Integrand size = 19, antiderivative size = 80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3}{8} b \sqrt {b x^2+c x^4}+\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^2}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {c}} \] Output:
3/8*b*(c*x^4+b*x^2)^(1/2)+1/4*(c*x^4+b*x^2)^(3/2)/x^2+3/8*b^2*arctanh(c^(1 /2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(1/2)
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{8} \sqrt {x^2 \left (b+c x^2\right )} \left (5 b+2 c x^2-\frac {3 b^2 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )}{\sqrt {c} x \sqrt {b+c x^2}}\right ) \] Input:
Integrate[(b*x^2 + c*x^4)^(3/2)/x^3,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(5*b + 2*c*x^2 - (3*b^2*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]])/(Sqrt[c]*x*Sqrt[b + c*x^2])))/8
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1424, 1131, 1131, 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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^4}dx^2\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx^2+\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^2}\right )\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2+\sqrt {b x^2+c x^4}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}+\sqrt {b x^2+c x^4}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}+\sqrt {b x^2+c x^4}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^2}\right )\) |
Input:
Int[(b*x^2 + c*x^4)^(3/2)/x^3,x]
Output:
((b*x^2 + c*x^4)^(3/2)/(2*x^2) + (3*b*(Sqrt[b*x^2 + c*x^4] + (b*ArcTanh[(S qrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c]))/4)/2
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_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
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 = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {\left (2 c \,x^{2}+5 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8}+\frac {3 b^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 \sqrt {c}\, x \sqrt {c \,x^{2}+b}}\) | \(76\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 x \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}+3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b x +3 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2}\right )}{8 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}}\) | \(84\) |
pseudoelliptic | \(\frac {4 c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}-3 \ln \left (2\right ) b^{2}+3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}+10 b \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{16 \sqrt {c}}\) | \(90\) |
Input:
int((c*x^4+b*x^2)^(3/2)/x^3,x,method=_RETURNVERBOSE)
Output:
1/8*(2*c*x^2+5*b)*(x^2*(c*x^2+b))^(1/2)+3/8*b^2*ln(c^(1/2)*x+(c*x^2+b)^(1/ 2))/c^(1/2)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.81 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, b^{2} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + 5 \, b c\right )}}{16 \, c}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + 5 \, b c\right )}}{8 \, c}\right ] \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^3,x, algorithm="fricas")
Output:
[1/16*(3*b^2*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2 *sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 + 5*b*c))/c, -1/8*(3*b^2*sqrt(-c)*arctan(s qrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 + 5*b*c))/c]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{3}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(3/2)/x**3,x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)/x**3, x)
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, \sqrt {c}} + \frac {3}{8} \, \sqrt {c x^{4} + b x^{2}} b + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{4 \, x^{2}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^3,x, algorithm="maxima")
Output:
3/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 3/8*sq rt(c*x^4 + b*x^2)*b + 1/4*(c*x^4 + b*x^2)^(3/2)/x^2
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=-\frac {3 \, b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {c}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, \sqrt {c}} + \frac {1}{8} \, {\left (2 \, c x^{2} \mathrm {sgn}\left (x\right ) + 5 \, b \mathrm {sgn}\left (x\right )\right )} \sqrt {c x^{2} + b} x \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^3,x, algorithm="giac")
Output:
-3/8*b^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/sqrt(c) + 3/16*b^2* log(abs(b))*sgn(x)/sqrt(c) + 1/8*(2*c*x^2*sgn(x) + 5*b*sgn(x))*sqrt(c*x^2 + b)*x
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^3} \,d x \] Input:
int((b*x^2 + c*x^4)^(3/2)/x^3,x)
Output:
int((b*x^2 + c*x^4)^(3/2)/x^3, x)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.76 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {5 \sqrt {c \,x^{2}+b}\, b c x +2 \sqrt {c \,x^{2}+b}\, c^{2} x^{3}+3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2}}{8 c} \] Input:
int((c*x^4+b*x^2)^(3/2)/x^3,x)
Output:
(5*sqrt(b + c*x**2)*b*c*x + 2*sqrt(b + c*x**2)*c**2*x**3 + 3*sqrt(c)*log(( sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2)/(8*c)