Integrand size = 32, antiderivative size = 103 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\frac {\left (b^2 d-a b e+a^2 f\right ) \sqrt {a+b x^2}}{b^3}+\frac {(b e-2 a f) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]
1/3*(-2*a*f+b*e)*(b*x^2+a)^(3/2)/b^3+1/5*f*(b*x^2+a)^(5/2)/b^3-c*arctanh(( b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)+(a^2*f-a*b*e+b^2*d)*(b*x^2+a)^(1/2)/b^3
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (8 a^2 f-2 a b \left (5 e+2 f x^2\right )+b^2 \left (15 d+5 e x^2+3 f x^4\right )\right )}{15 b^3}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]
(Sqrt[a + b*x^2]*(8*a^2*f - 2*a*b*(5*e + 2*f*x^2) + b^2*(15*d + 5*e*x^2 + 3*f*x^4)))/(15*b^3) - (c*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]
Time = 0.36 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2331, 2123, 2009}
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.
\(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2331 |
\(\displaystyle \frac {1}{2} \int \frac {f x^6+e x^4+d x^2+c}{x^2 \sqrt {b x^2+a}}dx^2\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {1}{2} \int \left (\frac {f \left (b x^2+a\right )^{3/2}}{b^2}+\frac {(b e-2 a f) \sqrt {b x^2+a}}{b^2}+\frac {f a^2-b e a+b^2 d}{b^2 \sqrt {b x^2+a}}+\frac {c}{x^2 \sqrt {b x^2+a}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \sqrt {a+b x^2} \left (a^2 f-a b e+b^2 d\right )}{b^3}-\frac {2 c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {2 \left (a+b x^2\right )^{3/2} (b e-2 a f)}{3 b^3}+\frac {2 f \left (a+b x^2\right )^{5/2}}{5 b^3}\right )\) |
((2*(b^2*d - a*b*e + a^2*f)*Sqrt[a + b*x^2])/b^3 + (2*(b*e - 2*a*f)*(a + b *x^2)^(3/2))/(3*b^3) + (2*f*(a + b*x^2)^(5/2))/(5*b^3) - (2*c*ArcTanh[Sqrt [a + b*x^2]/Sqrt[a]])/Sqrt[a])/2
3.2.46.3.1 Defintions of rubi rules used
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
Time = 3.52 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\frac {3 b^{3} c \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{2}+\sqrt {b \,x^{2}+a}\, \left (b \left (\frac {2 f \,x^{2}}{5}+e \right ) a^{\frac {3}{2}}-\frac {4 a^{\frac {5}{2}} f}{5}-\frac {3 \sqrt {a}\, b^{2} \left (\frac {1}{5} f \,x^{4}+\frac {1}{3} e \,x^{2}+d \right )}{2}\right )\right )}{3 \sqrt {a}\, b^{3}}\) | \(82\) |
default | \(f \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+e \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+\frac {d \sqrt {b \,x^{2}+a}}{b}-\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}\) | \(139\) |
-2/3*(3/2*b^3*c*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(b*x^2+a)^(1/2)*(b*(2/5*f *x^2+e)*a^(3/2)-4/5*a^(5/2)*f-3/2*a^(1/2)*b^2*(1/5*f*x^4+1/3*e*x^2+d)))/a^ (1/2)/b^3
Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.99 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} c \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{2} f x^{4} + 15 \, a b^{2} d - 10 \, a^{2} b e + 8 \, a^{3} f + {\left (5 \, a b^{2} e - 4 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, a b^{3}}, \frac {15 \, \sqrt {-a} b^{3} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, a b^{2} f x^{4} + 15 \, a b^{2} d - 10 \, a^{2} b e + 8 \, a^{3} f + {\left (5 \, a b^{2} e - 4 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a b^{3}}\right ] \]
[1/30*(15*sqrt(a)*b^3*c*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2 ) + 2*(3*a*b^2*f*x^4 + 15*a*b^2*d - 10*a^2*b*e + 8*a^3*f + (5*a*b^2*e - 4* a^2*b*f)*x^2)*sqrt(b*x^2 + a))/(a*b^3), 1/15*(15*sqrt(-a)*b^3*c*arctan(sqr t(-a)/sqrt(b*x^2 + a)) + (3*a*b^2*f*x^4 + 15*a*b^2*d - 10*a^2*b*e + 8*a^3* f + (5*a*b^2*e - 4*a^2*b*f)*x^2)*sqrt(b*x^2 + a))/(a*b^3)]
Time = 4.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.30 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\frac {\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + \frac {2 f \left (a + b x^{2}\right )^{\frac {5}{2}}}{5 b^{3}} + \frac {2 \left (a + b x^{2}\right )^{\frac {3}{2}} \left (- 2 a f + b e\right )}{3 b^{3}} + \frac {2 \sqrt {a + b x^{2}} \left (a^{2} f - a b e + b^{2} d\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {c \log {\left (x^{2} \right )} + d x^{2} + \frac {e x^{4}}{2} + \frac {f x^{6}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases}}{2} \]
Piecewise((2*c*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + 2*f*(a + b*x**2) **(5/2)/(5*b**3) + 2*(a + b*x**2)**(3/2)*(-2*a*f + b*e)/(3*b**3) + 2*sqrt( a + b*x**2)*(a**2*f - a*b*e + b**2*d)/b**3, Ne(b, 0)), ((c*log(x**2) + d*x **2 + e*x**4/2 + f*x**6/3)/sqrt(a), True))/2
Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x^{4}}{5 \, b} + \frac {\sqrt {b x^{2} + a} e x^{2}}{3 \, b} - \frac {4 \, \sqrt {b x^{2} + a} a f x^{2}}{15 \, b^{2}} - \frac {c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} d}{b} - \frac {2 \, \sqrt {b x^{2} + a} a e}{3 \, b^{2}} + \frac {8 \, \sqrt {b x^{2} + a} a^{2} f}{15 \, b^{3}} \]
1/5*sqrt(b*x^2 + a)*f*x^4/b + 1/3*sqrt(b*x^2 + a)*e*x^2/b - 4/15*sqrt(b*x^ 2 + a)*a*f*x^2/b^2 - c*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + sqrt(b*x^2 + a)*d/b - 2/3*sqrt(b*x^2 + a)*a*e/b^2 + 8/15*sqrt(b*x^2 + a)*a^2*f/b^3
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\frac {c \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {15 \, \sqrt {b x^{2} + a} b^{14} d + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{13} e - 15 \, \sqrt {b x^{2} + a} a b^{13} e + 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{12} f - 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{12} f + 15 \, \sqrt {b x^{2} + a} a^{2} b^{12} f}{15 \, b^{15}} \]
c*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/15*(15*sqrt(b*x^2 + a)*b^1 4*d + 5*(b*x^2 + a)^(3/2)*b^13*e - 15*sqrt(b*x^2 + a)*a*b^13*e + 3*(b*x^2 + a)^(5/2)*b^12*f - 10*(b*x^2 + a)^(3/2)*a*b^12*f + 15*sqrt(b*x^2 + a)*a^2 *b^12*f)/b^15
Time = 6.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx=\sqrt {b\,x^2+a}\,\left (\frac {8\,a^2\,f}{15\,b^3}+\frac {f\,x^4}{5\,b}-\frac {4\,a\,f\,x^2}{15\,b^2}\right )+\frac {d\,\sqrt {b\,x^2+a}}{b}-\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {e\,\sqrt {b\,x^2+a}\,\left (2\,a-b\,x^2\right )}{3\,b^2} \]