3.7.53 \(\int \frac {x^4}{a+c x^4} \, dx\) [653]

Optimal. Leaf size=190 \[ \frac {x}{c}+\frac {\sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{a} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} c^{5/4}}+\frac {\sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} c^{5/4}} \]

[Out]

x/c-1/4*a^(1/4)*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/c^(5/4)*2^(1/2)-1/4*a^(1/4)*arctan(1+c^(1/4)*x*2^(1/2)/a^
(1/4))/c^(5/4)*2^(1/2)+1/8*a^(1/4)*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/c^(5/4)*2^(1/2)-1/8*a^(1
/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/c^(5/4)*2^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {327, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\sqrt [4]{a} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{a} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} c^{5/4}}+\frac {\sqrt [4]{a} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} c^{5/4}}+\frac {x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a + c*x^4),x]

[Out]

x/c + (a^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*c^(5/4)) - (a^(1/4)*ArcTan[1 + (Sqrt[2]*c^(
1/4)*x)/a^(1/4)])/(2*Sqrt[2]*c^(5/4)) + (a^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sq
rt[2]*c^(5/4)) - (a^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*c^(5/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 327

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 173, normalized size = 0.91 \begin {gather*} \frac {8 \sqrt [4]{c} x+2 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-2 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+\sqrt {2} \sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )-\sqrt {2} \sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{8 c^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + c*x^4),x]

[Out]

(8*c^(1/4)*x + 2*Sqrt[2]*a^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - 2*Sqrt[2]*a^(1/4)*ArcTan[1 + (Sqrt[
2]*c^(1/4)*x)/a^(1/4)] + Sqrt[2]*a^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] - Sqrt[2]*a^(1
/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(8*c^(5/4))

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Maple [A]
time = 0.13, size = 108, normalized size = 0.57

method	result	size
risch	\(\frac {x}{c}-\frac {a \left (\munderset {\textit {\_R} =\RootOf \left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4 c^{2}}\)	\(34\)
default	\(\frac {x}{c}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 179, normalized size = 0.94 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}}} + \frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{c^{\frac {1}{4}}}}{8 \, c} + \frac {x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/sqrt
(sqrt(a)*sqrt(c)) + 2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*
sqrt(c)))/sqrt(sqrt(a)*sqrt(c)) + sqrt(2)*a^(1/4)*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/c^(1/
4) - sqrt(2)*a^(1/4)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/c^(1/4))/c + x/c

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Fricas [A]
time = 0.40, size = 119, normalized size = 0.63 \begin {gather*} -\frac {4 \, c \left (-\frac {a}{c^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {c^{4} x \left (-\frac {a}{c^{5}}\right )^{\frac {3}{4}} - \sqrt {c^{2} \sqrt {-\frac {a}{c^{5}}} + x^{2}} c^{4} \left (-\frac {a}{c^{5}}\right )^{\frac {3}{4}}}{a}\right ) + c \left (-\frac {a}{c^{5}}\right )^{\frac {1}{4}} \log \left (c \left (-\frac {a}{c^{5}}\right )^{\frac {1}{4}} + x\right ) - c \left (-\frac {a}{c^{5}}\right )^{\frac {1}{4}} \log \left (-c \left (-\frac {a}{c^{5}}\right )^{\frac {1}{4}} + x\right ) - 4 \, x}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.07, size = 22, normalized size = 0.12 \begin {gather*} \operatorname {RootSum} {\left (256 t^{4} c^{5} + a, \left ( t \mapsto t \log {\left (- 4 t c + x \right )} \right )\right )} + \frac {x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(256*_t**4*c**5 + a, Lambda(_t, _t*log(-4*_t*c + x))) + x/c

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Giac [A]
time = 0.77, size = 172, normalized size = 0.91 \begin {gather*} \frac {x}{c} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, c^{2}} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, c^{2}} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, c^{2}} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c - 1/4*sqrt(2)*(a*c^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/c^2 - 1/4*sqrt(2)*
(a*c^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/c^2 - 1/8*sqrt(2)*(a*c^3)^(1/4)*log(
x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/c^2 + 1/8*sqrt(2)*(a*c^3)^(1/4)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqr
t(a/c))/c^2

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Mupad [B]
time = 1.04, size = 48, normalized size = 0.25 \begin {gather*} \frac {x}{c}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{2\,c^{5/4}}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{2\,c^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + c*x^4),x)

[Out]

x/c - ((-a)^(1/4)*atan((c^(1/4)*x)/(-a)^(1/4)))/(2*c^(5/4)) - ((-a)^(1/4)*atanh((c^(1/4)*x)/(-a)^(1/4)))/(2*c^
(5/4))

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