Integrand size = 30, antiderivative size = 193 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 a^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 a b^2 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )} \] Output:
2*a^3*(d*x)^(1/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+6/5*a^2*b*(d*x)^(5/2)*(( b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+2/3*a*b^2*(d*x)^(9/2)*((b*x^2+a)^2)^(1/2)/ d^5/(b*x^2+a)+2/13*b^3*(d*x)^(13/2)*((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 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 x \sqrt {\left (a+b x^2\right )^2} \left (195 a^3+117 a^2 b x^2+65 a b^2 x^4+15 b^3 x^6\right )}{195 \sqrt {d x} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/Sqrt[d*x],x]
Output:
(2*x*Sqrt[(a + b*x^2)^2]*(195*a^3 + 117*a^2*b*x^2 + 65*a*b^2*x^4 + 15*b^3* x^6))/(195*Sqrt[d*x]*(a + b*x^2))
Time = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.52, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{\sqrt {d x}}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{\sqrt {d x}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^3 (d x)^{11/2}}{d^6}+\frac {3 a b^2 (d x)^{7/2}}{d^4}+\frac {3 a^2 b (d x)^{3/2}}{d^2}+\frac {a^3}{\sqrt {d x}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {2 a^3 \sqrt {d x}}{d}+\frac {6 a^2 b (d x)^{5/2}}{5 d^3}+\frac {2 a b^2 (d x)^{9/2}}{3 d^5}+\frac {2 b^3 (d x)^{13/2}}{13 d^7}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/Sqrt[d*x],x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((2*a^3*Sqrt[d*x])/d + (6*a^2*b*(d*x)^(5/ 2))/(5*d^3) + (2*a*b^2*(d*x)^(9/2))/(3*d^5) + (2*b^3*(d*x)^(13/2))/(13*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 = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(\frac {2 x \left (15 b^{3} x^{6}+65 b^{2} x^{4} a +117 a^{2} b \,x^{2}+195 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{195 \left (b \,x^{2}+a \right )^{3} \sqrt {d x}}\) | \(61\) |
risch | \(\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (15 b^{3} x^{6}+65 b^{2} x^{4} a +117 a^{2} b \,x^{2}+195 a^{3}\right ) x}{195 \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(61\) |
default | \(\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \sqrt {d x}\, \left (15 b^{3} x^{6}+65 b^{2} x^{4} a +117 a^{2} b \,x^{2}+195 a^{3}\right )}{195 d \left (b \,x^{2}+a \right )^{3}}\) | \(63\) |
orering | \(\frac {2 \left (15 b^{3} x^{6}+65 b^{2} x^{4} a +117 a^{2} b \,x^{2}+195 a^{3}\right ) x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{195 \left (b \,x^{2}+a \right )^{3} \sqrt {d x}}\) | \(70\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/195*x*(15*b^3*x^6+65*a*b^2*x^4+117*a^2*b*x^2+195*a^3)*((b*x^2+a)^2)^(3/2 )/(b*x^2+a)^3/(d*x)^(1/2)
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 \, {\left (15 \, b^{3} x^{6} + 65 \, a b^{2} x^{4} + 117 \, a^{2} b x^{2} + 195 \, a^{3}\right )} \sqrt {d x}}{195 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="fricas")
Output:
2/195*(15*b^3*x^6 + 65*a*b^2*x^4 + 117*a^2*b*x^2 + 195*a^3)*sqrt(d*x)/d
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{\sqrt {d x}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/(d*x)**(1/2),x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/sqrt(d*x), x)
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.45 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 \, {\left (5 \, {\left (9 \, b^{3} \sqrt {d} x^{3} + 13 \, a b^{2} \sqrt {d} x\right )} x^{\frac {7}{2}} + 26 \, {\left (5 \, a b^{2} \sqrt {d} x^{3} + 9 \, a^{2} b \sqrt {d} x\right )} x^{\frac {3}{2}} + \frac {117 \, {\left (a^{2} b \sqrt {d} x^{3} + 5 \, a^{3} \sqrt {d} x\right )}}{\sqrt {x}}\right )}}{585 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="maxima")
Output:
2/585*(5*(9*b^3*sqrt(d)*x^3 + 13*a*b^2*sqrt(d)*x)*x^(7/2) + 26*(5*a*b^2*sq rt(d)*x^3 + 9*a^2*b*sqrt(d)*x)*x^(3/2) + 117*(a^2*b*sqrt(d)*x^3 + 5*a^3*sq rt(d)*x)/sqrt(x))/d
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {d x} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 65 \, \sqrt {d x} a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 117 \, \sqrt {d x} a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 195 \, \sqrt {d x} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{195 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="giac")
Output:
2/195*(15*sqrt(d*x)*b^3*x^6*sgn(b*x^2 + a) + 65*sqrt(d*x)*a*b^2*x^4*sgn(b* x^2 + a) + 117*sqrt(d*x)*a^2*b*x^2*sgn(b*x^2 + a) + 195*sqrt(d*x)*a^3*sgn( b*x^2 + a))/d
Time = 18.79 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {6\,a^2\,x^3}{5}+\frac {2\,b^2\,x^7}{13}+\frac {2\,a^3\,x}{b}+\frac {2\,a\,b\,x^5}{3}\right )}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/(d*x)^(1/2),x)
Output:
((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((6*a^2*x^3)/5 + (2*b^2*x^7)/13 + (2*a^ 3*x)/b + (2*a*b*x^5)/3))/(x^2*(d*x)^(1/2) + (a*(d*x)^(1/2))/b)
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx=\frac {2 \sqrt {x}\, \sqrt {d}\, \left (15 b^{3} x^{6}+65 a \,b^{2} x^{4}+117 a^{2} b \,x^{2}+195 a^{3}\right )}{195 d} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x)
Output:
(2*sqrt(x)*sqrt(d)*(195*a**3 + 117*a**2*b*x**2 + 65*a*b**2*x**4 + 15*b**3* x**6))/(195*d)