Integrand size = 13, antiderivative size = 80 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {1}{2 a^3 x^2}-\frac {c x^2}{8 a^2 \left (a+c x^4\right )^2}-\frac {7 c x^2}{16 a^3 \left (a+c x^4\right )}-\frac {15 \sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}} \] Output:
-1/2/a^3/x^2-1/8*c*x^2/a^2/(c*x^4+a)^2-7/16*c*x^2/a^3/(c*x^4+a)-15/16*c^(1 /2)*arctan(c^(1/2)*x^2/a^(1/2))/a^(7/2)
Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\frac {-\frac {\sqrt {a} \left (8 a^2+25 a c x^4+15 c^2 x^8\right )}{x^2 \left (a+c x^4\right )^2}+15 \sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+15 \sqrt {c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{16 a^{7/2}} \] Input:
Integrate[1/(x^3*(a + c*x^4)^3),x]
Output:
(-((Sqrt[a]*(8*a^2 + 25*a*c*x^4 + 15*c^2*x^8))/(x^2*(a + c*x^4)^2)) + 15*S qrt[c]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 15*Sqrt[c]*ArcTan[1 + (Sq rt[2]*c^(1/4)*x)/a^(1/4)])/(16*a^(7/2))
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {807, 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.
\(\displaystyle \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (c x^4+a\right )^3}dx^2\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \int \frac {1}{x^4 \left (c x^4+a\right )^2}dx^2}{4 a}+\frac {1}{4 a x^2 \left (a+c x^4\right )^2}\right )\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (\frac {3 \int \frac {1}{x^4 \left (c x^4+a\right )}dx^2}{2 a}+\frac {1}{2 a x^2 \left (a+c x^4\right )}\right )}{4 a}+\frac {1}{4 a x^2 \left (a+c x^4\right )^2}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (\frac {3 \left (-\frac {c \int \frac {1}{c x^4+a}dx^2}{a}-\frac {1}{a x^2}\right )}{2 a}+\frac {1}{2 a x^2 \left (a+c x^4\right )}\right )}{4 a}+\frac {1}{4 a x^2 \left (a+c x^4\right )^2}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (\frac {3 \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x^2}\right )}{2 a}+\frac {1}{2 a x^2 \left (a+c x^4\right )}\right )}{4 a}+\frac {1}{4 a x^2 \left (a+c x^4\right )^2}\right )\) |
Input:
Int[1/(x^3*(a + c*x^4)^3),x]
Output:
(1/(4*a*x^2*(a + c*x^4)^2) + (5*(1/(2*a*x^2*(a + c*x^4)) + (3*(-(1/(a*x^2) ) - (Sqrt[c]*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*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 = 0.50 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72
method | result | size |
default | \(-\frac {1}{2 a^{3} x^{2}}-\frac {c \left (\frac {\frac {7}{8} c \,x^{6}+\frac {9}{8} a \,x^{2}}{\left (c \,x^{4}+a \right )^{2}}+\frac {15 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{8 \sqrt {a c}}\right )}{2 a^{3}}\) | \(58\) |
risch | \(\frac {-\frac {15 c^{2} x^{8}}{16 a^{3}}-\frac {25 c \,x^{4}}{16 a^{2}}-\frac {1}{2 a}}{x^{2} \left (c \,x^{4}+a \right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (-5 a^{7} \textit {\_R}^{2}-4 c \right ) x^{2}-a^{4} \textit {\_R} \right )\right )}{32}\) | \(82\) |
Input:
int(1/x^3/(c*x^4+a)^3,x,method=_RETURNVERBOSE)
Output:
-1/2/a^3/x^2-1/2/a^3*c*((7/8*c*x^6+9/8*a*x^2)/(c*x^4+a)^2+15/8/(a*c)^(1/2) *arctan(c*x^2/(a*c)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.68 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\left [-\frac {30 \, c^{2} x^{8} + 50 \, a c x^{4} - 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + 16 \, a^{2}}{32 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}, -\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} + 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (x^{2} \sqrt {\frac {c}{a}}\right ) + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}\right ] \] Input:
