Integrand size = 30, antiderivative size = 297 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 a^5 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^3 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{21/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{25/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{25 d^{11} \left (a+b x^2\right )} \] Output:
2/5*a^5*(d*x)^(5/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+10/9*a^4*b*(d*x)^(9/2) *((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+20/13*a^3*b^2*(d*x)^(13/2)*((b*x^2+a)^2 )^(1/2)/d^5/(b*x^2+a)+20/17*a^2*b^3*(d*x)^(17/2)*((b*x^2+a)^2)^(1/2)/d^7/( b*x^2+a)+10/21*a*b^4*(d*x)^(21/2)*((b*x^2+a)^2)^(1/2)/d^9/(b*x^2+a)+2/25*b ^5*(d*x)^(25/2)*((b*x^2+a)^2)^(1/2)/d^11/(b*x^2+a)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 x (d x)^{3/2} \sqrt {\left (a+b x^2\right )^2} \left (69615 a^5+193375 a^4 b x^2+267750 a^3 b^2 x^4+204750 a^2 b^3 x^6+82875 a b^4 x^8+13923 b^5 x^{10}\right )}{348075 \left (a+b x^2\right )} \] Input:
Integrate[(d*x)^(3/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(2*x*(d*x)^(3/2)*Sqrt[(a + b*x^2)^2]*(69615*a^5 + 193375*a^4*b*x^2 + 26775 0*a^3*b^2*x^4 + 204750*a^2*b^3*x^6 + 82875*a*b^4*x^8 + 13923*b^5*x^10))/(3 48075*(a + b*x^2))
Time = 0.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1384, 27, 244, 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.
\(\displaystyle \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 (d x)^{3/2} \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (d x)^{3/2} \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^5 (d x)^{23/2}}{d^{10}}+\frac {5 a b^4 (d x)^{19/2}}{d^8}+\frac {10 a^2 b^3 (d x)^{15/2}}{d^6}+\frac {10 a^3 b^2 (d x)^{11/2}}{d^4}+\frac {5 a^4 b (d x)^{7/2}}{d^2}+a^5 (d x)^{3/2}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {2 a^5 (d x)^{5/2}}{5 d}+\frac {10 a^4 b (d x)^{9/2}}{9 d^3}+\frac {20 a^3 b^2 (d x)^{13/2}}{13 d^5}+\frac {20 a^2 b^3 (d x)^{17/2}}{17 d^7}+\frac {10 a b^4 (d x)^{21/2}}{21 d^9}+\frac {2 b^5 (d x)^{25/2}}{25 d^{11}}\right )}{a+b x^2}\) |
Input:
Int[(d*x)^(3/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((2*a^5*(d*x)^(5/2))/(5*d) + (10*a^4*b*(d *x)^(9/2))/(9*d^3) + (20*a^3*b^2*(d*x)^(13/2))/(13*d^5) + (20*a^2*b^3*(d*x )^(17/2))/(17*d^7) + (10*a*b^4*(d*x)^(21/2))/(21*d^9) + (2*b^5*(d*x)^(25/2 ))/(25*d^11)))/(a + b*x^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28
method | result | size |
gosper | \(\frac {2 x \left (13923 x^{10} b^{5}+82875 a \,x^{8} b^{4}+204750 a^{2} x^{6} b^{3}+267750 a^{3} x^{4} b^{2}+193375 x^{2} a^{4} b +69615 a^{5}\right ) \left (d x \right )^{\frac {3}{2}} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{348075 \left (b \,x^{2}+a \right )^{5}}\) | \(83\) |
default | \(\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (d x \right )^{\frac {5}{2}} \left (13923 x^{10} b^{5}+82875 a \,x^{8} b^{4}+204750 a^{2} x^{6} b^{3}+267750 a^{3} x^{4} b^{2}+193375 x^{2} a^{4} b +69615 a^{5}\right )}{348075 d \left (b \,x^{2}+a \right )^{5}}\) | \(85\) |
risch | \(\frac {2 d^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{3} \left (13923 x^{10} b^{5}+82875 a \,x^{8} b^{4}+204750 a^{2} x^{6} b^{3}+267750 a^{3} x^{4} b^{2}+193375 x^{2} a^{4} b +69615 a^{5}\right )}{348075 \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(88\) |
orering | \(\frac {2 x \left (13923 x^{10} b^{5}+82875 a \,x^{8} b^{4}+204750 a^{2} x^{6} b^{3}+267750 a^{3} x^{4} b^{2}+193375 x^{2} a^{4} b +69615 a^{5}\right ) \left (d x \right )^{\frac {3}{2}} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{348075 \left (b \,x^{2}+a \right )^{5}}\) | \(92\) |
Input:
