\(\int \frac {1}{x^3 (a+c x^4)^3} \, dx\) [679]

Optimal result

Integrand size = 13, antiderivative size = 78 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {15 \sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}} \]

[Out]

-15/16/a^3/x^2+1/8/a/x^2/(c*x^4+a)^2+5/16/a^2/x^2/(c*x^4+a)-15/16*arctan(x^2*c^(1/2)/a^(1/2))*c^(1/2)/a^(7/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331, 211} \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15 \sqrt {c} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}}-\frac {15}{16 a^3 x^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2} \]

[In]

Int[1/(x^3*(a + c*x^4)^3),x]

[Out]

-15/(16*a^3*x^2) + 1/(8*a*x^2*(a + c*x^4)^2) + 5/(16*a^2*x^2*(a + c*x^4)) - (15*Sqrt[c]*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(16*a^(7/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )^3} \, dx,x,x^2\right ) \\ & = \frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5 \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )^2} \, dx,x,x^2\right )}{8 a} \\ & = \frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}+\frac {15 \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )} \, dx,x,x^2\right )}{16 a^2} \\ & = -\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {(15 c) \text {Subst}\left (\int \frac {1}{a+c x^2} \, dx,x,x^2\right )}{16 a^3} \\ & = -\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\frac {-\frac {\sqrt {a} \left (8 a^2+25 a c x^4+15 c^2 x^8\right )}{x^2 \left (a+c x^4\right )^2}+15 \sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+15 \sqrt {c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{16 a^{7/2}} \]

[In]

Integrate[1/(x^3*(a + c*x^4)^3),x]

[Out]

(-((Sqrt[a]*(8*a^2 + 25*a*c*x^4 + 15*c^2*x^8))/(x^2*(a + c*x^4)^2)) + 15*Sqrt[c]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x
)/a^(1/4)] + 15*Sqrt[c]*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(16*a^(7/2))

Maple [A] (verified)

Time = 3.99 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {1}{2 a^{3} x^{2}}-\frac {c \left (\frac {\frac {7}{8} c \,x^{6}+\frac {9}{8} a \,x^{2}}{\left (x^{4} c +a \right )^{2}}+\frac {15 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{8 \sqrt {a c}}\right )}{2 a^{3}}\)	\(58\)
risch	\(\frac {-\frac {15 c^{2} x^{8}}{16 a^{3}}-\frac {25 c \,x^{4}}{16 a^{2}}-\frac {1}{2 a}}{x^{2} \left (x^{4} c +a \right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (-5 \textit {\_R}^{2} a^{7}-4 c \right ) x^{2}-a^{4} \textit {\_R} \right )\right )}{32}\)	\(82\)

[In]

int(1/x^3/(c*x^4+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/a^3/x^2-1/2*c/a^3*((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)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.79 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\left [-\frac {30 \, c^{2} x^{8} + 50 \, a c x^{4} - 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + 16 \, a^{2}}{32 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}, -\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} - 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}\right ] \]

[In]

integrate(1/x^3/(c*x^4+a)^3,x, algorithm="fricas")

[Out]

[-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(a*sqrt(c/a)/(c*x^2)) + 8*a^2)/(a^3*c^2*x^10 + 2*a^4*c*x^6 + a^5
*x^2)]

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=\frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} - \frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} + \frac {- 8 a^{2} - 25 a c x^{4} - 15 c^{2} x^{8}}{16 a^{5} x^{2} + 32 a^{4} c x^{6} + 16 a^{3} c^{2} x^{10}} \]

[In]

integrate(1/x**3/(c*x**4+a)**3,x)

[Out]

15*sqrt(-c/a**7)*log(-a**4*sqrt(-c/a**7)/c + x**2)/32 - 15*sqrt(-c/a**7)*log(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)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}} - \frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} \]

[In]

integrate(1/x^3/(c*x^4+a)^3,x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} - \frac {7 \, c^{2} x^{6} + 9 \, a c x^{2}}{16 \, {\left (c x^{4} + a\right )}^{2} a^{3}} - \frac {1}{2 \, a^{3} x^{2}} \]

[In]

integrate(1/x^3/(c*x^4+a)^3,x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx=-\frac {\frac {1}{2\,a}+\frac {25\,c\,x^4}{16\,a^2}+\frac {15\,c^2\,x^8}{16\,a^3}}{a^2\,x^2+2\,a\,c\,x^6+c^2\,x^{10}}-\frac {15\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{16\,a^{7/2}} \]

[In]

int(1/(x^3*(a + c*x^4)^3),x)

[Out]

- (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))