Integrand size = 17, antiderivative size = 87 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {35}{24 b^3 x^3}+\frac {35 c}{8 b^4 x}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}+\frac {7}{8 b^2 x^3 \left (b+c x^2\right )}+\frac {35 c^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}} \]
-35/24/b^3/x^3+35/8*c/b^4/x+1/4/b/x^3/(c*x^2+b)^2+7/8/b^2/x^3/(c*x^2+b)+35 /8*c^(3/2)*arctan(x*c^(1/2)/b^(1/2))/b^(9/2)
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-8 b^3+56 b^2 c x^2+175 b c^2 x^4+105 c^3 x^6}{24 b^4 x^3 \left (b+c x^2\right )^2}+\frac {35 c^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}} \]
(-8*b^3 + 56*b^2*c*x^2 + 175*b*c^2*x^4 + 105*c^3*x^6)/(24*b^4*x^3*(b + c*x ^2)^2) + (35*c^(3/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))
Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {9, 253, 253, 264, 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.
\(\displaystyle \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^4 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {7 \int \frac {1}{x^4 \left (c x^2+b\right )^2}dx}{4 b}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{x^4 \left (c x^2+b\right )}dx}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {c \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^3 \left (b+c x^2\right )^2}\) |
1/(4*b*x^3*(b + c*x^2)^2) + (7*(1/(2*b*x^3*(b + c*x^2)) + (5*(-1/3*1/(b*x^ 3) - (c*(-(1/(b*x)) - (Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2)))/b))/ (2*b)))/(4*b)
3.3.17.3.1 Defintions of rubi rules used
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 = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {1}{3 b^{3} x^{3}}+\frac {3 c}{b^{4} x}+\frac {c^{2} \left (\frac {\frac {11}{8} c \,x^{3}+\frac {13}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {35 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{4}}\) | \(64\) |
risch | \(\frac {\frac {35 c^{3} x^{6}}{8 b^{4}}+\frac {175 c^{2} x^{4}}{24 b^{3}}+\frac {7 c \,x^{2}}{3 b^{2}}-\frac {1}{3 b}}{\left (c \,x^{2}+b \right )^{2} x^{3}}+\frac {35 \sqrt {-b c}\, c \ln \left (-c x -\sqrt {-b c}\right )}{16 b^{5}}-\frac {35 \sqrt {-b c}\, c \ln \left (-c x +\sqrt {-b c}\right )}{16 b^{5}}\) | \(102\) |
-1/3/b^3/x^3+3*c/b^4/x+1/b^4*c^2*((11/8*c*x^3+13/8*b*x)/(c*x^2+b)^2+35/8/( b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.74 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {210 \, c^{3} x^{6} + 350 \, b c^{2} x^{4} + 112 \, b^{2} c x^{2} - 16 \, b^{3} + 105 \, {\left (c^{3} x^{7} + 2 \, b c^{2} x^{5} + b^{2} c x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{48 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}, \frac {105 \, c^{3} x^{6} + 175 \, b c^{2} x^{4} + 56 \, b^{2} c x^{2} - 8 \, b^{3} + 105 \, {\left (c^{3} x^{7} + 2 \, b c^{2} x^{5} + b^{2} c x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}\right ] \]
[1/48*(210*c^3*x^6 + 350*b*c^2*x^4 + 112*b^2*c*x^2 - 16*b^3 + 105*(c^3*x^7 + 2*b*c^2*x^5 + b^2*c*x^3)*sqrt(-c/b)*log((c*x^2 + 2*b*x*sqrt(-c/b) - b)/ (c*x^2 + b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3), 1/24*(105*c^3*x^6 + 1 75*b*c^2*x^4 + 56*b^2*c*x^2 - 8*b^3 + 105*(c^3*x^7 + 2*b*c^2*x^5 + b^2*c*x ^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)]
Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.59 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {35 \sqrt {- \frac {c^{3}}{b^{9}}} \log {\left (- \frac {b^{5} \sqrt {- \frac {c^{3}}{b^{9}}}}{c^{2}} + x \right )}}{16} + \frac {35 \sqrt {- \frac {c^{3}}{b^{9}}} \log {\left (\frac {b^{5} \sqrt {- \frac {c^{3}}{b^{9}}}}{c^{2}} + x \right )}}{16} + \frac {- 8 b^{3} + 56 b^{2} c x^{2} + 175 b c^{2} x^{4} + 105 c^{3} x^{6}}{24 b^{6} x^{3} + 48 b^{5} c x^{5} + 24 b^{4} c^{2} x^{7}} \]
-35*sqrt(-c**3/b**9)*log(-b**5*sqrt(-c**3/b**9)/c**2 + x)/16 + 35*sqrt(-c* *3/b**9)*log(b**5*sqrt(-c**3/b**9)/c**2 + x)/16 + (-8*b**3 + 56*b**2*c*x** 2 + 175*b*c**2*x**4 + 105*c**3*x**6)/(24*b**6*x**3 + 48*b**5*c*x**5 + 24*b **4*c**2*x**7)
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {105 \, c^{3} x^{6} + 175 \, b c^{2} x^{4} + 56 \, b^{2} c x^{2} - 8 \, b^{3}}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}} + \frac {35 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} \]
1/24*(105*c^3*x^6 + 175*b*c^2*x^4 + 56*b^2*c*x^2 - 8*b^3)/(b^4*c^2*x^7 + 2 *b^5*c*x^5 + b^6*x^3) + 35/8*c^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4)
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {35 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} + \frac {11 \, c^{3} x^{3} + 13 \, b c^{2} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{4}} + \frac {9 \, c x^{2} - b}{3 \, b^{4} x^{3}} \]
35/8*c^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) + 1/8*(11*c^3*x^3 + 13*b*c^ 2*x)/((c*x^2 + b)^2*b^4) + 1/3*(9*c*x^2 - b)/(b^4*x^3)
Time = 12.91 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {7\,c\,x^2}{3\,b^2}-\frac {1}{3\,b}+\frac {175\,c^2\,x^4}{24\,b^3}+\frac {35\,c^3\,x^6}{8\,b^4}}{b^2\,x^3+2\,b\,c\,x^5+c^2\,x^7}+\frac {35\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,b^{9/2}} \]