Integrand size = 15, antiderivative size = 64 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}-\frac {x}{b^2 \sqrt {a+b x^2}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:
-1/3*x^3/b/(b*x^2+a)^(3/2)-x/b^2/(b*x^2+a)^(1/2)+arctanh(b^(1/2)*x/(b*x^2+ a)^(1/2))/b^(5/2)
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 a x-4 b x^3}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Input:
Integrate[x^4/(a + b*x^2)^(5/2),x]
Output:
(-3*a*x - 4*b*x^3)/(3*b^2*(a + b*x^2)^(3/2)) + (2*ArcTanh[(Sqrt[b]*x)/(-Sq rt[a] + Sqrt[a + b*x^2])])/b^(5/2)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {252, 252, 224, 219}
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.
\(\displaystyle \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\int \frac {x^2}{\left (b x^2+a\right )^{3/2}}dx}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {b x^2+a}}dx}{b}-\frac {x}{b \sqrt {a+b x^2}}}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{b}-\frac {x}{b \sqrt {a+b x^2}}}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {x}{b \sqrt {a+b x^2}}}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}\) |
Input:
Int[x^4/(a + b*x^2)^(5/2),x]
Output:
-1/3*x^3/(b*(a + b*x^2)^(3/2)) + (-(x/(b*Sqrt[a + b*x^2])) + ArcTanh[(Sqrt [b]*x)/Sqrt[a + b*x^2]]/b^(3/2))/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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]
Time = 0.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89
method | result | size |
pseudoelliptic | \(\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\frac {4 b^{\frac {3}{2}} x^{3}}{3}-x a \sqrt {b}}{b^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\) | \(57\) |
default | \(-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\) | \(59\) |
Input:
int(x^4/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
((b*x^2+a)^(3/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))-4/3*b^(3/2)*x^3-x*a*b^ (1/2))/b^(5/2)/(b*x^2+a)^(3/2)
Time = 0.08 (sec) , antiderivative size = 199, normalized size of antiderivative = 3.11 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (4 \, b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (4 \, b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:
integrate(x^4/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[1/6*(3*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(4*b^2*x^3 + 3*a*b*x)*sqrt(b*x^2 + a))/(b^5*x^4 + 2* a*b^4*x^2 + a^2*b^3), -1/3*(3*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-b)*arctan( sqrt(-b)*x/sqrt(b*x^2 + a)) + (4*b^2*x^3 + 3*a*b*x)*sqrt(b*x^2 + a))/(b^5* x^4 + 2*a*b^4*x^2 + a^2*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (54) = 108\).
Time = 1.63 (sec) , antiderivative size = 303, normalized size of antiderivative = 4.73 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 a^{\frac {39}{2}} b^{11} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} b^{12} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a^{19} b^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {4 a^{18} b^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:
integrate(x**4/(b*x**2+a)**(5/2),x)
Output:
3*a**(39/2)*b**11*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2) *b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2 /a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/ (3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqr t(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x **2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2 )*x**3/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x **2*sqrt(1 + b*x**2/a))
Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {\operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} \] Input:
integrate(x^4/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-1/3*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) - 1/3*x /(sqrt(b*x^2 + a)*b^2) + arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {x {\left (\frac {4 \, x^{2}}{b} + \frac {3 \, a}{b^{2}}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:
integrate(x^4/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
-1/3*x*(4*x^2/b + 3*a/b^2)/(b*x^2 + a)^(3/2) - log(abs(-sqrt(b)*x + sqrt(b *x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:
int(x^4/(a + b*x^2)^(5/2),x)
Output:
int(x^4/(a + b*x^2)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.16 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a b x -4 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}}{3 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^4/(b*x^2+a)^(5/2),x)
Output:
( - 3*sqrt(a + b*x**2)*a*b*x - 4*sqrt(a + b*x**2)*b**2*x**3 + 3*sqrt(b)*lo g((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2 + 6*sqrt(b)*log((sqrt(a + b *x**2) + sqrt(b)*x)/sqrt(a))*a*b*x**2 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**2*x**4)/(3*b**3*(a**2 + 2*a*b*x**2 + b**2*x**4))