frac {{\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} p^{2} - 3 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a p^{2} + 3 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} p^{2} + 5 \, {\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} p - 18 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a p + 21 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} p + 6 \, {\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} - 24 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a + 36 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2}}{2 \, {\left (b^{4} p^{3} + 9 \, b^{4} p^{2} + 26 \, b^{4} p + 24 \, b^{4}\right )}} - \frac {{\left (b x^{2} + a\right )}^{p + 1} a^{3}}{2 \, b^{4} {\left (p + 1\right )}} \] Input:
integrate(x^7*(b*x^2+a)^p,x, algorithm="giac")
Output:
1/2*((b*x^2 + a)^4*(b*x^2 + a)^p*p^2 - 3*(b*x^2 + a)^3*(b*x^2 + a)^p*a*p^2 + 3*(b*x^2 + a)^2*(b*x^2 + a)^p*a^2*p^2 + 5*(b*x^2 + a)^4*(b*x^2 + a)^p*p - 18*(b*x^2 + a)^3*(b*x^2 + a)^p*a*p + 21*(b*x^2 + a)^2*(b*x^2 + a)^p*a^2 *p + 6*(b*x^2 + a)^4*(b*x^2 + a)^p - 24*(b*x^2 + a)^3*(b*x^2 + a)^p*a + 36 *(b*x^2 + a)^2*(b*x^2 + a)^p*a^2)/(b^4*p^3 + 9*b^4*p^2 + 26*b^4*p + 24*b^4 ) - 1/2*(b*x^2 + a)^(p + 1)*a^3/(b^4*(p + 1))
Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.83 \[ \int x^7 \left (a+b x^2\right )^p \, dx={\left (b\,x^2+a\right )}^p\,\left (\frac {x^8\,\left (p^3+6\,p^2+11\,p+6\right )}{2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^4}{b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {3\,a^3\,p\,x^2}{b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^6\,\left (p^2+3\,p+2\right )}{2\,b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^2\,p\,x^4\,\left (p+1\right )}{2\,b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right ) \] Input:
int(x^7*(a + b*x^2)^p,x)
Output:
(a + b*x^2)^p*((x^8*(11*p + 6*p^2 + p^3 + 6))/(2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^4)/(b^4*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (3*a^3*p *x^2)/(b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^6*(3*p + p^2 + 2) )/(2*b*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^2*p*x^4*(p + 1))/(2*b^2 *(50*p + 35*p^2 + 10*p^3 + p^4 + 24)))
Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int x^7 \left (a+b x^2\right )^p \, dx=\frac {\left (b \,x^{2}+a \right )^{p} \left (b^{4} p^{3} x^{8}+6 b^{4} p^{2} x^{8}+a \,b^{3} p^{3} x^{6}+11 b^{4} p \,x^{8}+3 a \,b^{3} p^{2} x^{6}+6 b^{4} x^{8}+2 a \,b^{3} p \,x^{6}-3 a^{2} b^{2} p^{2} x^{4}-3 a^{2} b^{2} p \,x^{4}+6 a^{3} b p \,x^{2}-6 a^{4}\right )}{2 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:
int(x^7*(b*x^2+a)^p,x)
Output:
((a + b*x**2)**p*( - 6*a**4 + 6*a**3*b*p*x**2 - 3*a**2*b**2*p**2*x**4 - 3* a**2*b**2*p*x**4 + a*b**3*p**3*x**6 + 3*a*b**3*p**2*x**6 + 2*a*b**3*p*x**6 + b**4*p**3*x**8 + 6*b**4*p**2*x**8 + 11*b**4*p*x**8 + 6*b**4*x**8))/(2*b **4*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))