Integrand size = 17, antiderivative size = 72 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {1}{b^3 x}-\frac {c x}{4 b^2 \left (b+c x^2\right )^2}-\frac {7 c x}{8 b^3 \left (b+c x^2\right )}-\frac {15 \sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}} \] Output:
-1/b^3/x-1/4*c*x/b^2/(c*x^2+b)^2-7/8*c*x/b^3/(c*x^2+b)-15/8*c^(1/2)*arctan (c^(1/2)*x/b^(1/2))/b^(7/2)
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {8 b^2+25 b c x^2+15 c^2 x^4}{8 b^3 x \left (b+c x^2\right )^2}-\frac {15 \sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}} \] Input:
Integrate[x^4/(b*x^2 + c*x^4)^3,x]
Output:
-1/8*(8*b^2 + 25*b*c*x^2 + 15*c^2*x^4)/(b^3*x*(b + c*x^2)^2) - (15*Sqrt[c] *ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(7/2))
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {9, 253, 253, 264, 218}
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.
\(\displaystyle \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^2 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{x^2 \left (c x^2+b\right )^2}dx}{4 b}+\frac {1}{4 b x \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{2 b}+\frac {1}{2 b x \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{2 b}+\frac {1}{2 b x \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{2 b}+\frac {1}{2 b x \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x \left (b+c x^2\right )^2}\) |
Input:
Int[x^4/(b*x^2 + c*x^4)^3,x]
Output:
1/(4*b*x*(b + c*x^2)^2) + (5*(1/(2*b*x*(b + c*x^2)) + (3*(-(1/(b*x)) - (Sq rt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2)))/(2*b)))/(4*b)
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {1}{b^{3} x}-\frac {c \left (\frac {\frac {7}{8} c \,x^{3}+\frac {9}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{3}}\) | \(54\) |
risch | \(\frac {-\frac {15 c^{2} x^{4}}{8 b^{3}}-\frac {25 c \,x^{2}}{8 b^{2}}-\frac {1}{b}}{x \left (c \,x^{2}+b \right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+2 c \right ) x +b^{4} \textit {\_R} \right )\right )}{16}\) | \(79\) |
Input:
int(x^4/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
-1/b^3/x-1/b^3*c*((7/8*c*x^3+9/8*b*x)/(c*x^2+b)^2+15/8/(b*c)^(1/2)*arctan( c*x/(b*c)^(1/2)))
Time = 0.18 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.81 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=\left [-\frac {30 \, c^{2} x^{4} + 50 \, b c x^{2} - 15 \, {\left (c^{2} x^{5} + 2 \, b c x^{3} + b^{2} x\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) + 16 \, b^{2}}{16 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}}, -\frac {15 \, c^{2} x^{4} + 25 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + 2 \, b c x^{3} + b^{2} x\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 8 \, b^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}}\right ] \] Input:
integrate(x^4/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
[-1/16*(30*c^2*x^4 + 50*b*c*x^2 - 15*(c^2*x^5 + 2*b*c*x^3 + b^2*x)*sqrt(-c /b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)) + 16*b^2)/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x), -1/8*(15*c^2*x^4 + 25*b*c*x^2 + 15*(c^2*x^5 + 2*b *c*x^3 + b^2*x)*sqrt(c/b)*arctan(x*sqrt(c/b)) + 8*b^2)/(b^3*c^2*x^5 + 2*b^ 4*c*x^3 + b^5*x)]
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.61 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=\frac {15 \sqrt {- \frac {c}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c}{b^{7}}}}{c} + x \right )}}{16} - \frac {15 \sqrt {- \frac {c}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c}{b^{7}}}}{c} + x \right )}}{16} + \frac {- 8 b^{2} - 25 b c x^{2} - 15 c^{2} x^{4}}{8 b^{5} x + 16 b^{4} c x^{3} + 8 b^{3} c^{2} x^{5}} \] Input:
integrate(x**4/(c*x**4+b*x**2)**3,x)
Output:
15*sqrt(-c/b**7)*log(-b**4*sqrt(-c/b**7)/c + x)/16 - 15*sqrt(-c/b**7)*log( b**4*sqrt(-c/b**7)/c + x)/16 + (-8*b**2 - 25*b*c*x**2 - 15*c**2*x**4)/(8*b **5*x + 16*b**4*c*x**3 + 8*b**3*c**2*x**5)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {15 \, c^{2} x^{4} + 25 \, b c x^{2} + 8 \, b^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}} - \frac {15 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} \] Input:
integrate(x^4/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
-1/8*(15*c^2*x^4 + 25*b*c*x^2 + 8*b^2)/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x) - 15/8*c*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3)
Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {15 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} - \frac {7 \, c^{2} x^{3} + 9 \, b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{3}} - \frac {1}{b^{3} x} \] Input:
integrate(x^4/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
-15/8*c*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) - 1/8*(7*c^2*x^3 + 9*b*c*x)/ ((c*x^2 + b)^2*b^3) - 1/(b^3*x)
Time = 17.57 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {1}{b}+\frac {25\,c\,x^2}{8\,b^2}+\frac {15\,c^2\,x^4}{8\,b^3}}{b^2\,x+2\,b\,c\,x^3+c^2\,x^5}-\frac {15\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,b^{7/2}} \] Input:
int(x^4/(b*x^2 + c*x^4)^3,x)
Output:
- (1/b + (25*c*x^2)/(8*b^2) + (15*c^2*x^4)/(8*b^3))/(b^2*x + c^2*x^5 + 2*b *c*x^3) - (15*c^(1/2)*atan((c^(1/2)*x)/b^(1/2)))/(8*b^(7/2))
Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.68 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} x -30 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b c \,x^{3}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) c^{2} x^{5}-8 b^{3}-25 b^{2} c \,x^{2}-15 b \,c^{2} x^{4}}{8 b^{4} x \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^4/(c*x^4+b*x^2)^3,x)
Output:
( - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*x - 30*sqrt(c)*s qrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b*c*x**3 - 15*sqrt(c)*sqrt(b)*atan((c *x)/(sqrt(c)*sqrt(b)))*c**2*x**5 - 8*b**3 - 25*b**2*c*x**2 - 15*b*c**2*x** 4)/(8*b**4*x*(b**2 + 2*b*c*x**2 + c**2*x**4))