Integrand size = 30, antiderivative size = 295 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^7 \left (a+b x^2\right )}+\frac {2 a b^4 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^{11} \left (a+b x^2\right )} \] Output:
-2*a^5*((b*x^2+a)^2)^(1/2)/d/(d*x)^(1/2)/(b*x^2+a)+10/3*a^4*b*(d*x)^(3/2)* ((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+20/7*a^3*b^2*(d*x)^(7/2)*((b*x^2+a)^2)^( 1/2)/d^5/(b*x^2+a)+20/11*a^2*b^3*(d*x)^(11/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x ^2+a)+2/3*a*b^4*(d*x)^(15/2)*((b*x^2+a)^2)^(1/2)/d^9/(b*x^2+a)+2/19*b^5*(d *x)^(19/2)*((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 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{5/2} \left (4389 a^5-7315 a^4 b x^2-6270 a^3 b^2 x^4-3990 a^2 b^3 x^6-1463 a b^4 x^8-231 b^5 x^{10}\right )}{4389 (d x)^{3/2} \left (a+b x^2\right )^5} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(d*x)^(3/2),x]
Output:
(-2*x*((a + b*x^2)^2)^(5/2)*(4389*a^5 - 7315*a^4*b*x^2 - 6270*a^3*b^2*x^4 - 3990*a^2*b^3*x^6 - 1463*a*b^4*x^8 - 231*b^5*x^10))/(4389*(d*x)^(3/2)*(a + b*x^2)^5)
Time = 0.45 (sec) , antiderivative size = 141, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{(d x)^{3/2}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{(d x)^{3/2}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^5 (d x)^{17/2}}{d^{10}}+\frac {5 a b^4 (d x)^{13/2}}{d^8}+\frac {10 a^2 b^3 (d x)^{9/2}}{d^6}+\frac {10 a^3 b^2 (d x)^{5/2}}{d^4}+\frac {5 a^4 b \sqrt {d x}}{d^2}+\frac {a^5}{(d x)^{3/2}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {2 a^5}{d \sqrt {d x}}+\frac {10 a^4 b (d x)^{3/2}}{3 d^3}+\frac {20 a^3 b^2 (d x)^{7/2}}{7 d^5}+\frac {20 a^2 b^3 (d x)^{11/2}}{11 d^7}+\frac {2 a b^4 (d x)^{15/2}}{3 d^9}+\frac {2 b^5 (d x)^{19/2}}{19 d^{11}}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(d*x)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((-2*a^5)/(d*Sqrt[d*x]) + (10*a^4*b*(d*x) ^(3/2))/(3*d^3) + (20*a^3*b^2*(d*x)^(7/2))/(7*d^5) + (20*a^2*b^3*(d*x)^(11 /2))/(11*d^7) + (2*a*b^4*(d*x)^(15/2))/(3*d^9) + (2*b^5*(d*x)^(19/2))/(19* 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 = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28
method | result | size |
gosper | \(-\frac {2 x \left (-231 x^{10} b^{5}-1463 a \,x^{8} b^{4}-3990 a^{2} x^{6} b^{3}-6270 a^{3} x^{4} b^{2}-7315 x^{2} a^{4} b +4389 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{4389 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {3}{2}}}\) | \(83\) |
default | \(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (-231 x^{10} b^{5}-1463 a \,x^{8} b^{4}-3990 a^{2} x^{6} b^{3}-6270 a^{3} x^{4} b^{2}-7315 x^{2} a^{4} b +4389 a^{5}\right )}{4389 d \left (b \,x^{2}+a \right )^{5} \sqrt {d x}}\) | \(85\) |
risch | \(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-231 x^{10} b^{5}-1463 a \,x^{8} b^{4}-3990 a^{2} x^{6} b^{3}-6270 a^{3} x^{4} b^{2}-7315 x^{2} a^{4} b +4389 a^{5}\right )}{4389 d \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(85\) |
orering | \(-\frac {2 \left (-231 x^{10} b^{5}-1463 a \,x^{8} b^{4}-3990 a^{2} x^{6} b^{3}-6270 a^{3} x^{4} b^{2}-7315 x^{2} a^{4} b +4389 a^{5}\right ) x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{4389 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {3}{2}}}\) | \(92\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/4389*x*(-231*b^5*x^10-1463*a*b^4*x^8-3990*a^2*b^3*x^6-6270*a^3*b^2*x^4- 7315*a^4*b*x^2+4389*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5/(d*x)^(3/2)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (231 \, b^{5} x^{10} + 1463 \, a b^{4} x^{8} + 3990 \, a^{2} b^{3} x^{6} + 6270 \, a^{3} b^{2} x^{4} + 7315 \, a^{4} b x^{2} - 4389 \, a^{5}\right )} \sqrt {d x}}{4389 \, d^{2} x} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(3/2),x, algorithm="fricas")
