Integrand size = 32, antiderivative size = 118 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {c \sqrt {a+b x^2}}{5 a x^5}+\frac {(4 b c-5 a d) \sqrt {a+b x^2}}{15 a^2 x^3}-\frac {\left (8 b^2 c-10 a b d+15 a^2 e\right ) \sqrt {a+b x^2}}{15 a^3 x}+\frac {f \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}} \]
f*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(1/2)-1/5*c*(b*x^2+a)^(1/2)/a/x^5+1 /15*(-5*a*d+4*b*c)*(b*x^2+a)^(1/2)/a^2/x^3-1/15*(15*a^2*e-10*a*b*d+8*b^2*c )*(b*x^2+a)^(1/2)/a^3/x
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {\sqrt {a+b x^2} \left (8 b^2 c x^4-2 a b x^2 \left (2 c+5 d x^2\right )+a^2 \left (3 c+5 d x^2+15 e x^4\right )\right )}{15 a^3 x^5}-\frac {f \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}} \]
-1/15*(Sqrt[a + b*x^2]*(8*b^2*c*x^4 - 2*a*b*x^2*(2*c + 5*d*x^2) + a^2*(3*c + 5*d*x^2 + 15*e*x^4)))/(a^3*x^5) - (f*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2] ])/Sqrt[b]
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2338, 9, 1588, 358, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 2338 |
| \(\displaystyle -\frac {\int \frac {-5 a f x^5-5 a e x^3+(4 b c-5 a d) x}{x^5 \sqrt {b x^2+a}}dx}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle -\frac {\int \frac {-5 a f x^4-5 a e x^2+4 b c-5 a d}{x^4 \sqrt {b x^2+a}}dx}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 1588 |
| \(\displaystyle -\frac {-\frac {\int \frac {15 f x^2 a^2+15 e a^2-10 b d a+8 b^2 c}{x^2 \sqrt {b x^2+a}}dx}{3 a}-\frac {\sqrt {a+b x^2} (4 b c-5 a d)}{3 a x^3}}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle -\frac {-\frac {15 a^2 f \int \frac {1}{\sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} \left (15 a^2 e-10 a b d+8 b^2 c\right )}{a x}}{3 a}-\frac {\sqrt {a+b x^2} (4 b c-5 a d)}{3 a x^3}}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {-\frac {15 a^2 f \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-\frac {\sqrt {a+b x^2} \left (15 a^2 e-10 a b d+8 b^2 c\right )}{a x}}{3 a}-\frac {\sqrt {a+b x^2} (4 b c-5 a d)}{3 a x^3}}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {\frac {15 a^2 f \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {\sqrt {a+b x^2} \left (15 a^2 e-10 a b d+8 b^2 c\right )}{a x}}{3 a}-\frac {\sqrt {a+b x^2} (4 b c-5 a d)}{3 a x^3}}{5 a}-\frac {c \sqrt {a+b x^2}}{5 a x^5}\) |
-1/5*(c*Sqrt[a + b*x^2])/(a*x^5) - (-1/3*((4*b*c - 5*a*d)*Sqrt[a + b*x^2]) /(a*x^3) - (-(((8*b^2*c - 10*a*b*d + 15*a^2*e)*Sqrt[a + b*x^2])/(a*x)) + ( 15*a^2*f*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(3*a))/(5*a)
3.2.56.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, 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[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[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x_ Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, -1]
Int[((f_.)*(x_))^(m_)*((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, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f ^2*(m + 1)) Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x ) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 3.56 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 a^{2} e \,x^{4}-10 a b d \,x^{4}+8 b^{2} c \,x^{4}+5 a^{2} d \,x^{2}-4 a b c \,x^{2}+3 a^{2} c \right )}{15 a^{3} x^{5}}+\frac {f \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}\) | \(90\) |
| pseudoelliptic | \(\frac {f \,a^{3} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) x^{5}-\frac {\sqrt {b \,x^{2}+a}\, \left (-\frac {4 x^{2} \left (\frac {5 d \,x^{2}}{2}+c \right ) a \,b^{\frac {3}{2}}}{3}+\frac {8 b^{\frac {5}{2}} c \,x^{4}}{3}+a^{2} \sqrt {b}\, \left (5 e \,x^{4}+\frac {5}{3} d \,x^{2}+c \right )\right )}{5}}{\sqrt {b}\, x^{5} a^{3}}\) | \(96\) |
| default | \(\frac {f \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+d \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )-\frac {e \sqrt {b \,x^{2}+a}}{a x}+c \left (-\frac {\sqrt {b \,x^{2}+a}}{5 a \,x^{5}}-\frac {4 b \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )}{5 a}\right )\) | \(141\) |
-1/15*(b*x^2+a)^(1/2)*(15*a^2*e*x^4-10*a*b*d*x^4+8*b^2*c*x^4+5*a^2*d*x^2-4 *a*b*c*x^2+3*a^2*c)/a^3/x^5+f*ln(x*b^(1/2)+(b*x^2+a)^(1/2))/b^(1/2)
Time = 0.28 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.87 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} f x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c - {\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, a^{3} b x^{5}}, -\frac {15 \, a^{3} \sqrt {-b} f x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c - {\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a^{3} b x^{5}}\right ] \]
