Integrand size = 15, antiderivative size = 80 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=-\frac {a^3 \left (a+b x^2\right )^{11/2}}{11 b^4}+\frac {3 a^2 \left (a+b x^2\right )^{13/2}}{13 b^4}-\frac {a \left (a+b x^2\right )^{15/2}}{5 b^4}+\frac {\left (a+b x^2\right )^{17/2}}{17 b^4} \] Output:
-1/11*a^3*(b*x^2+a)^(11/2)/b^4+3/13*a^2*(b*x^2+a)^(13/2)/b^4-1/5*a*(b*x^2+ a)^(15/2)/b^4+1/17*(b*x^2+a)^(17/2)/b^4
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\frac {\left (a+b x^2\right )^{11/2} \left (-16 a^3+88 a^2 b x^2-286 a b^2 x^4+715 b^3 x^6\right )}{12155 b^4} \] Input:
Integrate[x^7*(a + b*x^2)^(9/2),x]
Output:
((a + b*x^2)^(11/2)*(-16*a^3 + 88*a^2*b*x^2 - 286*a*b^2*x^4 + 715*b^3*x^6) )/(12155*b^4)
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int x^7 \left (a+b x^2\right )^{9/2} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^6 \left (b x^2+a\right )^{9/2}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right )^{15/2}}{b^3}-\frac {3 a \left (b x^2+a\right )^{13/2}}{b^3}+\frac {3 a^2 \left (b x^2+a\right )^{11/2}}{b^3}-\frac {a^3 \left (b x^2+a\right )^{9/2}}{b^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 a^3 \left (a+b x^2\right )^{11/2}}{11 b^4}+\frac {6 a^2 \left (a+b x^2\right )^{13/2}}{13 b^4}+\frac {2 \left (a+b x^2\right )^{17/2}}{17 b^4}-\frac {2 a \left (a+b x^2\right )^{15/2}}{5 b^4}\right )\) |
Input:
Int[x^7*(a + b*x^2)^(9/2),x]
Output:
((-2*a^3*(a + b*x^2)^(11/2))/(11*b^4) + (6*a^2*(a + b*x^2)^(13/2))/(13*b^4 ) - (2*a*(a + b*x^2)^(15/2))/(5*b^4) + (2*(a + b*x^2)^(17/2))/(17*b^4))/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 = 0.36 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}} \left (-715 b^{3} x^{6}+286 a \,b^{2} x^{4}-88 a^{2} b \,x^{2}+16 a^{3}\right )}{12155 b^{4}}\) | \(47\) |
pseudoelliptic | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}} \left (-715 b^{3} x^{6}+286 a \,b^{2} x^{4}-88 a^{2} b \,x^{2}+16 a^{3}\right )}{12155 b^{4}}\) | \(47\) |
orering | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}} \left (-715 b^{3} x^{6}+286 a \,b^{2} x^{4}-88 a^{2} b \,x^{2}+16 a^{3}\right )}{12155 b^{4}}\) | \(47\) |
default | \(\frac {x^{6} \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{17 b}-\frac {6 a \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{15 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{13 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{143 b^{2}}\right )}{15 b}\right )}{17 b}\) | \(82\) |
trager | \(-\frac {\left (-715 b^{8} x^{16}-3289 a \,b^{7} x^{14}-5808 a^{2} b^{6} x^{12}-4714 a^{3} b^{5} x^{10}-1515 a^{4} b^{4} x^{8}-5 a^{5} b^{3} x^{6}+6 a^{6} b^{2} x^{4}-8 a^{7} b \,x^{2}+16 a^{8}\right ) \sqrt {b \,x^{2}+a}}{12155 b^{4}}\) | \(102\) |
risch | \(-\frac {\left (-715 b^{8} x^{16}-3289 a \,b^{7} x^{14}-5808 a^{2} b^{6} x^{12}-4714 a^{3} b^{5} x^{10}-1515 a^{4} b^{4} x^{8}-5 a^{5} b^{3} x^{6}+6 a^{6} b^{2} x^{4}-8 a^{7} b \,x^{2}+16 a^{8}\right ) \sqrt {b \,x^{2}+a}}{12155 b^{4}}\) | \(102\) |
Input:
int(x^7*(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/12155*(b*x^2+a)^(11/2)*(-715*b^3*x^6+286*a*b^2*x^4-88*a^2*b*x^2+16*a^3) /b^4
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\frac {{\left (715 \, b^{8} x^{16} + 3289 \, a b^{7} x^{14} + 5808 \, a^{2} b^{6} x^{12} + 4714 \, a^{3} b^{5} x^{10} + 1515 \, a^{4} b^{4} x^{8} + 5 \, a^{5} b^{3} x^{6} - 6 \, a^{6} b^{2} x^{4} + 8 \, a^{7} b x^{2} - 16 \, a^{8}\right )} \sqrt {b x^{2} + a}}{12155 \, b^{4}} \] Input:
integrate(x^7*(b*x^2+a)^(9/2),x, algorithm="fricas")
Output:
1/12155*(715*b^8*x^16 + 3289*a*b^7*x^14 + 5808*a^2*b^6*x^12 + 4714*a^3*b^5 *x^10 + 1515*a^4*b^4*x^8 + 5*a^5*b^3*x^6 - 6*a^6*b^2*x^4 + 8*a^7*b*x^2 - 1 6*a^8)*sqrt(b*x^2 + a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (70) = 140\).
