Integrand size = 15, antiderivative size = 94 \[ \int x^4 \sqrt {a+b x^2} \, dx=-\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:
-1/16*a^2*x*(b*x^2+a)^(1/2)/b^2+1/24*a*x^3*(b*x^2+a)^(1/2)/b+1/6*x^5*(b*x^ 2+a)^(1/2)+1/16*a^3*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int x^4 \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-3 a^2 x+2 a b x^3+8 b^2 x^5\right )}{48 b^2}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Input:
Integrate[x^4*Sqrt[a + b*x^2],x]
Output:
(Sqrt[a + b*x^2]*(-3*a^2*x + 2*a*b*x^3 + 8*b^2*x^5))/(48*b^2) + (a^3*ArcTa nh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(5/2))
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {248, 262, 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.
\(\displaystyle \int x^4 \sqrt {a+b x^2} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{6} a \int \frac {x^4}{\sqrt {b x^2+a}}dx+\frac {1}{6} x^5 \sqrt {a+b x^2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\) |
Input:
Int[x^4*Sqrt[a + b*x^2],x]
Output:
(x^5*Sqrt[a + b*x^2])/6 + (a*((x^3*Sqrt[a + b*x^2])/(4*b) - (3*a*((x*Sqrt[ a + b*x^2])/(2*b) - (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))) /(4*b)))/6
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*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.66
method | result | size |
risch | \(-\frac {x \left (-8 b^{2} x^{4}-2 a b \,x^{2}+3 a^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b^{2}}+\frac {a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\) | \(62\) |
default | \(\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\) | \(82\) |
pseudoelliptic | \(\frac {8 b^{\frac {5}{2}} \sqrt {b \,x^{2}+a}\, x^{5}+2 a \,b^{\frac {3}{2}} x^{3} \sqrt {b \,x^{2}+a}-3 a^{2} x \sqrt {b}\, \sqrt {b \,x^{2}+a}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{3}}{48 b^{\frac {5}{2}}}\) | \(82\) |
Input:
int(x^4*(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/48*x*(-8*b^2*x^4-2*a*b*x^2+3*a^2)*(b*x^2+a)^(1/2)/b^2+1/16*a^3/b^(5/2)* ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.55 \[ \int x^4 \sqrt {a+b x^2} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 2 \, a b^{2} x^{3} - 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} x^{5} + 2 \, a b^{2} x^{3} - 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \] Input:
integrate(x^4*(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[1/96*(3*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*( 8*b^3*x^5 + 2*a*b^2*x^3 - 3*a^2*b*x)*sqrt(b*x^2 + a))/b^3, -1/48*(3*a^3*sq rt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*x^5 + 2*a*b^2*x^3 - 3*a ^2*b*x)*sqrt(b*x^2 + a))/b^3]
Time = 4.53 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int x^4 \sqrt {a+b x^2} \, dx=- \frac {a^{\frac {5}{2}} x}{16 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {3}{2}} x^{3}}{48 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 \sqrt {a} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {5}{2}}} + \frac {b x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:
integrate(x**4*(b*x**2+a)**(1/2),x)
Output:
-a**(5/2)*x/(16*b**2*sqrt(1 + b*x**2/a)) - a**(3/2)*x**3/(48*b*sqrt(1 + b* x**2/a)) + 5*sqrt(a)*x**5/(24*sqrt(1 + b*x**2/a)) + a**3*asinh(sqrt(b)*x/s qrt(a))/(16*b**(5/2)) + b*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int x^4 \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{3}}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} a^{2} x}{16 \, b^{2}} + \frac {a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/6*(b*x^2 + a)^(3/2)*x^3/b - 1/8*(b*x^2 + a)^(3/2)*a*x/b^2 + 1/16*sqrt(b* x^2 + a)*a^2*x/b^2 + 1/16*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68 \[ \int x^4 \sqrt {a+b x^2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} + \frac {a}{b}\right )} x^{2} - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/48*(2*(4*x^2 + a/b)*x^2 - 3*a^2/b^2)*sqrt(b*x^2 + a)*x - 1/16*a^3*log(ab s(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int x^4 \sqrt {a+b x^2} \, dx=\int x^4\,\sqrt {b\,x^2+a} \,d x \] Input:
int(x^4*(a + b*x^2)^(1/2),x)
Output:
int(x^4*(a + b*x^2)^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.85 \[ \int x^4 \sqrt {a+b x^2} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{2} b x +2 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}}{48 b^{3}} \] Input:
int(x^4*(b*x^2+a)^(1/2),x)
Output:
( - 3*sqrt(a + b*x**2)*a**2*b*x + 2*sqrt(a + b*x**2)*a*b**2*x**3 + 8*sqrt( a + b*x**2)*b**3*x**5 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a**3)/(48*b**3)