Integrand size = 15, antiderivative size = 68 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {x^3}{b \sqrt {a+b x^2}}+\frac {3 x \sqrt {a+b x^2}}{2 b^2}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:
-x^3/b/(b*x^2+a)^(1/2)+3/2*x*(b*x^2+a)^(1/2)/b^2-3/2*a*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 = 0.97 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {3 a x+b x^3}{2 b^2 \sqrt {a+b x^2}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Input:
Integrate[x^4/(a + b*x^2)^(3/2),x]
Output:
(3*a*x + b*x^3)/(2*b^2*Sqrt[a + b*x^2]) - (3*a*ArcTanh[(Sqrt[b]*x)/(-Sqrt[ a] + Sqrt[a + b*x^2])])/b^(5/2)
Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {252, 262, 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.
\(\displaystyle \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {3 \int \frac {x^2}{\sqrt {b x^2+a}}dx}{b}-\frac {x^3}{b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3 \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{b}-\frac {x^3}{b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {3 \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{b}-\frac {x^3}{b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{b}-\frac {x^3}{b \sqrt {a+b x^2}}\) |
Input:
Int[x^4/(a + b*x^2)^(3/2),x]
Output:
-(x^3/(b*Sqrt[a + b*x^2])) + (3*((x*Sqrt[a + b*x^2])/(2*b) - (a*ArcTanh[(S qrt[b]*x)/Sqrt[a + b*x^2]])/(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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Time = 0.39 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x \sqrt {b \,x^{2}+a}}{2 b^{2}}+\frac {a x}{b^{2} \sqrt {b \,x^{2}+a}}-\frac {3 a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}}\) | \(54\) |
pseudoelliptic | \(\frac {b^{\frac {3}{2}} x^{3}-3 \sqrt {b \,x^{2}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a +3 x a \sqrt {b}}{2 b^{\frac {5}{2}} \sqrt {b \,x^{2}+a}}\) | \(59\) |
default | \(\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\) | \(61\) |
Input:
int(x^4/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*x*(b*x^2+a)^(1/2)/b^2+a/b^2*x/(b*x^2+a)^(1/2)-3/2*a/b^(5/2)*ln(b^(1/2) *x+(b*x^2+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.34 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \] Input:
integrate(x^4/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
[1/4*(3*(a*b*x^2 + a^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(b^2*x^3 + 3*a*b*x)*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3), 1/2*(3*( a*b*x^2 + a^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (b^2*x^3 + 3* a*b*x)*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3)]
Time = 1.84 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:
integrate(x**4/(b*x**2+a)**(3/2),x)
Output:
3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2* b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1 + b*x**2/a))
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {3 \, a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
1/2*x^3/(sqrt(b*x^2 + a)*b) + 3/2*a*x/(sqrt(b*x^2 + a)*b^2) - 3/2*a*arcsin h(b*x/sqrt(a*b))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x {\left (\frac {x^{2}}{b} + \frac {3 \, a}{b^{2}}\right )}}{2 \, \sqrt {b x^{2} + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
1/2*x*(x^2/b + 3*a/b^2)/sqrt(b*x^2 + a) + 3/2*a*log(abs(-sqrt(b)*x + sqrt( b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:
int(x^4/(a + b*x^2)^(3/2),x)
Output:
int(x^4/(a + b*x^2)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.68 \[ \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {12 \sqrt {b \,x^{2}+a}\, a b x +4 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+9 \sqrt {b}\, a^{2}+9 \sqrt {b}\, a b \,x^{2}}{8 b^{3} \left (b \,x^{2}+a \right )} \] Input:
int(x^4/(b*x^2+a)^(3/2),x)
Output:
(12*sqrt(a + b*x**2)*a*b*x + 4*sqrt(a + b*x**2)*b**2*x**3 - 12*sqrt(b)*log ((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2 - 12*sqrt(b)*log((sqrt(a + b *x**2) + sqrt(b)*x)/sqrt(a))*a*b*x**2 + 9*sqrt(b)*a**2 + 9*sqrt(b)*a*b*x** 2)/(8*b**3*(a + b*x**2))