Integrand size = 11, antiderivative size = 84 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}} \] Output:
5/16*a^2*x*(b*x^2+a)^(1/2)+5/24*a*x*(b*x^2+a)^(3/2)+1/6*x*(b*x^2+a)^(5/2)+ 5/16*a^3*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(1/2)
Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {1}{48} \sqrt {a+b x^2} \left (33 a^2 x+26 a b x^3+8 b^2 x^5\right )-\frac {5 a^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 \sqrt {b}} \] Input:
Integrate[(a + b*x^2)^(5/2),x]
Output:
(Sqrt[a + b*x^2]*(33*a^2*x + 26*a*b*x^3 + 8*b^2*x^5))/48 - (5*a^3*Log[-(Sq rt[b]*x) + Sqrt[a + b*x^2]])/(16*Sqrt[b])
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {211, 211, 211, 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.
\(\displaystyle \int \left (a+b x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{6} a \int \left (b x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\) |
Input:
Int[(a + b*x^2)^(5/2),x]
Output:
(x*(a + b*x^2)^(5/2))/6 + (5*a*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/6
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
Time = 0.36 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {x \left (8 b^{2} x^{4}+26 a b \,x^{2}+33 a^{2}\right ) \sqrt {b \,x^{2}+a}}{48}+\frac {5 a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}\) | \(59\) |
pseudoelliptic | \(\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{3}}{16 \sqrt {b}}+\frac {11 x \left (\frac {8 b^{\frac {5}{2}} x^{4}}{33}+\frac {26 a \,b^{\frac {3}{2}} x^{2}}{33}+a^{2} \sqrt {b}\right ) \sqrt {b \,x^{2}+a}}{16 \sqrt {b}}\) | \(67\) |
default | \(\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\) | \(68\) |
Input:
int((b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/48*x*(8*b^2*x^4+26*a*b*x^2+33*a^2)*(b*x^2+a)^(1/2)+5/16*a^3*ln(b^(1/2)*x +(b*x^2+a)^(1/2))/b^(1/2)
Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.74 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b}\right ] \] Input:
integrate((b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[1/96*(15*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 + 26*a*b^2*x^3 + 33*a^2*b*x)*sqrt(b*x^2 + a))/b, -1/48*(15*a^3* sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*x^5 + 26*a*b^2*x^3 + 33*a^2*b*x)*sqrt(b*x^2 + a))/b]
Time = 2.56 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.15 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {11 a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{16} + \frac {13 a^{\frac {3}{2}} b x^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{24} + \frac {\sqrt {a} b^{2} x^{5} \sqrt {1 + \frac {b x^{2}}{a}}}{6} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} \] Input:
integrate((b*x**2+a)**(5/2),x)
Output:
11*a**(5/2)*x*sqrt(1 + b*x**2/a)/16 + 13*a**(3/2)*b*x**3*sqrt(1 + b*x**2/a )/24 + sqrt(a)*b**2*x**5*sqrt(1 + b*x**2/a)/6 + 5*a**3*asinh(sqrt(b)*x/sqr t(a))/(16*sqrt(b))
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x + \frac {5}{16} \, \sqrt {b x^{2} + a} a^{2} x + \frac {5 \, a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} \] Input:
integrate((b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
1/6*(b*x^2 + a)^(5/2)*x + 5/24*(b*x^2 + a)^(3/2)*a*x + 5/16*sqrt(b*x^2 + a )*a^2*x + 5/16*a^3*arcsinh(b*x/sqrt(a*b))/sqrt(b)
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \left (a+b x^2\right )^{5/2} \, dx=-\frac {5 \, a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} x^{2} + 13 \, a b\right )} x^{2} + 33 \, a^{2}\right )} \sqrt {b x^{2} + a} x \] Input:
integrate((b*x^2+a)^(5/2),x, algorithm="giac")
Output:
-5/16*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + 1/48*(2*(4*b^2* x^2 + 13*a*b)*x^2 + 33*a^2)*sqrt(b*x^2 + a)*x
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \] Input:
int((a + b*x^2)^(5/2),x)
Output:
(x*(a + b*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(5/2)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {33 \sqrt {b \,x^{2}+a}\, a^{2} b x +26 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}}{48 b} \] Input:
int((b*x^2+a)^(5/2),x)
Output:
(33*sqrt(a + b*x**2)*a**2*b*x + 26*sqrt(a + b*x**2)*a*b**2*x**3 + 8*sqrt(a + b*x**2)*b**3*x**5 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a**3)/(48*b)