Integrand size = 30, antiderivative size = 195 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2 a^3 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^7 \left (a+b x^2\right )} \] Output:
2/3*a^3*(d*x)^(3/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+6/7*a^2*b*(d*x)^(7/2)* ((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+6/11*a*b^2*(d*x)^(11/2)*((b*x^2+a)^2)^(1 /2)/d^5/(b*x^2+a)+2/15*b^3*(d*x)^(15/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x^2+a)
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.34 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2 x \sqrt {d x} \sqrt {\left (a+b x^2\right )^2} \left (385 a^3+495 a^2 b x^2+315 a b^2 x^4+77 b^3 x^6\right )}{1155 \left (a+b x^2\right )} \] Input:
Integrate[Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
(2*x*Sqrt[d*x]*Sqrt[(a + b*x^2)^2]*(385*a^3 + 495*a^2*b*x^2 + 315*a*b^2*x^ 4 + 77*b^3*x^6))/(1155*(a + b*x^2))
Time = 0.39 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.53, 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.
\(\displaystyle \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^3 \sqrt {d x} \left (b x^2+a\right )^3dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \sqrt {d x} \left (b x^2+a\right )^3dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^3 (d x)^{13/2}}{d^6}+\frac {3 a b^2 (d x)^{9/2}}{d^4}+\frac {3 a^2 b (d x)^{5/2}}{d^2}+a^3 \sqrt {d x}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {2 a^3 (d x)^{3/2}}{3 d}+\frac {6 a^2 b (d x)^{7/2}}{7 d^3}+\frac {6 a b^2 (d x)^{11/2}}{11 d^5}+\frac {2 b^3 (d x)^{15/2}}{15 d^7}\right )}{a+b x^2}\) |
Input:
Int[Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((2*a^3*(d*x)^(3/2))/(3*d) + (6*a^2*b*(d* x)^(7/2))/(7*d^3) + (6*a*b^2*(d*x)^(11/2))/(11*d^5) + (2*b^3*(d*x)^(15/2)) /(15*d^7)))/(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 = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {2 x \left (77 b^{3} x^{6}+315 b^{2} x^{4} a +495 a^{2} b \,x^{2}+385 a^{3}\right ) \sqrt {d x}\, {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{1155 \left (b \,x^{2}+a \right )^{3}}\) | \(61\) |
default | \(\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (d x \right )^{\frac {3}{2}} \left (77 b^{3} x^{6}+315 b^{2} x^{4} a +495 a^{2} b \,x^{2}+385 a^{3}\right )}{1155 d \left (b \,x^{2}+a \right )^{3}}\) | \(63\) |
risch | \(\frac {2 d \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2} \left (77 b^{3} x^{6}+315 b^{2} x^{4} a +495 a^{2} b \,x^{2}+385 a^{3}\right )}{1155 \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(64\) |
orering | \(\frac {2 x \left (77 b^{3} x^{6}+315 b^{2} x^{4} a +495 a^{2} b \,x^{2}+385 a^{3}\right ) \sqrt {d x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{1155 \left (b \,x^{2}+a \right )^{3}}\) | \(70\) |
Input:
int((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/1155*x*(77*b^3*x^6+315*a*b^2*x^4+495*a^2*b*x^2+385*a^3)*(d*x)^(1/2)*((b* x^2+a)^2)^(3/2)/(b*x^2+a)^3
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.21 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2}{1155} \, {\left (77 \, b^{3} x^{7} + 315 \, a b^{2} x^{5} + 495 \, a^{2} b x^{3} + 385 \, a^{3} x\right )} \sqrt {d x} \] Input:
integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
2/1155*(77*b^3*x^7 + 315*a*b^2*x^5 + 495*a^2*b*x^3 + 385*a^3*x)*sqrt(d*x)
\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int \sqrt {d x} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((d*x)**(1/2)*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(sqrt(d*x)*((a + b*x**2)**2)**(3/2), x)
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.43 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2}{165} \, {\left (11 \, b^{3} \sqrt {d} x^{3} + 15 \, a b^{2} \sqrt {d} x\right )} x^{\frac {9}{2}} + \frac {4}{77} \, {\left (7 \, a b^{2} \sqrt {d} x^{3} + 11 \, a^{2} b \sqrt {d} x\right )} x^{\frac {5}{2}} + \frac {2}{21} \, {\left (3 \, a^{2} b \sqrt {d} x^{3} + 7 \, a^{3} \sqrt {d} x\right )} \sqrt {x} \] Input:
integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
2/165*(11*b^3*sqrt(d)*x^3 + 15*a*b^2*sqrt(d)*x)*x^(9/2) + 4/77*(7*a*b^2*sq rt(d)*x^3 + 11*a^2*b*sqrt(d)*x)*x^(5/2) + 2/21*(3*a^2*b*sqrt(d)*x^3 + 7*a^ 3*sqrt(d)*x)*sqrt(x)
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.44 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2}{15} \, \sqrt {d x} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {6}{11} \, \sqrt {d x} a b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {6}{7} \, \sqrt {d x} a^{2} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2}{3} \, \sqrt {d x} a^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
2/15*sqrt(d*x)*b^3*x^7*sgn(b*x^2 + a) + 6/11*sqrt(d*x)*a*b^2*x^5*sgn(b*x^2 + a) + 6/7*sqrt(d*x)*a^2*b*x^3*sgn(b*x^2 + a) + 2/3*sqrt(d*x)*a^3*x*sgn(b *x^2 + a)
Timed out. \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int \sqrt {d\,x}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \] Input:
int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.20 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {2 \sqrt {x}\, \sqrt {d}\, x \left (77 b^{3} x^{6}+315 a \,b^{2} x^{4}+495 a^{2} b \,x^{2}+385 a^{3}\right )}{1155} \] Input:
int((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
(2*sqrt(x)*sqrt(d)*x*(385*a**3 + 495*a**2*b*x**2 + 315*a*b**2*x**4 + 77*b* *3*x**6))/1155