Output:
2/4389*(231*b^5*x^10 + 1463*a*b^4*x^8 + 3990*a^2*b^3*x^6 + 6270*a^3*b^2*x^ 4 + 7315*a^4*b*x^2 - 4389*a^5)*sqrt(d*x)/(d^2*x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{\left (d x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/(d*x)**(3/2),x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/(d*x)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (77 \, {\left (15 \, b^{5} \sqrt {d} x^{3} + 19 \, a b^{4} \sqrt {d} x\right )} x^{\frac {13}{2}} + 532 \, {\left (11 \, a b^{4} \sqrt {d} x^{3} + 15 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {9}{2}} + 1710 \, {\left (7 \, a^{2} b^{3} \sqrt {d} x^{3} + 11 \, a^{3} b^{2} \sqrt {d} x\right )} x^{\frac {5}{2}} + 4180 \, {\left (3 \, a^{3} b^{2} \sqrt {d} x^{3} + 7 \, a^{4} b \sqrt {d} x\right )} \sqrt {x} + \frac {7315 \, {\left (a^{4} b \sqrt {d} x^{3} - 3 \, a^{5} \sqrt {d} x\right )}}{x^{\frac {3}{2}}}\right )}}{21945 \, d^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(3/2),x, algorithm="maxima")
Output:
2/21945*(77*(15*b^5*sqrt(d)*x^3 + 19*a*b^4*sqrt(d)*x)*x^(13/2) + 532*(11*a *b^4*sqrt(d)*x^3 + 15*a^2*b^3*sqrt(d)*x)*x^(9/2) + 1710*(7*a^2*b^3*sqrt(d) *x^3 + 11*a^3*b^2*sqrt(d)*x)*x^(5/2) + 4180*(3*a^3*b^2*sqrt(d)*x^3 + 7*a^4 *b*sqrt(d)*x)*sqrt(x) + 7315*(a^4*b*sqrt(d)*x^3 - 3*a^5*sqrt(d)*x)/x^(3/2) )/d^2
Time = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {4389 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {d x}} - \frac {231 \, \sqrt {d x} b^{5} d^{189} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + 1463 \, \sqrt {d x} a b^{4} d^{189} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 3990 \, \sqrt {d x} a^{2} b^{3} d^{189} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 6270 \, \sqrt {d x} a^{3} b^{2} d^{189} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 7315 \, \sqrt {d x} a^{4} b d^{189} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{190}}\right )}}{4389 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(3/2),x, algorithm="giac")
Output:
-2/4389*(4389*a^5*sgn(b*x^2 + a)/sqrt(d*x) - (231*sqrt(d*x)*b^5*d^189*x^9* sgn(b*x^2 + a) + 1463*sqrt(d*x)*a*b^4*d^189*x^7*sgn(b*x^2 + a) + 3990*sqrt (d*x)*a^2*b^3*d^189*x^5*sgn(b*x^2 + a) + 6270*sqrt(d*x)*a^3*b^2*d^189*x^3* sgn(b*x^2 + a) + 7315*sqrt(d*x)*a^4*b*d^189*x*sgn(b*x^2 + a))/d^190)/d
Time = 18.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=\frac {2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (3803\,a^4+3512\,a^3\,b\,x^2+2758\,a^2\,b^2\,x^4+1232\,a\,b^3\,x^6+231\,b^4\,x^8\right )}{4389\,d\,\sqrt {d\,x}}-\frac {16384\,a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4389\,d\,\sqrt {d\,x}\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/(d*x)^(3/2),x)
Output:
(2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*(3803*a^4 + 231*b^4*x^8 + 3512*a^3*b* x^2 + 1232*a*b^3*x^6 + 2758*a^2*b^2*x^4))/(4389*d*(d*x)^(1/2)) - (16384*a^ 5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4389*d*(d*x)^(1/2)*(a + b*x^2))
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{3/2}} \, dx=\frac {2 \sqrt {d}\, \left (231 b^{5} x^{10}+1463 a \,b^{4} x^{8}+3990 a^{2} b^{3} x^{6}+6270 a^{3} b^{2} x^{4}+7315 a^{4} b \,x^{2}-4389 a^{5}\right )}{4389 \sqrt {x}\, d^{2}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(3/2),x)
Output:
(2*sqrt(d)*( - 4389*a**5 + 7315*a**4*b*x**2 + 6270*a**3*b**2*x**4 + 3990*a **2*b**3*x**6 + 1463*a*b**4*x**8 + 231*b**5*x**10))/(4389*sqrt(x)*d**2)