int((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/348075*x*(13923*b^5*x^10+82875*a*b^4*x^8+204750*a^2*b^3*x^6+267750*a^3*b ^2*x^4+193375*a^4*b*x^2+69615*a^5)*(d*x)^(3/2)*((b*x^2+a)^2)^(5/2)/(b*x^2+ a)^5
Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.24 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{348075} \, {\left (13923 \, b^{5} d x^{12} + 82875 \, a b^{4} d x^{10} + 204750 \, a^{2} b^{3} d x^{8} + 267750 \, a^{3} b^{2} d x^{6} + 193375 \, a^{4} b d x^{4} + 69615 \, a^{5} d x^{2}\right )} \sqrt {d x} \] Input:
integrate((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
2/348075*(13923*b^5*d*x^12 + 82875*a*b^4*d*x^10 + 204750*a^2*b^3*d*x^8 + 2 67750*a^3*b^2*d*x^6 + 193375*a^4*b*d*x^4 + 69615*a^5*d*x^2)*sqrt(d*x)
\[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int \left (d x\right )^{\frac {3}{2}} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((d*x)**(3/2)*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral((d*x)**(3/2)*((a + b*x**2)**2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.49 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{525} \, {\left (21 \, b^{5} d^{\frac {3}{2}} x^{3} + 25 \, a b^{4} d^{\frac {3}{2}} x\right )} x^{\frac {19}{2}} + \frac {8}{357} \, {\left (17 \, a b^{4} d^{\frac {3}{2}} x^{3} + 21 \, a^{2} b^{3} d^{\frac {3}{2}} x\right )} x^{\frac {15}{2}} + \frac {12}{221} \, {\left (13 \, a^{2} b^{3} d^{\frac {3}{2}} x^{3} + 17 \, a^{3} b^{2} d^{\frac {3}{2}} x\right )} x^{\frac {11}{2}} + \frac {8}{117} \, {\left (9 \, a^{3} b^{2} d^{\frac {3}{2}} x^{3} + 13 \, a^{4} b d^{\frac {3}{2}} x\right )} x^{\frac {7}{2}} + \frac {2}{45} \, {\left (5 \, a^{4} b d^{\frac {3}{2}} x^{3} + 9 \, a^{5} d^{\frac {3}{2}} x\right )} x^{\frac {3}{2}} \] Input:
integrate((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
2/525*(21*b^5*d^(3/2)*x^3 + 25*a*b^4*d^(3/2)*x)*x^(19/2) + 8/357*(17*a*b^4 *d^(3/2)*x^3 + 21*a^2*b^3*d^(3/2)*x)*x^(15/2) + 12/221*(13*a^2*b^3*d^(3/2) *x^3 + 17*a^3*b^2*d^(3/2)*x)*x^(11/2) + 8/117*(9*a^3*b^2*d^(3/2)*x^3 + 13* a^4*b*d^(3/2)*x)*x^(7/2) + 2/45*(5*a^4*b*d^(3/2)*x^3 + 9*a^5*d^(3/2)*x)*x^ (3/2)
Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.46 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{348075} \, {\left (13923 \, \sqrt {d x} b^{5} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + 82875 \, \sqrt {d x} a b^{4} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 204750 \, \sqrt {d x} a^{2} b^{3} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 267750 \, \sqrt {d x} a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 193375 \, \sqrt {d x} a^{4} b x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 69615 \, \sqrt {d x} a^{5} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} d \] Input:
integrate((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
2/348075*(13923*sqrt(d*x)*b^5*x^12*sgn(b*x^2 + a) + 82875*sqrt(d*x)*a*b^4* x^10*sgn(b*x^2 + a) + 204750*sqrt(d*x)*a^2*b^3*x^8*sgn(b*x^2 + a) + 267750 *sqrt(d*x)*a^3*b^2*x^6*sgn(b*x^2 + a) + 193375*sqrt(d*x)*a^4*b*x^4*sgn(b*x ^2 + a) + 69615*sqrt(d*x)*a^5*x^2*sgn(b*x^2 + a))*d
Timed out. \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int {\left (d\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int((d*x)^(3/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int((d*x)^(3/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.22 \[ \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 \sqrt {x}\, \sqrt {d}\, d \,x^{2} \left (13923 b^{5} x^{10}+82875 a \,b^{4} x^{8}+204750 a^{2} b^{3} x^{6}+267750 a^{3} b^{2} x^{4}+193375 a^{4} b \,x^{2}+69615 a^{5}\right )}{348075} \] Input:
int((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(2*sqrt(x)*sqrt(d)*d*x**2*(69615*a**5 + 193375*a**4*b*x**2 + 267750*a**3*b **2*x**4 + 204750*a**2*b**3*x**6 + 82875*a*b**4*x**8 + 13923*b**5*x**10))/ 348075