[1/30*(15*a^3*sqrt(b)*f*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a ) - 2*((8*b^3*c - 10*a*b^2*d + 15*a^2*b*e)*x^4 + 3*a^2*b*c - (4*a*b^2*c - 5*a^2*b*d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^5), -1/15*(15*a^3*sqrt(-b)*f*x^5 *arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + ((8*b^3*c - 10*a*b^2*d + 15*a^2*b*e) *x^4 + 3*a^2*b*c - (4*a*b^2*c - 5*a^2*b*d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^ 5)]
Time = 1.49 (sec) , antiderivative size = 427, normalized size of antiderivative = 3.62 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=- \frac {3 a^{4} b^{\frac {9}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {2 a^{3} b^{\frac {11}{2}} c x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {3 a^{2} b^{\frac {13}{2}} c x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {12 a b^{\frac {15}{2}} c x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {8 b^{\frac {17}{2}} c x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} + f \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \wedge b \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a}} & \text {otherwise} \end {cases}\right ) - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a x^{2}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{2}} + 1}}{a} + \frac {2 b^{\frac {3}{2}} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{2}} \]
-3*a**4*b**(9/2)*c*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5* x**6 + 15*a**3*b**6*x**8) - 2*a**3*b**(11/2)*c*x**2*sqrt(a/(b*x**2) + 1)/( 15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*a**2*b**(13 /2)*c*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 1 5*a**3*b**6*x**8) - 12*a*b**(15/2)*c*x**6*sqrt(a/(b*x**2) + 1)/(15*a**5*b* *4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 8*b**(17/2)*c*x**8*sqrt (a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x** 8) + f*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0 ) & Ne(b, 0)), (x*log(x)/sqrt(b*x**2), Ne(b, 0)), (x/sqrt(a), True)) - sqr t(b)*d*sqrt(a/(b*x**2) + 1)/(3*a*x**2) - sqrt(b)*e*sqrt(a/(b*x**2) + 1)/a + 2*b**(3/2)*d*sqrt(a/(b*x**2) + 1)/(3*a**2)
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=\frac {f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {8 \, \sqrt {b x^{2} + a} b^{2} c}{15 \, a^{3} x} + \frac {2 \, \sqrt {b x^{2} + a} b d}{3 \, a^{2} x} - \frac {\sqrt {b x^{2} + a} e}{a x} + \frac {4 \, \sqrt {b x^{2} + a} b c}{15 \, a^{2} x^{3}} - \frac {\sqrt {b x^{2} + a} d}{3 \, a x^{3}} - \frac {\sqrt {b x^{2} + a} c}{5 \, a x^{5}} \]
f*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 8/15*sqrt(b*x^2 + a)*b^2*c/(a^3*x) + 2/ 3*sqrt(b*x^2 + a)*b*d/(a^2*x) - sqrt(b*x^2 + a)*e/(a*x) + 4/15*sqrt(b*x^2 + a)*b*c/(a^2*x^3) - 1/3*sqrt(b*x^2 + a)*d/(a*x^3) - 1/5*sqrt(b*x^2 + a)*c /(a*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (100) = 200\).
Time = 0.33 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.70 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {f \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{2 \, \sqrt {b}} + \frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} \sqrt {b} e + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {3}{2}} d - 60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a \sqrt {b} e + 80 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} b^{\frac {5}{2}} c - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a b^{\frac {3}{2}} d + 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} \sqrt {b} e - 40 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a b^{\frac {5}{2}} c + 50 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{2} b^{\frac {3}{2}} d - 60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} \sqrt {b} e + 8 \, a^{2} b^{\frac {5}{2}} c - 10 \, a^{3} b^{\frac {3}{2}} d + 15 \, a^{4} \sqrt {b} e\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]
-1/2*f*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/sqrt(b) + 2/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^8*sqrt(b)*e + 30*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(3/2 )*d - 60*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*sqrt(b)*e + 80*(sqrt(b)*x - sqr t(b*x^2 + a))^4*b^(5/2)*c - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b^(3/2)*d + 90*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*sqrt(b)*e - 40*(sqrt(b)*x - sqrt (b*x^2 + a))^2*a*b^(5/2)*c + 50*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(3/2 )*d - 60*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*sqrt(b)*e + 8*a^2*b^(5/2)*c - 10*a^3*b^(3/2)*d + 15*a^4*sqrt(b)*e)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a )^5
Time = 6.65 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \sqrt {a+b x^2}} \, dx=\frac {f\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {e\,\sqrt {b\,x^2+a}}{a\,x}-\frac {d\,\sqrt {b\,x^2+a}\,\left (a-2\,b\,x^2\right )}{3\,a^2\,x^3}-\frac {c\,\sqrt {b\,x^2+a}\,\left (3\,a^2-4\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,x^5} \]