Integrand size = 19, antiderivative size = 110 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=-\frac {3 c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{8 b^2}-\frac {c x^3 \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 b \left (a+b x^2\right )}+\frac {3 c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{5/2}} \] Output:
-3/8*c*x*(c/(b*x^2+a)^2)^(1/2)/b^2-1/4*c*x^3*(c/(b*x^2+a)^2)^(1/2)/b/(b*x^ 2+a)+3/8*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(1/ 2)/b^(5/2)
Time = 1.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (-\sqrt {a} \sqrt {b} x \left (3 a+5 b x^2\right )+3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{8 \sqrt {a} b^{5/2}} \] Input:
Integrate[x^4*(c/(a + b*x^2)^2)^(3/2),x]
Output:
((c/(a + b*x^2)^2)^(3/2)*(a + b*x^2)*(-(Sqrt[a]*Sqrt[b]*x*(3*a + 5*b*x^2)) + 3*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]))/(8*Sqrt[a]*b^(5/2))
Time = 0.39 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2045, 27, 252, 252, 218}
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.
\(\displaystyle \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {a^3 x^4}{\left (b x^2+a\right )^3}dx}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {x^4}{\left (b x^2+a\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {3 \int \frac {x^2}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 b}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle c \left (a+b x^2\right ) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}}\) |
Input:
Int[x^4*(c/(a + b*x^2)^2)^(3/2),x]
Output:
c*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*(-1/4*x^3/(b*(a + b*x^2)^2) + (3*(-1/2 *x/(b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*Sqrt[a]*b^(3/2))))/(4* b))
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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
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 = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {\left (-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{2} x^{4}+5 \sqrt {a b}\, b \,x^{3}-6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a b \,x^{2}+3 \sqrt {a b}\, a x -3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{8 b^{2} \sqrt {a b}}\) | \(99\) |
risch | \(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (-\frac {5 x^{3}}{8 b}-\frac {3 a x}{8 b^{2}}\right )}{b \,x^{2}+a}-\frac {3 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \ln \left (b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, b^{2}}+\frac {3 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, b^{2}}\) | \(129\) |
Input:
int(x^4*(c/(b*x^2+a)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/8*(-3*arctan(b*x/(a*b)^(1/2))*b^2*x^4+5*(a*b)^(1/2)*b*x^3-6*arctan(b*x/ (a*b)^(1/2))*a*b*x^2+3*(a*b)^(1/2)*a*x-3*arctan(b*x/(a*b)^(1/2))*a^2)*(b*x ^2+a)*(c/(b*x^2+a)^2)^(3/2)/b^2/(a*b)^(1/2)
Time = 0.09 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.59 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b c x^{2} + a c\right )} \sqrt {-\frac {c}{a b}} \log \left (\frac {b c x^{2} - a c + 2 \, {\left (a b^{2} x^{3} + a^{2} b x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}} \sqrt {-\frac {c}{a b}}}{b x^{2} + a}\right ) - 2 \, {\left (5 \, b c x^{3} + 3 \, a c x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{16 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, \frac {3 \, {\left (b c x^{2} + a c\right )} \sqrt {\frac {c}{a b}} \arctan \left (\frac {{\left (b^{2} x^{3} + a b x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}} \sqrt {\frac {c}{a b}}}{c}\right ) - {\left (5 \, b c x^{3} + 3 \, a c x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}}\right ] \] Input:
integrate(x^4*(c/(b*x^2+a)^2)^(3/2),x, algorithm="fricas")
Output:
[1/16*(3*(b*c*x^2 + a*c)*sqrt(-c/(a*b))*log((b*c*x^2 - a*c + 2*(a*b^2*x^3 + a^2*b*x)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))*sqrt(-c/(a*b)))/(b*x^2 + a) ) - 2*(5*b*c*x^3 + 3*a*c*x)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^3*x^2 + a*b^2), 1/8*(3*(b*c*x^2 + a*c)*sqrt(c/(a*b))*arctan((b^2*x^3 + a*b*x)*sq rt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))*sqrt(c/(a*b))/c) - (5*b*c*x^3 + 3*a*c*x) *sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^3*x^2 + a*b^2)]
\[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int x^{4} \left (\frac {c}{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**4*(c/(b*x**2+a)**2)**(3/2),x)
Output:
Integral(x**4*(c/(a**2 + 2*a*b*x**2 + b**2*x**4))**(3/2), x)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=-\frac {5 \, b c^{\frac {3}{2}} x^{3} + 3 \, a c^{\frac {3}{2}} x}{8 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} + \frac {3 \, c^{\frac {3}{2}} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2}} \] Input:
integrate(x^4*(c/(b*x^2+a)^2)^(3/2),x, algorithm="maxima")
Output:
-1/8*(5*b*c^(3/2)*x^3 + 3*a*c^(3/2)*x)/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2) + 3/8*c^(3/2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2)
Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {1}{8} \, c^{\frac {3}{2}} {\left (\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {a b} b^{2}} - \frac {5 \, b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a x \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{2} b^{2}}\right )} \] Input:
integrate(x^4*(c/(b*x^2+a)^2)^(3/2),x, algorithm="giac")
Output:
1/8*c^(3/2)*(3*arctan(b*x/sqrt(a*b))*sgn(b*x^2 + a)/(sqrt(a*b)*b^2) - (5*b *x^3*sgn(b*x^2 + a) + 3*a*x*sgn(b*x^2 + a))/((b*x^2 + a)^2*b^2))
Timed out. \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int x^4\,{\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2} \,d x \] Input:
int(x^4*(c/(a + b*x^2)^2)^(3/2),x)
Output:
int(x^4*(c/(a + b*x^2)^2)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int x^4 \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}-3 a^{2} b x -5 a \,b^{2} x^{3}\right )}{8 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^4*(c/(b*x^2+a)^2)^(3/2),x)
Output:
(sqrt(c)*c*(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2 + 6*sqrt( b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 + 3*sqrt(b)*sqrt(a)*atan ((b*x)/(sqrt(b)*sqrt(a)))*b**2*x**4 - 3*a**2*b*x - 5*a*b**2*x**3))/(8*a*b* *3*(a**2 + 2*a*b*x**2 + b**2*x**4))