integrate(1/x^3/(c*x^4+a)^3,x, algorithm="fricas")
Output:
[-1/32*(30*c^2*x^8 + 50*a*c*x^4 - 15*(c^2*x^10 + 2*a*c*x^6 + a^2*x^2)*sqrt (-c/a)*log((c*x^4 - 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) + 16*a^2)/(a^3*c^ 2*x^10 + 2*a^4*c*x^6 + a^5*x^2), -1/16*(15*c^2*x^8 + 25*a*c*x^4 + 15*(c^2* x^10 + 2*a*c*x^6 + a^2*x^2)*sqrt(c/a)*arctan(x^2*sqrt(c/a)) + 8*a^2)/(a^3* c^2*x^10 + 2*a^4*c*x^6 + a^5*x^2)]
Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.51 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} - \frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} + \frac {- 8 a^{2} - 25 a c x^{4} - 15 c^{2} x^{8}}{16 a^{5} x^{2} + 32 a^{4} c x^{6} + 16 a^{3} c^{2} x^{10}} \] Input:
integrate(1/x**3/(c*x**4+a)**3,x)
Output:
15*sqrt(-c/a**7)*log(-a**4*sqrt(-c/a**7)/c + x**2)/32 - 15*sqrt(-c/a**7)*l og(a**4*sqrt(-c/a**7)/c + x**2)/32 + (-8*a**2 - 25*a*c*x**4 - 15*c**2*x**8 )/(16*a**5*x**2 + 32*a**4*c*x**6 + 16*a**3*c**2*x**10)
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}} - \frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} \] Input:
integrate(1/x^3/(c*x^4+a)^3,x, algorithm="maxima")
Output:
-1/16*(15*c^2*x^8 + 25*a*c*x^4 + 8*a^2)/(a^3*c^2*x^10 + 2*a^4*c*x^6 + a^5* x^2) - 15/16*c*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^3)
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} - \frac {7 \, c^{2} x^{6} + 9 \, a c x^{2}}{16 \, {\left (c x^{4} + a\right )}^{2} a^{3}} - \frac {1}{2 \, a^{3} x^{2}} \] Input:
integrate(1/x^3/(c*x^4+a)^3,x, algorithm="giac")
Output:
-15/16*c*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^3) - 1/16*(7*c^2*x^6 + 9*a*c *x^2)/((c*x^4 + a)^2*a^3) - 1/2/(a^3*x^2)
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {\frac {1}{2\,a}+\frac {25\,c\,x^4}{16\,a^2}+\frac {15\,c^2\,x^8}{16\,a^3}}{a^2\,x^2+2\,a\,c\,x^6+c^2\,x^{10}}-\frac {15\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{16\,a^{7/2}} \] Input:
int(1/(x^3*(a + c*x^4)^3),x)
Output:
- (1/(2*a) + (25*c*x^4)/(16*a^2) + (15*c^2*x^8)/(16*a^3))/(a^2*x^2 + c^2*x ^10 + 2*a*c*x^6) - (15*c^(1/2)*atan((c^(1/2)*x^2)/a^(1/2)))/(16*a^(7/2))
Time = 0.22 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.55 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\frac {15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a^{2} x^{2}+30 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a c \,x^{6}+15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{2} x^{10}+15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a^{2} x^{2}+30 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a c \,x^{6}+15 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{2} x^{10}-8 a^{3}-25 a^{2} c \,x^{4}-15 a \,c^{2} x^{8}}{16 a^{4} x^{2} \left (c^{2} x^{8}+2 a c \,x^{4}+a^{2}\right )} \] Input:
int(1/x^3/(c*x^4+a)^3,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**2*x**2 + 30*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*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**2*x**2 + 30*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*sq rt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**2*x**10 - 8*a**3 - 25*a**2*c*x**4 - 15*a*c**2*x**8)/(16*a**4*x**2*(a**2 + 2*a*c*x**4 + c**2*x**8))