Integrand size = 30, antiderivative size = 191 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {6 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {2 a b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )} \] Output:
-2/5*a^3*((b*x^2+a)^2)^(1/2)/d/(d*x)^(5/2)/(b*x^2+a)-6*a^2*b*((b*x^2+a)^2) ^(1/2)/d^3/(d*x)^(1/2)/(b*x^2+a)+2*a*b^2*(d*x)^(3/2)*((b*x^2+a)^2)^(1/2)/d ^5/(b*x^2+a)+2/7*b^3*(d*x)^(7/2)*((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 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{3/2} \left (7 a^3+105 a^2 b x^2-35 a b^2 x^4-5 b^3 x^6\right )}{35 (d x)^{7/2} \left (a+b x^2\right )^3} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(7/2),x]
Output:
(-2*x*((a + b*x^2)^2)^(3/2)*(7*a^3 + 105*a^2*b*x^2 - 35*a*b^2*x^4 - 5*b^3* x^6))/(35*(d*x)^(7/2)*(a + b*x^2)^3)
Time = 0.39 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{(d x)^{7/2}}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{(d x)^{7/2}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^3}{(d x)^{7/2}}+\frac {3 b a^2}{d^2 (d x)^{3/2}}+\frac {3 b^2 \sqrt {d x} a}{d^4}+\frac {b^3 (d x)^{5/2}}{d^6}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {2 a^3}{5 d (d x)^{5/2}}-\frac {6 a^2 b}{d^3 \sqrt {d x}}+\frac {2 a b^2 (d x)^{3/2}}{d^5}+\frac {2 b^3 (d x)^{7/2}}{7 d^7}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(7/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((-2*a^3)/(5*d*(d*x)^(5/2)) - (6*a^2*b)/( d^3*Sqrt[d*x]) + (2*a*b^2*(d*x)^(3/2))/d^5 + (2*b^3*(d*x)^(7/2))/(7*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 | \(-\frac {2 x \left (-5 b^{3} x^{6}-35 b^{2} x^{4} a +105 a^{2} b \,x^{2}+7 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{35 \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {7}{2}}}\) | \(61\) |
default | \(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (-5 b^{3} x^{6}-35 b^{2} x^{4} a +105 a^{2} b \,x^{2}+7 a^{3}\right )}{35 d \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {5}{2}}}\) | \(63\) |
risch | \(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 b^{3} x^{6}-35 b^{2} x^{4} a +105 a^{2} b \,x^{2}+7 a^{3}\right )}{35 d^{3} \left (b \,x^{2}+a \right ) x^{2} \sqrt {d x}}\) | \(66\) |
orering | \(-\frac {2 \left (-5 b^{3} x^{6}-35 b^{2} x^{4} a +105 a^{2} b \,x^{2}+7 a^{3}\right ) x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{35 \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {7}{2}}}\) | \(70\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/35*x*(-5*b^3*x^6-35*a*b^2*x^4+105*a^2*b*x^2+7*a^3)*((b*x^2+a)^2)^(3/2)/ (b*x^2+a)^3/(d*x)^(7/2)
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{6} + 35 \, a b^{2} x^{4} - 105 \, a^{2} b x^{2} - 7 \, a^{3}\right )} \sqrt {d x}}{35 \, d^{4} x^{3}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(7/2),x, algorithm="fricas")
Output:
2/35*(5*b^3*x^6 + 35*a*b^2*x^4 - 105*a^2*b*x^2 - 7*a^3)*sqrt(d*x)/(d^4*x^3 )
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{\left (d x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/(d*x)**(7/2),x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/(d*x)**(7/2), x)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.45 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, {\left (3 \, b^{3} \sqrt {d} x^{3} + 7 \, a b^{2} \sqrt {d} x\right )} \sqrt {x} + \frac {70 \, {\left (a b^{2} \sqrt {d} x^{3} - 3 \, a^{2} b \sqrt {d} x\right )}}{x^{\frac {3}{2}}} - \frac {21 \, {\left (5 \, a^{2} b \sqrt {d} x^{3} + a^{3} \sqrt {d} x\right )}}{x^{\frac {7}{2}}}\right )}}{105 \, d^{4}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(7/2),x, algorithm="maxima")
Output:
2/105*(5*(3*b^3*sqrt(d)*x^3 + 7*a*b^2*sqrt(d)*x)*sqrt(x) + 70*(a*b^2*sqrt( d)*x^3 - 3*a^2*b*sqrt(d)*x)/x^(3/2) - 21*(5*a^2*b*sqrt(d)*x^3 + a^3*sqrt(d )*x)/x^(7/2))/d^4
Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.56 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {7 \, {\left (15 \, a^{2} b d^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} d^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{\sqrt {d x} d^{2} x^{2}} - \frac {5 \, {\left (\sqrt {d x} b^{3} d^{21} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 7 \, \sqrt {d x} a b^{2} d^{21} x \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{d^{21}}\right )}}{35 \, d^{4}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(7/2),x, algorithm="giac")
Output:
-2/35*(7*(15*a^2*b*d^3*x^2*sgn(b*x^2 + a) + a^3*d^3*sgn(b*x^2 + a))/(sqrt( d*x)*d^2*x^2) - 5*(sqrt(d*x)*b^3*d^21*x^3*sgn(b*x^2 + a) + 7*sqrt(d*x)*a*b ^2*d^21*x*sgn(b*x^2 + a))/d^21)/d^4
Time = 18.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.48 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,a^3}{5\,b\,d^3}+\frac {6\,a^2\,x^2}{d^3}-\frac {2\,b^2\,x^6}{7\,d^3}-\frac {2\,a\,b\,x^4}{d^3}\right )}{x^4\,\sqrt {d\,x}+\frac {a\,x^2\,\sqrt {d\,x}}{b}} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/(d*x)^(7/2),x)
Output:
-((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((2*a^3)/(5*b*d^3) + (6*a^2*x^2)/d^3 - (2*b^2*x^6)/(7*d^3) - (2*a*b*x^4)/d^3))/(x^4*(d*x)^(1/2) + (a*x^2*(d*x)^( 1/2))/b)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{(d x)^{7/2}} \, dx=\frac {2 \sqrt {d}\, \left (5 b^{3} x^{6}+35 a \,b^{2} x^{4}-105 a^{2} b \,x^{2}-7 a^{3}\right )}{35 \sqrt {x}\, d^{4} x^{2}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(7/2),x)
Output:
(2*sqrt(d)*( - 7*a**3 - 105*a**2*b*x**2 + 35*a*b**2*x**4 + 5*b**3*x**6))/( 35*sqrt(x)*d**4*x**2)