Integrand size = 13, antiderivative size = 70 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=-\frac {1}{6 a^2 x^6}+\frac {c}{a^3 x^2}+\frac {c^2 x^2}{4 a^3 \left (a+c x^4\right )}+\frac {5 c^{3/2} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{7/2}} \] Output:
-1/6/a^2/x^6+c/a^3/x^2+1/4*c^2*x^2/a^3/(c*x^4+a)+5/4*c^(3/2)*arctan(c^(1/2 )*x^2/a^(1/2))/a^(7/2)
Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.49 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\frac {\frac {\sqrt {a} \left (-2 a^2+10 a c x^4+15 c^2 x^8\right )}{x^6 \left (a+c x^4\right )}-15 c^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-15 c^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{12 a^{7/2}} \] Input:
Integrate[1/(x^7*(a + c*x^4)^2),x]
Output:
((Sqrt[a]*(-2*a^2 + 10*a*c*x^4 + 15*c^2*x^8))/(x^6*(a + c*x^4)) - 15*c^(3/ 2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - 15*c^(3/2)*ArcTan[1 + (Sqrt[2 ]*c^(1/4)*x)/a^(1/4)])/(12*a^(7/2))
Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {807, 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.
\(\displaystyle \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^8 \left (c x^4+a\right )^2}dx^2\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \int \frac {1}{x^8 \left (c x^4+a\right )}dx^2}{2 a}+\frac {1}{2 a x^6 \left (a+c x^4\right )}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (-\frac {c \int \frac {1}{x^4 \left (c x^4+a\right )}dx^2}{a}-\frac {1}{3 a x^6}\right )}{2 a}+\frac {1}{2 a x^6 \left (a+c x^4\right )}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^4+a}dx^2}{a}-\frac {1}{a x^2}\right )}{a}-\frac {1}{3 a x^6}\right )}{2 a}+\frac {1}{2 a x^6 \left (a+c x^4\right )}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x^2}\right )}{a}-\frac {1}{3 a x^6}\right )}{2 a}+\frac {1}{2 a x^6 \left (a+c x^4\right )}\right )\) |
Input:
Int[1/(x^7*(a + c*x^4)^2),x]
Output:
(1/(2*a*x^6*(a + c*x^4)) + (5*(-1/3*1/(a*x^6) - (c*(-(1/(a*x^2)) - (Sqrt[c ]*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/a^(3/2)))/a))/(2*a))/2
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Time = 0.48 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {1}{6 a^{2} x^{6}}+\frac {c}{a^{3} x^{2}}+\frac {c^{2} \left (\frac {x^{2}}{2 c \,x^{4}+2 a}+\frac {5 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \sqrt {a c}}\right )}{2 a^{3}}\) | \(59\) |
risch | \(\frac {\frac {5 c^{2} x^{8}}{4 a^{3}}+\frac {5 c \,x^{4}}{6 a^{2}}-\frac {1}{6 a}}{x^{6} \left (c \,x^{4}+a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2}+c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (5 a^{7} \textit {\_R}^{2}+4 c^{3}\right ) x^{2}-a^{4} c \textit {\_R} \right )\right )}{8}\) | \(87\) |
Input:
int(1/x^7/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/6/a^2/x^6+c/a^3/x^2+1/2*c^2/a^3*(1/2*x^2/(c*x^4+a)+5/2/(a*c)^(1/2)*arct an(c*x^2/(a*c)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.51 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\left [\frac {30 \, c^{2} x^{8} + 20 \, a c x^{4} + 15 \, {\left (c^{2} x^{10} + a c x^{6}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} + 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) - 4 \, a^{2}}{24 \, {\left (a^{3} c x^{10} + a^{4} x^{6}\right )}}, \frac {15 \, c^{2} x^{8} + 10 \, a c x^{4} + 15 \, {\left (c^{2} x^{10} + a c x^{6}\right )} \sqrt {\frac {c}{a}} \arctan \left (x^{2} \sqrt {\frac {c}{a}}\right ) - 2 \, a^{2}}{12 \, {\left (a^{3} c x^{10} + a^{4} x^{6}\right )}}\right ] \] Input:
