Integrand size = 17, antiderivative size = 101 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^2}+\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c}+\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{5/2}} \] Output:
-3/128*b^2*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^2+1/16*(2*c*x^2+b)*(c*x^4+b*x ^2)^(3/2)/c+3/128*b^4*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(5/2)
Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.20 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (-3 b^3+2 b^2 c x^2+24 b c^2 x^4+16 c^3 x^6\right )+6 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{128 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x*(b*x^2 + c*x^4)^(3/2),x]
Output:
(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(-3*b^3 + 2*b^2*c*x^2 + 24*b *c^2*x^4 + 16*c^3*x^6) + 6*b^4*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c* x^2])]))/(128*c^(5/2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.39 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1424, 1087, 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.
\(\displaystyle \int x \left (b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \left (c x^4+b x^2\right )^{3/2}dx^2\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^4+b x^2}dx^2}{16 c}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{16 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{16 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{16 c}\right )\) |
Input:
Int[x*(b*x^2 + c*x^4)^(3/2),x]
Output:
(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(8*c) - (3*b^2*(((b + 2*c*x^2)*Sqrt [b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/ (4*c^(3/2))))/(16*c))/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 = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{4}}{256}+\frac {3 \left (\frac {32 c^{\frac {7}{2}} x^{6}}{3}+16 c^{\frac {5}{2}} b \,x^{4}+\frac {4 c^{\frac {3}{2}} b^{2} x^{2}}{3}-2 \sqrt {c}\, b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256}-\frac {3 \ln \left (2\right ) b^{4}}{256}}{c^{\frac {5}{2}}}\) | \(100\) |
risch | \(-\frac {\left (-16 c^{3} x^{6}-24 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+3 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{2}}+\frac {3 b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\) | \(101\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (16 x^{3} \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}}-8 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b x +2 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x +3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{3} x +3 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4}\right )}{128 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}}}\) | \(122\) |
Input:
int(x*(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
3/256*(ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/2)+b)/c^(1/2))*b^4+(32/3*c ^(7/2)*x^6+16*c^(5/2)*b*x^4+4/3*c^(3/2)*b^2*x^2-2*c^(1/2)*b^3)*(x^2*(c*x^2 +b))^(1/2)-ln(2)*b^4)/c^(5/2)
Time = 0.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.87 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {3 \, b^{4} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} - 3 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{256 \, c^{3}}, -\frac {3 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} - 3 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{128 \, c^{3}}\right ] \] Input:
integrate(x*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/256*(3*b^4*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(16*c^4*x^6 + 24*b*c^3*x^4 + 2*b^2*c^2*x^2 - 3*b^3*c)*sqrt(c*x^4 + b*x^2 ))/c^3, -1/128*(3*b^4*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (16*c^4*x^6 + 24*b*c^3*x^4 + 2*b^2*c^2*x^2 - 3*b^3*c)*sqrt(c*x^4 + b*x^2))/c^3]
\[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\int x \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x*(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x*(x**2*(b + c*x**2))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{8} \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{64 \, c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{128 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{16 \, c} \] Input:
integrate(x*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
1/8*(c*x^4 + b*x^2)^(3/2)*x^2 - 3/64*sqrt(c*x^4 + b*x^2)*b^2*x^2/c + 3/256 *b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 3/128*sqrt (c*x^4 + b*x^2)*b^3/c^2 + 1/16*(c*x^4 + b*x^2)^(3/2)*b/c
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 \, b^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, c^{\frac {5}{2}}} + \frac {3 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {5}{2}}} + \frac {1}{128} \, {\left (2 \, {\left (4 \, {\left (2 \, c x^{2} \mathrm {sgn}\left (x\right ) + 3 \, b \mathrm {sgn}\left (x\right )\right )} x^{2} + \frac {b^{2} \mathrm {sgn}\left (x\right )}{c}\right )} x^{2} - \frac {3 \, b^{3} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} \sqrt {c x^{2} + b} x \] Input:
integrate(x*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
-3/128*b^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(5/2) + 3/256*b ^4*log(abs(b))*sgn(x)/c^(5/2) + 1/128*(2*(4*(2*c*x^2*sgn(x) + 3*b*sgn(x))* x^2 + b^2*sgn(x)/c)*x^2 - 3*b^3*sgn(x)/c^2)*sqrt(c*x^2 + b)*x
Time = 18.90 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}\,\left (c\,x^2+\frac {b}{2}\right )}{8\,c}-\frac {3\,b^2\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{32\,c} \] Input:
int(x*(b*x^2 + c*x^4)^(3/2),x)
Output:
((b*x^2 + c*x^4)^(3/2)*(b/2 + c*x^2))/(8*c) - (3*b^2*((b/(4*c) + x^2/2)*(b *x^2 + c*x^4)^(1/2) - (b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/ 2)))/(8*c^(3/2))))/(32*c)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int x \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {-3 \sqrt {c \,x^{2}+b}\, b^{3} c x +2 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{3}+24 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{5}+16 \sqrt {c \,x^{2}+b}\, c^{4} x^{7}+3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{4}}{128 c^{3}} \] Input:
int(x*(c*x^4+b*x^2)^(3/2),x)
Output:
( - 3*sqrt(b + c*x**2)*b**3*c*x + 2*sqrt(b + c*x**2)*b**2*c**2*x**3 + 24*s qrt(b + c*x**2)*b*c**3*x**5 + 16*sqrt(b + c*x**2)*c**4*x**7 + 3*sqrt(c)*lo g((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**4)/(128*c**3)