Time = 1.24 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.55 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\begin {cases} - \frac {16 a^{8} \sqrt {a + b x^{2}}}{12155 b^{4}} + \frac {8 a^{7} x^{2} \sqrt {a + b x^{2}}}{12155 b^{3}} - \frac {6 a^{6} x^{4} \sqrt {a + b x^{2}}}{12155 b^{2}} + \frac {a^{5} x^{6} \sqrt {a + b x^{2}}}{2431 b} + \frac {303 a^{4} x^{8} \sqrt {a + b x^{2}}}{2431} + \frac {4714 a^{3} b x^{10} \sqrt {a + b x^{2}}}{12155} + \frac {528 a^{2} b^{2} x^{12} \sqrt {a + b x^{2}}}{1105} + \frac {23 a b^{3} x^{14} \sqrt {a + b x^{2}}}{85} + \frac {b^{4} x^{16} \sqrt {a + b x^{2}}}{17} & \text {for}\: b \neq 0 \\\frac {a^{\frac {9}{2}} x^{8}}{8} & \text {otherwise} \end {cases} \] Input:
integrate(x**7*(b*x**2+a)**(9/2),x)
Output:
Piecewise((-16*a**8*sqrt(a + b*x**2)/(12155*b**4) + 8*a**7*x**2*sqrt(a + b *x**2)/(12155*b**3) - 6*a**6*x**4*sqrt(a + b*x**2)/(12155*b**2) + a**5*x** 6*sqrt(a + b*x**2)/(2431*b) + 303*a**4*x**8*sqrt(a + b*x**2)/2431 + 4714*a **3*b*x**10*sqrt(a + b*x**2)/12155 + 528*a**2*b**2*x**12*sqrt(a + b*x**2)/ 1105 + 23*a*b**3*x**14*sqrt(a + b*x**2)/85 + b**4*x**16*sqrt(a + b*x**2)/1 7, Ne(b, 0)), (a**(9/2)*x**8/8, True))
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}} x^{6}}{17 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a x^{4}}{85 \, b^{2}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a^{2} x^{2}}{1105 \, b^{3}} - \frac {16 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a^{3}}{12155 \, b^{4}} \] Input:
integrate(x^7*(b*x^2+a)^(9/2),x, algorithm="maxima")
Output:
1/17*(b*x^2 + a)^(11/2)*x^6/b - 2/85*(b*x^2 + a)^(11/2)*a*x^4/b^2 + 8/1105 *(b*x^2 + a)^(11/2)*a^2*x^2/b^3 - 16/12155*(b*x^2 + a)^(11/2)*a^3/b^4
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\frac {715 \, {\left (b x^{2} + a\right )}^{\frac {17}{2}} - 2431 \, {\left (b x^{2} + a\right )}^{\frac {15}{2}} a + 2805 \, {\left (b x^{2} + a\right )}^{\frac {13}{2}} a^{2} - 1105 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a^{3}}{12155 \, b^{4}} \] Input:
integrate(x^7*(b*x^2+a)^(9/2),x, algorithm="giac")
Output:
1/12155*(715*(b*x^2 + a)^(17/2) - 2431*(b*x^2 + a)^(15/2)*a + 2805*(b*x^2 + a)^(13/2)*a^2 - 1105*(b*x^2 + a)^(11/2)*a^3)/b^4
Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.21 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\sqrt {b\,x^2+a}\,\left (\frac {303\,a^4\,x^8}{2431}-\frac {16\,a^8}{12155\,b^4}+\frac {b^4\,x^{16}}{17}+\frac {4714\,a^3\,b\,x^{10}}{12155}+\frac {23\,a\,b^3\,x^{14}}{85}+\frac {a^5\,x^6}{2431\,b}-\frac {6\,a^6\,x^4}{12155\,b^2}+\frac {8\,a^7\,x^2}{12155\,b^3}+\frac {528\,a^2\,b^2\,x^{12}}{1105}\right ) \] Input:
int(x^7*(a + b*x^2)^(9/2),x)
Output:
(a + b*x^2)^(1/2)*((303*a^4*x^8)/2431 - (16*a^8)/(12155*b^4) + (b^4*x^16)/ 17 + (4714*a^3*b*x^10)/12155 + (23*a*b^3*x^14)/85 + (a^5*x^6)/(2431*b) - ( 6*a^6*x^4)/(12155*b^2) + (8*a^7*x^2)/(12155*b^3) + (528*a^2*b^2*x^12)/1105 )
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int x^7 \left (a+b x^2\right )^{9/2} \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (715 b^{8} x^{16}+3289 a \,b^{7} x^{14}+5808 a^{2} b^{6} x^{12}+4714 a^{3} b^{5} x^{10}+1515 a^{4} b^{4} x^{8}+5 a^{5} b^{3} x^{6}-6 a^{6} b^{2} x^{4}+8 a^{7} b \,x^{2}-16 a^{8}\right )}{12155 b^{4}} \] Input:
int(x^7*(b*x^2+a)^(9/2),x)
Output:
(sqrt(a + b*x**2)*( - 16*a**8 + 8*a**7*b*x**2 - 6*a**6*b**2*x**4 + 5*a**5* b**3*x**6 + 1515*a**4*b**4*x**8 + 4714*a**3*b**5*x**10 + 5808*a**2*b**6*x* *12 + 3289*a*b**7*x**14 + 715*b**8*x**16))/(12155*b**4)