Integrand size = 32, antiderivative size = 114 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=\frac {f \sqrt {a+b x^2}}{b}-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 b c-4 a d) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {\left (3 b^2 c-4 a b d+8 a^2 e\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
-1/8*(8*a^2*e-4*a*b*d+3*b^2*c)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+f* (b*x^2+a)^(1/2)/b-1/4*c*(b*x^2+a)^(1/2)/a/x^4+1/8*(-4*a*d+3*b*c)*(b*x^2+a) ^(1/2)/a^2/x^2
Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a b c+3 b^2 c x^2-4 a b d x^2+8 a^2 f x^4\right )}{8 a^2 b x^4}+\frac {\left (-3 b^2 c+4 a b d-8 a^2 e\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
(Sqrt[a + b*x^2]*(-2*a*b*c + 3*b^2*c*x^2 - 4*a*b*d*x^2 + 8*a^2*f*x^4))/(8* a^2*b*x^4) + ((-3*b^2*c + 4*a*b*d - 8*a^2*e)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[ a]])/(8*a^(5/2))
Time = 0.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2331, 2124, 27, 1192, 1471, 25, 299, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 2331 |
| \(\displaystyle \frac {1}{2} \int \frac {f x^6+e x^4+d x^2+c}{x^6 \sqrt {b x^2+a}}dx^2\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {-4 a f x^4-4 a e x^2+3 b c-4 a d}{2 x^4 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {-4 a f x^4-4 a e x^2+3 b c-4 a d}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {-4 a f x^8-4 a (b e-2 a f) x^4+3 b^3 c-4 a b^2 d+4 a^2 b e-4 a^3 f}{\left (a-x^4\right )^2}d\sqrt {b x^2+a}}{2 a b}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {b^2 \sqrt {a+b x^2} (3 b c-4 a d)}{2 a \left (a-x^4\right )}-\frac {\int -\frac {8 a^2 f x^4+3 b^3 c-4 a b^2 d+8 a^2 b e-8 a^3 f}{a-x^4}d\sqrt {b x^2+a}}{2 a}}{2 a b}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {8 a^2 f x^4+3 b^3 c-4 a b^2 d+8 a^2 b e-8 a^3 f}{a-x^4}d\sqrt {b x^2+a}}{2 a}+\frac {b^2 \sqrt {a+b x^2} (3 b c-4 a d)}{2 a \left (a-x^4\right )}}{2 a b}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {b \left (8 a^2 e-4 a b d+3 b^2 c\right ) \int \frac {1}{a-x^4}d\sqrt {b x^2+a}-8 a^2 f \sqrt {a+b x^2}}{2 a}+\frac {b^2 \sqrt {a+b x^2} (3 b c-4 a d)}{2 a \left (a-x^4\right )}}{2 a b}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (8 a^2 e-4 a b d+3 b^2 c\right )}{\sqrt {a}}-8 a^2 f \sqrt {a+b x^2}}{2 a}+\frac {b^2 \sqrt {a+b x^2} (3 b c-4 a d)}{2 a \left (a-x^4\right )}}{2 a b}-\frac {c \sqrt {a+b x^2}}{2 a x^4}\right )\) |
(-1/2*(c*Sqrt[a + b*x^2])/(a*x^4) - ((b^2*(3*b*c - 4*a*d)*Sqrt[a + b*x^2]) /(2*a*(a - x^4)) + (-8*a^2*f*Sqrt[a + b*x^2] + (b*(3*b^2*c - 4*a*b*d + 8*a ^2*e)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a])/(2*a))/(2*a*b))/2
3.2.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : > With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px , a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - a*d))), x] + Simp[1/((m + 1)*(b*c - a*d)) Int[(a + b*x)^(m + 1)*(c + d*x )^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] || ! ILtQ[n, -1])
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 = 3.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83
| method | result | size |
| pseudoelliptic | \(\frac {-\left (a^{2} e -\frac {1}{2} a b d +\frac {3}{8} b^{2} c \right ) b \,x^{4} a^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\sqrt {b \,x^{2}+a}\, \left (a^{2} f \,x^{4}-\frac {b \left (2 d \,x^{2}+c \right ) a}{4}+\frac {3 b^{2} c \,x^{2}}{8}\right ) a^{\frac {5}{2}}}{b \,a^{\frac {9}{2}} x^{4}}\) | \(95\) |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (4 a d \,x^{2}-3 c b \,x^{2}+2 a c \right )}{8 a^{2} x^{4}}+\frac {\frac {8 a^{2} f \sqrt {b \,x^{2}+a}}{b}-\frac {\left (8 a^{2} e -4 a b d +3 b^{2} c \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}}{8 a^{2}}\) | \(108\) |
| default | \(\frac {f \sqrt {b \,x^{2}+a}}{b}-\frac {e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}+d \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )+c \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\) | \(167\) |
