Integrand size = 15, antiderivative size = 81 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x \sqrt {b x^2+c x^4}}-\frac {3 \sqrt {b x^2+c x^4}}{2 b^2 x^3}+\frac {3 c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}} \] Output:
1/b/x/(c*x^4+b*x^2)^(1/2)-3/2*(c*x^4+b*x^2)^(1/2)/b^2/x^3+3/2*c*arctanh(b^ (1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(5/2)
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-\sqrt {b} \left (b+3 c x^2\right )+3 c x^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{2 b^{5/2} x \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(b*x^2 + c*x^4)^(-3/2),x]
Output:
(-(Sqrt[b]*(b + 3*c*x^2)) + 3*c*x^2*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2 ]/Sqrt[b]])/(2*b^(5/2)*x*Sqrt[x^2*(b + c*x^2)])
Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1401, 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.
\(\displaystyle \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1401 |
\(\displaystyle \frac {3 \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx}{b}+\frac {1}{b x \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle \frac {3 \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{b}+\frac {1}{b x \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {3 \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{b}+\frac {1}{b x \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{b}+\frac {1}{b x \sqrt {b x^2+c x^4}}\) |
Input:
Int[(b*x^2 + c*x^4)^(-3/2),x]
Output:
1/(b*x*Sqrt[b*x^2 + c*x^4]) + (3*(-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*Ar cTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))))/b
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[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[-(b*x^2 + c*x^4)^ (p + 1)/(2*b*(p + 1)*x), x] + Simp[(4*p + 3)/(2*b*(p + 1)) Int[(b*x^2 + c *x^4)^(p + 1)/x^2, x], x] /; FreeQ[{b, c}, x] && !IntegerQ[p] && LtQ[p, -1 ]
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 = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {x \left (c \,x^{2}+b \right ) \left (3 b^{\frac {3}{2}} c \,x^{2}-3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {c \,x^{2}+b}\, b c \,x^{2}+b^{\frac {5}{2}}\right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {7}{2}}}\) | \(77\) |
risch | \(-\frac {c \,x^{2}+b}{2 b^{2} x \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (-\frac {c}{b^{2} \sqrt {c \,x^{2}+b}}+\frac {3 c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{2 b^{\frac {5}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(99\) |
Input:
int(1/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*x*(c*x^2+b)*(3*b^(3/2)*c*x^2-3*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c *x^2+b)^(1/2)*b*c*x^2+b^(5/2))/(c*x^4+b*x^2)^(3/2)/b^(7/2)
Time = 0.08 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.40 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} + b^{2}\right )}}{4 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, -\frac {3 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} + b^{2}\right )}}{2 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \] Input:
integrate(1/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/4*(3*(c^2*x^5 + b*c*x^3)*sqrt(b)*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)*(3*b*c*x^2 + b^2))/(b^3*c*x^5 + b^4*x^3), -1/2*(3*(c^2*x^5 + b*c*x^3)*sqrt(-b)*arctan(sqrt(c*x^4 + b*x^2 )*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2)*(3*b*c*x^2 + b^2))/(b^3*c*x^5 + b^ 4*x^3)]
\[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral((b*x**2 + c*x**4)**(-3/2), x)
\[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((c*x^4 + b*x^2)^(-3/2), x)
Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {3 \, c \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{2 \, \sqrt {-b} b^{2} \mathrm {sgn}\left (x\right )} - \frac {3 \, {\left (c x^{2} + b\right )} c - 2 \, b c}{2 \, {\left ({\left (c x^{2} + b\right )}^{\frac {3}{2}} - \sqrt {c x^{2} + b} b\right )} b^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
-3/2*c*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2*sgn(x)) - 1/2*(3*(c* x^2 + b)*c - 2*b*c)/(((c*x^2 + b)^(3/2) - sqrt(c*x^2 + b)*b)*b^2*sgn(x))
Time = 17.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {b}{c\,x^2}\right )}{5\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \] Input:
int(1/(b*x^2 + c*x^4)^(3/2),x)
Output:
-(x*(b/(c*x^2) + 1)^(3/2)*hypergeom([3/2, 5/2], 7/2, -b/(c*x^2)))/(5*(b*x^ 2 + c*x^4)^(3/2))
Time = 0.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.12 \[ \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, b^{2}-3 \sqrt {c \,x^{2}+b}\, b c \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b c \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{2} x^{4}}{2 b^{3} x^{2} \left (c \,x^{2}+b \right )} \] Input:
int(1/(c*x^4+b*x^2)^(3/2),x)
Output:
( - sqrt(b + c*x**2)*b**2 - 3*sqrt(b + c*x**2)*b*c*x**2 - 3*sqrt(b)*log((s qrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*b*c*x**2 - 3*sqrt(b)*log(( sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4 + 3*sqrt(b)*log ((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b*c*x**2 + 3*sqrt(b)*lo g((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*c**2*x**4)/(2*b**3*x** 2*(b + c*x**2))