Integrand size = 15, antiderivative size = 108 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {3 c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{8 a^2}+\frac {c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 a \left (a+b x^2\right )}+\frac {3 c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b}} \] Output:
3/8*c*x*(c/(b*x^2+a)^2)^(1/2)/a^2+1/4*c*x*(c/(b*x^2+a)^2)^(1/2)/a/(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^(5/2)/ b^(1/2)
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.77 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (\sqrt {a} \sqrt {b} x \left (5 a+3 b x^2\right )+3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{8 a^{5/2} \sqrt {b}} \] Input:
Integrate[(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*(5*a + 3*b*x^2) + 3*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]))/(8*a^(5/2)*Sqrt[b])
Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2045, 215, 215, 216}
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.
\(\displaystyle \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^3}dx}{a^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {3}{4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^2}dx+\frac {a^2 x}{4 \left (a+b x^2\right )^2}\right )}{a^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\frac {b x^2}{a}+1}dx+\frac {a x}{2 \left (a+b x^2\right )}\right )+\frac {a^2 x}{4 \left (a+b x^2\right )^2}\right )}{a^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \left (\frac {a^2 x}{4 \left (a+b x^2\right )^2}+\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b}}+\frac {a x}{2 \left (a+b x^2\right )}\right )\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{a^3}\) |
Input:
Int[(c/(a + b*x^2)^2)^(3/2),x]
Output:
(c*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*((a^2*x)/(4*(a + b*x^2)^2) + (3*((a*x )/(2*(a + b*x^2)) + (Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[b])))/4) )/a^3
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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 = 0.15 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\left (3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{2} x^{4}+3 \sqrt {a b}\, b \,x^{3}+6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a b \,x^{2}+5 \sqrt {a b}\, a x +3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{8 \sqrt {a b}\, a^{2}}\) | \(99\) |
risch | \(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (\frac {3 b \,x^{3}}{8 a^{2}}+\frac {5 x}{8 a}\right )}{b \,x^{2}+a}-\frac {3 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \ln \left (b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, a^{2}}+\frac {3 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, a^{2}}\) | \(129\) |
Input:
int((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+3*(a*b)^(1/2)*b*x^3+6*arctan(b*x/(a *b)^(1/2))*a*b*x^2+5*(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)/(a*b)^(1/2)/a^2
Time = 0.09 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.61 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b c x^{2} + a c\right )} \sqrt {-\frac {c}{a b}} \log \left (\frac {b c x^{2} - a c + 2 \, {\left (a b^{2} x^{3} + a^{2} b x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}} \sqrt {-\frac {c}{a b}}}{b x^{2} + a}\right ) + 2 \, {\left (3 \, b c x^{3} + 5 \, a c x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{16 \, {\left (a^{2} b x^{2} + a^{3}\right )}}, \frac {3 \, {\left (b c x^{2} + a c\right )} \sqrt {\frac {c}{a b}} \arctan \left (\frac {{\left (b^{2} x^{3} + a b x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}} \sqrt {\frac {c}{a b}}}{c}\right ) + {\left (3 \, b c x^{3} + 5 \, a c x\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{8 \, {\left (a^{2} b x^{2} + a^{3}\right )}}\right ] \] Input:
integrate((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*(3*b*c*x^3 + 5*a*c*x)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^2*b*x^ 2 + a^3), 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) + (3*b*c*x^3 + 5*a*c*x) *sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^2*b*x^2 + a^3)]
\[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int \left (\frac {c}{\left (a + b x^{2}\right )^{2}}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((c/(b*x**2+a)**2)**(3/2),x)
Output:
Integral((c/(a + b*x**2)**2)**(3/2), x)
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.62 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {3 \, b c^{\frac {3}{2}} x^{3} + 5 \, a c^{\frac {3}{2}} x}{8 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} + \frac {3 \, c^{\frac {3}{2}} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2}} \] Input:
integrate((c/(b*x^2+a)^2)^(3/2),x, algorithm="maxima")
Output:
1/8*(3*b*c^(3/2)*x^3 + 5*a*c^(3/2)*x)/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) + 3/8*c^(3/2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2)
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {1}{8} \, c^{\frac {3}{2}} {\left (\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{3} + 5 \, a x}{{\left (b x^{2} + a\right )}^{2} a^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate((c/(b*x^2+a)^2)^(3/2),x, algorithm="giac")
Output:
1/8*c^(3/2)*(3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) + (3*b*x^3 + 5*a*x)/( (b*x^2 + a)^2*a^2))*sgn(b*x^2 + a)
Timed out. \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int {\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2} \,d x \] Input:
int((c/(a + b*x^2)^2)^(3/2),x)
Output:
int((c/(a + b*x^2)^2)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07 \[ \int \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}+5 a^{2} b x +3 a \,b^{2} x^{3}\right )}{8 a^{3} b \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int((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 + 5*a**2*b*x + 3*a*b**2*x**3))/(8*a**3 *b*(a**2 + 2*a*b*x**2 + b**2*x**4))