Integrand size = 19, antiderivative size = 77 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{2} c \sqrt {b x^2+c x^4}-\frac {b \sqrt {b x^2+c x^4}}{x^2}+\frac {3}{2} b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \] Output:
1/2*c*(c*x^4+b*x^2)^(1/2)-b*(c*x^4+b*x^2)^(1/2)/x^2+3/2*b*c^(1/2)*arctanh( c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {-2 b^2-b c x^2+c^2 x^4+6 b \sqrt {c} x \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{2 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(b*x^2 + c*x^4)^(3/2)/x^5,x]
Output:
(-2*b^2 - b*c*x^2 + c^2*x^4 + 6*b*Sqrt[c]*x*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[ c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])])/(2*Sqrt[x^2*(b + c*x^2)])
Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1424, 1125, 25, 27, 1160, 1091, 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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^6}dx^2\) |
\(\Big \downarrow \) 1125 |
\(\displaystyle \frac {1}{2} \left (-\int -\frac {c \left (c x^2+2 b\right )}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\int \frac {c \left (c x^2+2 b\right )}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (c \int \frac {c x^2+2 b}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {3}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (c \left (3 b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )\) |
Input:
Int[(b*x^2 + c*x^4)^(3/2)/x^5,x]
Output:
((-2*b*Sqrt[b*x^2 + c*x^4])/x^2 + c*(Sqrt[b*x^2 + c*x^4] + (3*b*ArcTanh[(S qrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c]))/2
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[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 2)*(d + e*x))), x] - Simp[e^(2*m + 2) Int[(1/Sqrt[a + b*x + c*x^2])*Expan dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x ), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && EqQ[m + p, -3/2]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {\left (-c \,x^{2}+2 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{2}}+\frac {3 b \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x \sqrt {c \,x^{2}+b}}\) | \(77\) |
pseudoelliptic | \(\frac {3 x^{2} b \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) \sqrt {c}-2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-c \,x^{2}+2 b \right )}{4 x^{2}}\) | \(77\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{2}+3 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{2}-2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}+3 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c x \right )}{2 x^{4} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \sqrt {c}}\) | \(107\) |
Input:
int((c*x^4+b*x^2)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/2*(-c*x^2+2*b)/x^2*(x^2*(c*x^2+b))^(1/2)+3/2*b*c^(1/2)*ln(c^(1/2)*x+(c* x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.81 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\left [\frac {3 \, b \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{4 \, x^{2}}, -\frac {3 \, b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{2 \, x^{2}}\right ] \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="fricas")
Output:
[1/4*(3*b*sqrt(c)*x^2*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*sqrt(c*x^4 + b*x^2)*(c*x^2 - 2*b))/x^2, -1/2*(3*b*sqrt(-c)*x^2*arctan(sq rt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sqrt(c*x^4 + b*x^2)*(c*x^2 - 2*b ))/x^2]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{5}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(3/2)/x**5,x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)/x**5, x)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {3}{4} \, b \sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{2 \, x^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{2 \, x^{4}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="maxima")
Output:
3/4*b*sqrt(c)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 3/2*sqrt( c*x^4 + b*x^2)*b/x^2 + 1/2*(c*x^4 + b*x^2)^(3/2)/x^4
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b} c x \mathrm {sgn}\left (x\right ) - \frac {3}{4} \, b \sqrt {c} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, b^{2} \sqrt {c} \mathrm {sgn}\left (x\right )}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="giac")
Output:
1/2*sqrt(c*x^2 + b)*c*x*sgn(x) - 3/4*b*sqrt(c)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 2*b^2*sqrt(c)*sgn(x)/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^5} \,d x \] Input:
int((b*x^2 + c*x^4)^(3/2)/x^5,x)
Output:
int((b*x^2 + c*x^4)^(3/2)/x^5, x)
Time = 0.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {-8 \sqrt {c \,x^{2}+b}\, b +4 \sqrt {c \,x^{2}+b}\, c \,x^{2}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b x -9 \sqrt {c}\, b x}{8 x} \] Input:
int((c*x^4+b*x^2)^(3/2)/x^5,x)
Output:
( - 8*sqrt(b + c*x**2)*b + 4*sqrt(b + c*x**2)*c*x**2 + 12*sqrt(c)*log((sqr t(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b*x - 9*sqrt(c)*b*x)/(8*x)