Integrand size = 19, antiderivative size = 84 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=-\frac {\sqrt {b x^2+c x^4}}{4 x^5}-\frac {c \sqrt {b x^2+c x^4}}{8 b x^3}+\frac {c^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{3/2}} \] Output:
-1/4*(c*x^4+b*x^2)^(1/2)/x^5-1/8*c*(c*x^4+b*x^2)^(1/2)/b/x^3+1/8*c^2*arcta nh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(3/2)
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-\sqrt {b} \sqrt {b+c x^2} \left (2 b+c x^2\right )+c^2 x^4 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{8 b^{3/2} x^5 \sqrt {b+c x^2}} \] Input:
Integrate[Sqrt[b*x^2 + c*x^4]/x^6,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-(Sqrt[b]*Sqrt[b + c*x^2]*(2*b + c*x^2)) + c^2*x^4 *ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(8*b^(3/2)*x^5*Sqrt[b + c*x^2])
Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1425, 1430, 1400, 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.
\(\displaystyle \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {1}{4} c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle \frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {1}{4} c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\) |
Input:
Int[Sqrt[b*x^2 + c*x^4]/x^6,x]
Output:
-1/4*Sqrt[b*x^2 + c*x^4]/x^5 + (c*(-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*A rcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))))/4
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[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[m + 2*p + 1, 0 ]
Time = 0.79 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.05
method | result | size |
risch | \(-\frac {\left (c \,x^{2}+2 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x^{5} b}+\frac {c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 b^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\) | \(88\) |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (\sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{2} x^{4}-\sqrt {c \,x^{2}+b}\, c^{2} x^{4}+\left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{2}-2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \right )}{8 x^{5} \sqrt {c \,x^{2}+b}\, b^{2}}\) | \(106\) |
Input:
int((c*x^4+b*x^2)^(1/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/8*(c*x^2+2*b)/x^5/b*(x^2*(c*x^2+b))^(1/2)+1/8*c^2/b^(3/2)*ln((2*b+2*b^( 1/2)*(c*x^2+b)^(1/2))/x)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\left [\frac {\sqrt {b} c^{2} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (b c x^{2} + 2 \, b^{2}\right )}}{16 \, b^{2} x^{5}}, -\frac {\sqrt {-b} c^{2} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (b c x^{2} + 2 \, b^{2}\right )}}{8 \, b^{2} x^{5}}\right ] \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="fricas")
Output:
[1/16*(sqrt(b)*c^2*x^5*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b) )/x^3) - 2*sqrt(c*x^4 + b*x^2)*(b*c*x^2 + 2*b^2))/(b^2*x^5), -1/8*(sqrt(-b )*c^2*x^5*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2) *(b*c*x^2 + 2*b^2))/(b^2*x^5)]
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{6}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(1/2)/x**6,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))/x**6, x)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{6}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="maxima")
Output:
integrate(sqrt(c*x^4 + b*x^2)/x^6, x)
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=-\frac {\frac {c^{3} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{3} \mathrm {sgn}\left (x\right ) + \sqrt {c x^{2} + b} b c^{3} \mathrm {sgn}\left (x\right )}{b c^{2} x^{4}}}{8 \, c} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="giac")
Output:
-1/8*(c^3*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + ((c*x^2 + b)^(3/2)*c^3*sgn(x) + sqrt(c*x^2 + b)*b*c^3*sgn(x))/(b*c^2*x^4))/c
Timed out. \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^2}}{x^6} \,d x \] Input:
int((b*x^2 + c*x^4)^(1/2)/x^6,x)
Output:
int((b*x^2 + c*x^4)^(1/2)/x^6, x)
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b^{2}-\sqrt {c \,x^{2}+b}\, b c \,x^{2}-\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}}{8 b^{2} x^{4}} \] Input:
int((c*x^4+b*x^2)^(1/2)/x^6,x)
Output:
( - 2*sqrt(b + c*x**2)*b**2 - sqrt(b + c*x**2)*b*c*x**2 - sqrt(b)*log((sqr t(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4 + sqrt(b)*log((sqr t(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4)/(8*b**2*x**4)