Integrand size = 17, antiderivative size = 68 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{3/2}} \] Output:
1/8*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c-1/8*b^2*arctanh(c^(1/2)*x^2/(c*x^4+b *x^2)^(1/2))/c^(3/2)
Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.43 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (b+2 c x^2\right )+2 b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{8 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x*Sqrt[b*x^2 + c*x^4],x]
Output:
(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(b + 2*c*x^2) + 2*b^2*ArcTan h[(Sqrt[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])]))/(8*c^(3/2)*Sqrt[x^2*(b + c*x^ 2)])
Time = 0.34 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1424, 1087, 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.
\(\displaystyle \int x \sqrt {b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \sqrt {c x^4+b x^2}dx^2\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )\) |
Input:
Int[x*Sqrt[b*x^2 + c*x^4],x]
Output:
(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x^2)/Sq rt[b*x^2 + c*x^4]])/(4*c^(3/2)))/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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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[(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 = 77, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 c}-\frac {b^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 c^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\) | \(77\) |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (2 x \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}-\sqrt {c}\, \sqrt {c \,x^{2}+b}\, b x -\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2}\right )}{8 x \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}}}\) | \(84\) |
pseudoelliptic | \(\frac {4 c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\ln \left (2\right ) b^{2}-\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}+2 b \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{16 c^{\frac {3}{2}}}\) | \(89\) |
Input:
int(x*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(2*c*x^2+b)*(x^2*(c*x^2+b))^(1/2)/c-1/8*b^2/c^(3/2)*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 = 140, normalized size of antiderivative = 2.06 \[ \int x \sqrt {b x^2+c x^4} \, dx=\left [\frac {b^{2} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + b c\right )}}{16 \, c^{2}}, \frac {b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + b c\right )}}{8 \, c^{2}}\right ] \] Input:
integrate(x*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/16*(b^2*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*s qrt(c*x^4 + b*x^2)*(2*c^2*x^2 + b*c))/c^2, 1/8*(b^2*sqrt(-c)*arctan(sqrt(c *x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 + b*c ))/c^2]
\[ \int x \sqrt {b x^2+c x^4} \, dx=\int x \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \] Input:
integrate(x*(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x*sqrt(x**2*(b + c*x**2)), x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {1}{4} \, \sqrt {c x^{4} + b x^{2}} x^{2} - \frac {b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{4} + b x^{2}} b}{8 \, c} \] Input:
integrate(x*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/4*sqrt(c*x^4 + b*x^2)*x^2 - 1/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b* x^2)*sqrt(c))/c^(3/2) + 1/8*sqrt(c*x^4 + b*x^2)*b/c
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {1}{8} \, \sqrt {c x^{2} + b} {\left (2 \, x^{2} \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{c}\right )} x + \frac {b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, c^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {3}{2}}} \] Input:
integrate(x*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(c*x^2 + b)*(2*x^2*sgn(x) + b*sgn(x)/c)*x + 1/8*b^2*log(abs(-sqrt( c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(3/2) - 1/16*b^2*log(abs(b))*sgn(x)/c^(3 /2)
Time = 18.00 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}}{2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{3/2}} \] Input:
int(x*(b*x^2 + c*x^4)^(1/2),x)
Output:
((b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2))/2 - (b^2*log((b/2 + c*x^2)/c^(1/ 2) + (b*x^2 + c*x^4)^(1/2)))/(16*c^(3/2))
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int x \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c \,x^{2}+b}\, b c x +2 \sqrt {c \,x^{2}+b}\, c^{2} x^{3}-\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2}}{8 c^{2}} \] Input:
int(x*(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*b*c*x + 2*sqrt(b + c*x**2)*c**2*x**3 - sqrt(c)*log((sqrt (b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2)/(8*c**2)