integrate(1/x^7/(c*x^4+a)^2,x, algorithm="fricas")
Output:
[1/24*(30*c^2*x^8 + 20*a*c*x^4 + 15*(c^2*x^10 + a*c*x^6)*sqrt(-c/a)*log((c *x^4 + 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) - 4*a^2)/(a^3*c*x^10 + a^4*x^6 ), 1/12*(15*c^2*x^8 + 10*a*c*x^4 + 15*(c^2*x^10 + a*c*x^6)*sqrt(c/a)*arcta n(x^2*sqrt(c/a)) - 2*a^2)/(a^3*c*x^10 + a^4*x^6)]
Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=- \frac {5 \sqrt {- \frac {c^{3}}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {c^{3}}{a^{7}}}}{c^{2}} + x^{2} \right )}}{8} + \frac {5 \sqrt {- \frac {c^{3}}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {c^{3}}{a^{7}}}}{c^{2}} + x^{2} \right )}}{8} + \frac {- 2 a^{2} + 10 a c x^{4} + 15 c^{2} x^{8}}{12 a^{4} x^{6} + 12 a^{3} c x^{10}} \] Input:
integrate(1/x**7/(c*x**4+a)**2,x)
Output:
-5*sqrt(-c**3/a**7)*log(-a**4*sqrt(-c**3/a**7)/c**2 + x**2)/8 + 5*sqrt(-c* *3/a**7)*log(a**4*sqrt(-c**3/a**7)/c**2 + x**2)/8 + (-2*a**2 + 10*a*c*x**4 + 15*c**2*x**8)/(12*a**4*x**6 + 12*a**3*c*x**10)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\frac {15 \, c^{2} x^{8} + 10 \, a c x^{4} - 2 \, a^{2}}{12 \, {\left (a^{3} c x^{10} + a^{4} x^{6}\right )}} + \frac {5 \, c^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} a^{3}} \] Input:
integrate(1/x^7/(c*x^4+a)^2,x, algorithm="maxima")
Output:
1/12*(15*c^2*x^8 + 10*a*c*x^4 - 2*a^2)/(a^3*c*x^10 + a^4*x^6) + 5/4*c^2*ar ctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^3)
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\frac {c^{2} x^{2}}{4 \, {\left (c x^{4} + a\right )} a^{3}} + \frac {5 \, c^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} a^{3}} + \frac {6 \, c x^{4} - a}{6 \, a^{3} x^{6}} \] Input:
integrate(1/x^7/(c*x^4+a)^2,x, algorithm="giac")
Output:
1/4*c^2*x^2/((c*x^4 + a)*a^3) + 5/4*c^2*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c) *a^3) + 1/6*(6*c*x^4 - a)/(a^3*x^6)
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\frac {\frac {5\,c\,x^4}{6\,a^2}-\frac {1}{6\,a}+\frac {5\,c^2\,x^8}{4\,a^3}}{c\,x^{10}+a\,x^6}+\frac {5\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{4\,a^{7/2}} \] Input:
int(1/(x^7*(a + c*x^4)^2),x)
Output:
((5*c*x^4)/(6*a^2) - 1/(6*a) + (5*c^2*x^8)/(4*a^3))/(a*x^6 + c*x^10) + (5* c^(3/2)*atan((c^(1/2)*x^2)/a^(1/2)))/(4*a^(7/2))
Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.79 \[ \int \frac {1}{x^7 \left (a+c x^4\right )^2} \, dx=\frac {-15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a c \,x^{6}-15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{2} x^{10}-15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a c \,x^{6}-15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{2} x^{10}-2 a^{3}+10 a^{2} c \,x^{4}+15 a \,c^{2} x^{8}}{12 a^{4} x^{6} \left (c \,x^{4}+a \right )} \] Input:
int(1/x^7/(c*x^4+a)^2,x)
Output:
( - 15*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c** (1/4)*a**(1/4)*sqrt(2)))*a*c*x**6 - 15*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**( 1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**2*x**10 - 15*s qrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a* *(1/4)*sqrt(2)))*a*c*x**6 - 15*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqr t(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**2*x**10 - 2*a**3 + 10* a**2*c*x**4 + 15*a*c**2*x**8)/(12*a**4*x**6*(a + c*x**4))