3.7.77 \(\int \frac {x}{(a+c x^4)^3} \, dx\) [677]

Optimal. Leaf size=68 \[ \frac {x^2}{8 a \left (a+c x^4\right )^2}+\frac {3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {281, 205, 211} \begin {gather*} \frac {3 \text {ArcTan}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {c}}+\frac {3 x^2}{16 a^2 \left (a+c x^4\right )}+\frac {x^2}{8 a \left (a+c x^4\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + c*x^4)^3,x]

[Out]

x^2/(8*a*(a + c*x^4)^2) + (3*x^2)/(16*a^2*(a + c*x^4)) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(16*a^(5/2)*Sqrt[c]
)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 58, normalized size = 0.85 \begin {gather*} \frac {1}{16} \left (\frac {5 a x^2+3 c x^6}{a^2 \left (a+c x^4\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{a^{5/2} \sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + c*x^4)^3,x]

[Out]

((5*a*x^2 + 3*c*x^6)/(a^2*(a + c*x^4)^2) + (3*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(a^(5/2)*Sqrt[c]))/16

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Maple [A]
time = 0.18, size = 63, normalized size = 0.93

method	result	size
default	\(\frac {x^{2}}{8 a \left (x^{4} c +a \right )^{2}}+\frac {\frac {3 x^{2}}{16 a \left (x^{4} c +a \right )}+\frac {3 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{16 a \sqrt {a c}}}{a}\)	\(63\)
risch	\(\frac {\frac {3 c \,x^{6}}{16 a^{2}}+\frac {5 x^{2}}{16 a}}{\left (x^{4} c +a \right )^{2}}-\frac {3 \ln \left (x^{2} \sqrt {-a c}-a \right )}{32 \sqrt {-a c}\, a^{2}}+\frac {3 \ln \left (x^{2} \sqrt {-a c}+a \right )}{32 \sqrt {-a c}\, a^{2}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^2/a/(c*x^4+a)^2+3/8/a*(1/2*x^2/a/(c*x^4+a)+1/2/a/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2)))

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Maxima [A]
time = 0.50, size = 62, normalized size = 0.91 \begin {gather*} \frac {3 \, c x^{6} + 5 \, a x^{2}}{16 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )}} + \frac {3 \, \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*c*x^6 + 5*a*x^2)/(a^2*c^2*x^8 + 2*a^3*c*x^4 + a^4) + 3/16*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^2)

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Fricas [A]
time = 0.39, size = 196, normalized size = 2.88 \begin {gather*} \left [\frac {6 \, a c^{2} x^{6} + 10 \, a^{2} c x^{2} - 3 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{4} - 2 \, \sqrt {-a c} x^{2} - a}{c x^{4} + a}\right )}{32 \, {\left (a^{3} c^{3} x^{8} + 2 \, a^{4} c^{2} x^{4} + a^{5} c\right )}}, \frac {3 \, a c^{2} x^{6} + 5 \, a^{2} c x^{2} - 3 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c}}{c x^{2}}\right )}{16 \, {\left (a^{3} c^{3} x^{8} + 2 \, a^{4} c^{2} x^{4} + a^{5} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(6*a*c^2*x^6 + 10*a^2*c*x^2 - 3*(c^2*x^8 + 2*a*c*x^4 + a^2)*sqrt(-a*c)*log((c*x^4 - 2*sqrt(-a*c)*x^2 - a
)/(c*x^4 + a)))/(a^3*c^3*x^8 + 2*a^4*c^2*x^4 + a^5*c), 1/16*(3*a*c^2*x^6 + 5*a^2*c*x^2 - 3*(c^2*x^8 + 2*a*c*x^
4 + a^2)*sqrt(a*c)*arctan(sqrt(a*c)/(c*x^2)))/(a^3*c^3*x^8 + 2*a^4*c^2*x^4 + a^5*c)]

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Sympy [A]
time = 0.22, size = 110, normalized size = 1.62 \begin {gather*} - \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (- a^{3} \sqrt {- \frac {1}{a^{5} c}} + x^{2} \right )}}{32} + \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (a^{3} \sqrt {- \frac {1}{a^{5} c}} + x^{2} \right )}}{32} + \frac {5 a x^{2} + 3 c x^{6}}{16 a^{4} + 32 a^{3} c x^{4} + 16 a^{2} c^{2} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a**5*c))*log(-a**3*sqrt(-1/(a**5*c)) + x**2)/32 + 3*sqrt(-1/(a**5*c))*log(a**3*sqrt(-1/(a**5*c)) +
x**2)/32 + (5*a*x**2 + 3*c*x**6)/(16*a**4 + 32*a**3*c*x**4 + 16*a**2*c**2*x**8)

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Giac [A]
time = 0.56, size = 49, normalized size = 0.72 \begin {gather*} \frac {3 \, \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{2}} + \frac {3 \, c x^{6} + 5 \, a x^{2}}{16 \, {\left (c x^{4} + a\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*a^2) + 1/16*(3*c*x^6 + 5*a*x^2)/((c*x^4 + a)^2*a^2)

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Mupad [B]
time = 1.00, size = 59, normalized size = 0.87 \begin {gather*} \frac {\frac {5\,x^2}{16\,a}+\frac {3\,c\,x^6}{16\,a^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{16\,a^{5/2}\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + c*x^4)^3,x)

[Out]

((5*x^2)/(16*a) + (3*c*x^6)/(16*a^2))/(a^2 + c^2*x^8 + 2*a*c*x^4) + (3*atan((c^(1/2)*x^2)/a^(1/2)))/(16*a^(5/2
)*c^(1/2))

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