(-(a^2*e-1/2*a*b*d+3/8*b^2*c)*b*x^4*a^2*arctanh((b*x^2+a)^(1/2)/a^(1/2))+( b*x^2+a)^(1/2)*(a^2*f*x^4-1/4*b*(2*d*x^2+c)*a+3/8*b^2*c*x^2)*a^(5/2))/b/a^ (9/2)/x^4
Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.94 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=\left [\frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c + {\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{3} b x^{4}}, \frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c + {\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{3} b x^{4}}\right ] \]
[1/16*((3*b^3*c - 4*a*b^2*d + 8*a^2*b*e)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt( b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*a^3*f*x^4 - 2*a^2*b*c + (3*a*b^2*c - 4*a^2*b*d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4), 1/8*((3*b^3*c - 4*a*b^2*d + 8*a^2*b*e)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*a^3*f*x^4 - 2*a^2*b*c + (3*a*b^2*c - 4*a^2*b*d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4)]
Time = 33.69 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.70 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=f \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{b} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \sqrt {a}} & \text {otherwise} \end {cases}\right ) - \frac {c}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {\sqrt {b} c}{8 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} + \frac {3 b^{\frac {3}{2}} c}{8 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{\sqrt {a}} + \frac {b d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} - \frac {3 b^{2} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {5}{2}}} \]
f*Piecewise((sqrt(a + b*x**2)/b, Ne(b, 0)), (x**2/(2*sqrt(a)), True)) - c/ (4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + sqrt(b)*c/(8*a*x**3*sqrt(a/(b*x**2 ) + 1)) - sqrt(b)*d*sqrt(a/(b*x**2) + 1)/(2*a*x) + 3*b**(3/2)*c/(8*a**2*x* sqrt(a/(b*x**2) + 1)) - e*asinh(sqrt(a)/(sqrt(b)*x))/sqrt(a) + b*d*asinh(s qrt(a)/(sqrt(b)*x))/(2*a**(3/2)) - 3*b**2*c*asinh(sqrt(a)/(sqrt(b)*x))/(8* a**(5/2))
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.12 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=-\frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {b d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {e \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} f}{b} + \frac {3 \, \sqrt {b x^{2} + a} b c}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} d}{2 \, a x^{2}} - \frac {\sqrt {b x^{2} + a} c}{4 \, a x^{4}} \]
-3/8*b^2*c*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 1/2*b*d*arcsinh(a/(sqrt (a*b)*abs(x)))/a^(3/2) - e*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + sqrt(b* x^2 + a)*f/b + 3/8*sqrt(b*x^2 + a)*b*c/(a^2*x^2) - 1/2*sqrt(b*x^2 + a)*d/( a*x^2) - 1/4*sqrt(b*x^2 + a)*c/(a*x^4)
Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.23 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=\frac {8 \, \sqrt {b x^{2} + a} f + \frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} c - 5 \, \sqrt {b x^{2} + a} a b^{3} c - 4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2} d + 4 \, \sqrt {b x^{2} + a} a^{2} b^{2} d}{a^{2} b^{2} x^{4}}}{8 \, b} \]
1/8*(8*sqrt(b*x^2 + a)*f + (3*b^3*c - 4*a*b^2*d + 8*a^2*b*e)*arctan(sqrt(b *x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x^2 + a)^(3/2)*b^3*c - 5*sqrt(b *x^2 + a)*a*b^3*c - 4*(b*x^2 + a)^(3/2)*a*b^2*d + 4*sqrt(b*x^2 + a)*a^2*b^ 2*d)/(a^2*b^2*x^4))/b
Time = 6.99 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx=\frac {f\,\sqrt {b\,x^2+a}}{b}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {5\,c\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,c\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {d\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {b\,d\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {3\,b^2\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}} \]