Integrand size = 30, antiderivative size = 191 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^7 \left (a+b x^2\right )} \] Output:
-2*a^3*((b*x^2+a)^2)^(1/2)/d/(d*x)^(1/2)/(b*x^2+a)+2*a^2*b*(d*x)^(3/2)*((b *x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+6/7*a*b^2*(d*x)^(7/2)*((b*x^2+a)^2)^(1/2)/d ^5/(b*x^2+a)+2/11*b^3*(d*x)^(11/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x^2+a)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{3/2} \left (77 a^3-77 a^2 b x^2-33 a b^2 x^4-7 b^3 x^6\right )}{77 (d x)^{3/2} \left (a+b x^2\right )^3} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(3/2),x]
Output:
(-2*x*((a + b*x^2)^2)^(3/2)*(77*a^3 - 77*a^2*b*x^2 - 33*a*b^2*x^4 - 7*b^3* x^6))/(77*(d*x)^(3/2)*(a + b*x^2)^3)
Time = 0.43 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{(d x)^{3/2}}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{(d x)^{3/2}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^3 (d x)^{9/2}}{d^6}+\frac {3 a b^2 (d x)^{5/2}}{d^4}+\frac {3 a^2 b \sqrt {d x}}{d^2}+\frac {a^3}{(d x)^{3/2}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {2 a^3}{d \sqrt {d x}}+\frac {2 a^2 b (d x)^{3/2}}{d^3}+\frac {6 a b^2 (d x)^{7/2}}{7 d^5}+\frac {2 b^3 (d x)^{11/2}}{11 d^7}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((-2*a^3)/(d*Sqrt[d*x]) + (2*a^2*b*(d*x)^ (3/2))/d^3 + (6*a*b^2*(d*x)^(7/2))/(7*d^5) + (2*b^3*(d*x)^(11/2))/(11*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 = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {2 x \left (-7 b^{3} x^{6}-33 b^{2} x^{4} a -77 a^{2} b \,x^{2}+77 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{77 \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {3}{2}}}\) | \(61\) |
default | \(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (-7 b^{3} x^{6}-33 b^{2} x^{4} a -77 a^{2} b \,x^{2}+77 a^{3}\right )}{77 d \left (b \,x^{2}+a \right )^{3} \sqrt {d x}}\) | \(63\) |
risch | \(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-7 b^{3} x^{6}-33 b^{2} x^{4} a -77 a^{2} b \,x^{2}+77 a^{3}\right )}{77 d \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(63\) |
orering | \(-\frac {2 \left (-7 b^{3} x^{6}-33 b^{2} x^{4} a -77 a^{2} b \,x^{2}+77 a^{3}\right ) x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{77 \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {3}{2}}}\) | \(70\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/77*x*(-7*b^3*x^6-33*a*b^2*x^4-77*a^2*b*x^2+77*a^3)*((b*x^2+a)^2)^(3/2)/ (b*x^2+a)^3/(d*x)^(3/2)
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (7 \, b^{3} x^{6} + 33 \, a b^{2} x^{4} + 77 \, a^{2} b x^{2} - 77 \, a^{3}\right )} \sqrt {d x}}{77 \, d^{2} x} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(3/2),x, algorithm="fricas")
Output:
2/77*(7*b^3*x^6 + 33*a*b^2*x^4 + 77*a^2*b*x^2 - 77*a^3)*sqrt(d*x)/(d^2*x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{\left (d x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/(d*x)**(3/2),x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/(d*x)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, {\left (7 \, b^{3} \sqrt {d} x^{3} + 11 \, a b^{2} \sqrt {d} x\right )} x^{\frac {5}{2}} + 22 \, {\left (3 \, a b^{2} \sqrt {d} x^{3} + 7 \, a^{2} b \sqrt {d} x\right )} \sqrt {x} + \frac {77 \, {\left (a^{2} b \sqrt {d} x^{3} - 3 \, a^{3} \sqrt {d} x\right )}}{x^{\frac {3}{2}}}\right )}}{231 \, d^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(3/2),x, algorithm="maxima")
Output:
2/231*(3*(7*b^3*sqrt(d)*x^3 + 11*a*b^2*sqrt(d)*x)*x^(5/2) + 22*(3*a*b^2*sq rt(d)*x^3 + 7*a^2*b*sqrt(d)*x)*sqrt(x) + 77*(a^2*b*sqrt(d)*x^3 - 3*a^3*sqr t(d)*x)/x^(3/2))/d^2
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {77 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {d x}} - \frac {7 \, \sqrt {d x} b^{3} d^{65} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 33 \, \sqrt {d x} a b^{2} d^{65} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 77 \, \sqrt {d x} a^{2} b d^{65} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{66}}\right )}}{77 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(3/2),x, algorithm="giac")
Output:
-2/77*(77*a^3*sgn(b*x^2 + a)/sqrt(d*x) - (7*sqrt(d*x)*b^3*d^65*x^5*sgn(b*x ^2 + a) + 33*sqrt(d*x)*a*b^2*d^65*x^3*sgn(b*x^2 + a) + 77*sqrt(d*x)*a^2*b* d^65*x*sgn(b*x^2 + a))/d^66)/d
Time = 18.87 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,a^2\,x^2}{d}-\frac {2\,a^3}{b\,d}+\frac {2\,b^2\,x^6}{11\,d}+\frac {6\,a\,b\,x^4}{7\,d}\right )}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/(d*x)^(3/2),x)
Output:
((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((2*a^2*x^2)/d - (2*a^3)/(b*d) + (2*b^2 *x^6)/(11*d) + (6*a*b*x^4)/(7*d)))/(x^2*(d*x)^(1/2) + (a*(d*x)^(1/2))/b)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {2 \sqrt {d}\, \left (7 b^{3} x^{6}+33 a \,b^{2} x^{4}+77 a^{2} b \,x^{2}-77 a^{3}\right )}{77 \sqrt {x}\, d^{2}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(3/2),x)
Output:
(2*sqrt(d)*( - 77*a**3 + 77*a**2*b*x**2 + 33*a*b**2*x**4 + 7*b**3*x**6))/( 77*sqrt(x)*d**2)