Integrand size = 32, antiderivative size = 117 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=-\frac {c \sqrt {a+b x^2}}{a x}+\frac {(4 b e-3 a f) x \sqrt {a+b x^2}}{8 b^2}+\frac {f x^3 \sqrt {a+b x^2}}{4 b}+\frac {\left (8 b^2 d-4 a b e+3 a^2 f\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]
1/8*(3*a^2*f-4*a*b*e+8*b^2*d)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)-c *(b*x^2+a)^(1/2)/a/x+1/8*(-3*a*f+4*b*e)*x*(b*x^2+a)^(1/2)/b^2+1/4*f*x^3*(b *x^2+a)^(1/2)/b
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-8 b^2 c+4 a b e x^2-3 a^2 f x^2+2 a b f x^4\right )}{8 a b^2 x}+\frac {\left (-8 b^2 d+4 a b e-3 a^2 f\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \]
(Sqrt[a + b*x^2]*(-8*b^2*c + 4*a*b*e*x^2 - 3*a^2*f*x^2 + 2*a*b*f*x^4))/(8* a*b^2*x) + ((-8*b^2*d + 4*a*b*e - 3*a^2*f)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x ^2]])/(8*b^(5/2))
Time = 0.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2338, 9, 25, 1473, 27, 299, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 2338 |
| \(\displaystyle -\frac {\int -\frac {a f x^5+a e x^3+a d x}{x \sqrt {b x^2+a}}dx}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle -\frac {\int -\frac {a f x^4+a e x^2+a d}{\sqrt {b x^2+a}}dx}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a f x^4+a e x^2+a d}{\sqrt {b x^2+a}}dx}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 1473 |
| \(\displaystyle \frac {\frac {\int \frac {a \left ((4 b e-3 a f) x^2+4 b d\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {a f x^3 \sqrt {a+b x^2}}{4 b}}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a \int \frac {(4 b e-3 a f) x^2+4 b d}{\sqrt {b x^2+a}}dx}{4 b}+\frac {a f x^3 \sqrt {a+b x^2}}{4 b}}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {a \left (\frac {\left (3 a^2 f-4 a b e+8 b^2 d\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} (4 b e-3 a f)}{2 b}\right )}{4 b}+\frac {a f x^3 \sqrt {a+b x^2}}{4 b}}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {a \left (\frac {\left (3 a^2 f-4 a b e+8 b^2 d\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} (4 b e-3 a f)}{2 b}\right )}{4 b}+\frac {a f x^3 \sqrt {a+b x^2}}{4 b}}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 f-4 a b e+8 b^2 d\right )}{2 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 b e-3 a f)}{2 b}\right )}{4 b}+\frac {a f x^3 \sqrt {a+b x^2}}{4 b}}{a}-\frac {c \sqrt {a+b x^2}}{a x}\) |
-((c*Sqrt[a + b*x^2])/(a*x)) + ((a*f*x^3*Sqrt[a + b*x^2])/(4*b) + (a*(((4* b*e - 3*a*f)*x*Sqrt[a + b*x^2])/(2*b) + ((8*b^2*d - 4*a*b*e + 3*a^2*f)*Arc Tanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/(4*b))/a
3.2.54.3.1 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_)*(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[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[((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_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) , x] + Simp[1/(e*(4*p + 2*q + 1)) Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, 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[(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 = 3.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (-2 a b f \,x^{4}+3 a^{2} f \,x^{2}-4 a b e \,x^{2}+8 b^{2} c \right )}{8 b^{2} a x}+\frac {\left (3 a^{2} f -4 a e b +8 b^{2} d \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\) | \(93\) |
| pseudoelliptic | \(\frac {3 b^{2} x \left (a^{2} f -\frac {4}{3} a e b +\frac {8}{3} b^{2} d \right ) a \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-8 \sqrt {b \,x^{2}+a}\, \left (b^{2} c -\frac {x^{2} a \left (\frac {f \,x^{2}}{2}+e \right ) b}{2}+\frac {3 a^{2} f \,x^{2}}{8}\right ) b^{\frac {5}{2}}}{8 x \,b^{\frac {9}{2}} a}\) | \(98\) |
| default | \(\frac {d \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+f \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+e \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )-\frac {c \sqrt {b \,x^{2}+a}}{a x}\) | \(145\) |
