Integrand size = 19, antiderivative size = 95 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {a c \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{b^3}-\frac {a^2 c \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 b^3 \left (a+b x^2\right )}+\frac {c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3} \] Output:
a*c*(c/(b*x^2+a)^2)^(1/2)/b^3-1/4*a^2*c*(c/(b*x^2+a)^2)^(1/2)/b^3/(b*x^2+a )+1/2*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)*ln(b*x^2+a)/b^3
Time = 1.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (a \left (3 a+4 b x^2\right )+2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )\right )}{4 b^3} \] Input:
Integrate[x^5*(c/(a + b*x^2)^2)^(3/2),x]
Output:
((c/(a + b*x^2)^2)^(3/2)*(a + b*x^2)*(a*(3*a + 4*b*x^2) + 2*(a + b*x^2)^2* Log[a + b*x^2]))/(4*b^3)
Time = 0.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2045, 27, 243, 49, 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.
\(\displaystyle \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {a^3 x^5}{\left (b x^2+a\right )^3}dx}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {x^5}{\left (b x^2+a\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {x^4}{\left (b x^2+a\right )^3}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \left (\frac {a^2}{b^2 \left (b x^2+a\right )^3}-\frac {2 a}{b^2 \left (b x^2+a\right )^2}+\frac {1}{b^2 \left (b x^2+a\right )}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (-\frac {a^2}{2 b^3 \left (a+b x^2\right )^2}+\frac {2 a}{b^3 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{b^3}\right )\) |
Input:
Int[x^5*(c/(a + b*x^2)^2)^(3/2),x]
Output:
(c*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*(-1/2*a^2/(b^3*(a + b*x^2)^2) + (2*a) /(b^3*(a + b*x^2)) + Log[a + b*x^2]/b^3))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (\frac {a \,x^{2}}{b^{2}}+\frac {3 a^{2}}{4 b^{3}}\right )}{b \,x^{2}+a}+\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (b \,x^{2}+a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\) | \(77\) |
default | \(\frac {\left (2 \ln \left (b \,x^{2}+a \right ) b^{2} x^{4}+4 \ln \left (b \,x^{2}+a \right ) a b \,x^{2}+4 a b \,x^{2}+2 \ln \left (b \,x^{2}+a \right ) a^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{4 b^{3}}\) | \(83\) |
Input:
int(x^5*(c/(b*x^2+a)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
c*(c/(b*x^2+a)^2)^(1/2)/(b*x^2+a)*(a*x^2/b^2+3/4*a^2/b^3)+1/2*c*(c/(b*x^2+ a)^2)^(1/2)*(b*x^2+a)*ln(b*x^2+a)/b^3
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {{\left (4 \, a b c x^{2} + 3 \, a^{2} c + 2 \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )} \log \left (b x^{2} + a\right )\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \] Input:
integrate(x^5*(c/(b*x^2+a)^2)^(3/2),x, algorithm="fricas")
Output:
1/4*(4*a*b*c*x^2 + 3*a^2*c + 2*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*log(b*x^2 + a))*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))/(b^4*x^2 + a*b^3)
\[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int x^{5} \left (\frac {c}{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**5*(c/(b*x**2+a)**2)**(3/2),x)
Output:
Integral(x**5*(c/(a**2 + 2*a*b*x**2 + b**2*x**4))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {4 \, a b c^{\frac {3}{2}} x^{2} + 3 \, a^{2} c^{\frac {3}{2}}}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {c^{\frac {3}{2}} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \] Input:
integrate(x^5*(c/(b*x^2+a)^2)^(3/2),x, algorithm="maxima")
Output:
1/4*(4*a*b*c^(3/2)*x^2 + 3*a^2*c^(3/2))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) + 1/2*c^(3/2)*log(b*x^2 + a)/b^3
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {1}{4} \, c^{\frac {3}{2}} {\left (\frac {2 \, \log \left ({\left | b x^{2} + a \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{b^{3}} - \frac {3 \, b x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a x^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{2} b^{2}}\right )} \] Input:
integrate(x^5*(c/(b*x^2+a)^2)^(3/2),x, algorithm="giac")
Output:
1/4*c^(3/2)*(2*log(abs(b*x^2 + a))*sgn(b*x^2 + a)/b^3 - (3*b*x^4*sgn(b*x^2 + a) + 2*a*x^2*sgn(b*x^2 + a))/((b*x^2 + a)^2*b^2))
Timed out. \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int x^5\,{\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2} \,d x \] Input:
int(x^5*(c/(a + b*x^2)^2)^(3/2),x)
Output:
int(x^5*(c/(a + b*x^2)^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int x^5 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b \,x^{2}+2 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{2} x^{4}+a^{2}-2 b^{2} x^{4}\right )}{4 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^5*(c/(b*x^2+a)^2)^(3/2),x)
Output:
(sqrt(c)*c*(2*log(a + b*x**2)*a**2 + 4*log(a + b*x**2)*a*b*x**2 + 2*log(a + b*x**2)*b**2*x**4 + a**2 - 2*b**2*x**4))/(4*b**3*(a**2 + 2*a*b*x**2 + b* *2*x**4))