-1/8*(b*x^2+a)^(1/2)*(-2*a*b*f*x^4+3*a^2*f*x^2-4*a*b*e*x^2+8*b^2*c)/b^2/a/ x+1/8*(3*a^2*f-4*a*b*e+8*b^2*d)/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.27 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.85 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=\left [\frac {{\left (8 \, a b^{2} d - 4 \, a^{2} b e + 3 \, a^{3} f\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, a b^{2} f x^{4} - 8 \, b^{3} c + {\left (4 \, a b^{2} e - 3 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a b^{3} x}, -\frac {{\left (8 \, a b^{2} d - 4 \, a^{2} b e + 3 \, a^{3} f\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, a b^{2} f x^{4} - 8 \, b^{3} c + {\left (4 \, a b^{2} e - 3 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a b^{3} x}\right ] \]
[1/16*((8*a*b^2*d - 4*a^2*b*e + 3*a^3*f)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b *x^2 + a)*sqrt(b)*x - a) + 2*(2*a*b^2*f*x^4 - 8*b^3*c + (4*a*b^2*e - 3*a^2 *b*f)*x^2)*sqrt(b*x^2 + a))/(a*b^3*x), -1/8*((8*a*b^2*d - 4*a^2*b*e + 3*a^ 3*f)*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*a*b^2*f*x^4 - 8*b^ 3*c + (4*a*b^2*e - 3*a^2*b*f)*x^2)*sqrt(b*x^2 + a))/(a*b^3*x)]
Time = 0.92 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.10 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=d \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \wedge b \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {x \sqrt {a + b x^{2}}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \sqrt {a}} & \text {otherwise} \end {cases}\right ) + f \left (\begin {cases} \frac {3 a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} - \frac {3 a x \sqrt {a + b x^{2}}}{8 b^{2}} + \frac {x^{3} \sqrt {a + b x^{2}}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{5}}{5 \sqrt {a}} & \text {otherwise} \end {cases}\right ) - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{2}} + 1}}{a} \]
d*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0) & N e(b, 0)), (x*log(x)/sqrt(b*x**2), Ne(b, 0)), (x/sqrt(a), True)) + e*Piecew ise((-a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2*b) + x*sqrt(a + b*x**2)/(2*b), Ne(b , 0)), (x**3/(3*sqrt(a)), True)) + f*Piecewise((3*a**2*Piecewise((log(2*sq rt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2) , True))/(8*b**2) - 3*a*x*sqrt(a + b*x**2)/(8*b**2) + x**3*sqrt(a + b*x**2 )/(4*b), Ne(b, 0)), (x**5/(5*sqrt(a)), True)) - sqrt(b)*c*sqrt(a/(b*x**2) + 1)/a
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} e x}{2 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a f x}{8 \, b^{2}} + \frac {d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a e \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {3 \, a^{2} f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a} c}{a x} \]
1/4*sqrt(b*x^2 + a)*f*x^3/b + 1/2*sqrt(b*x^2 + a)*e*x/b - 3/8*sqrt(b*x^2 + a)*a*f*x/b^2 + d*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/2*a*e*arcsinh(b*x/sqr t(a*b))/b^(3/2) + 3/8*a^2*f*arcsinh(b*x/sqrt(a*b))/b^(5/2) - sqrt(b*x^2 + a)*c/(a*x)
Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, f x^{2}}{b} + \frac {4 \, b^{2} e - 3 \, a b f}{b^{3}}\right )} x + \frac {2 \, \sqrt {b} c}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} - \frac {{\left (8 \, b^{2} d - 4 \, a b e + 3 \, a^{2} f\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{16 \, b^{\frac {5}{2}}} \]
1/8*sqrt(b*x^2 + a)*(2*f*x^2/b + (4*b^2*e - 3*a*b*f)/b^3)*x + 2*sqrt(b)*c/ ((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) - 1/16*(8*b^2*d - 4*a*b*e + 3*a^2*f) *log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/b^(5/2)
Timed out. \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \sqrt {a+b x^2}} \, dx=\int \frac {f\,x^6+e\,x^4+d\,x^2+c}{x^2\,\sqrt {b\,x^2+a}} \,d x \]