3.1096 \(\int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=78 \[ -\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}}+\frac {x \left (a+b x^4\right )^{3/4}}{4 b} \]

[Out]

1/4*x*(b*x^4+a)^(3/4)/b-1/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)-1/8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4
))/b^(5/4)

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Rubi [A]  time = 0.02, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 240, 212, 206, 203} \[ -\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}}+\frac {x \left (a+b x^4\right )^{3/4}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^4)^(1/4),x]

[Out]

(x*(a + b*x^4)^(3/4))/(4*b) - (a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(5/4)) - (a*ArcTanh[(b^(1/4)*x)/(
a + b*x^4)^(1/4)])/(8*b^(5/4))

Rule 203

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

Rule 206

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

Rule 212

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

Rule 240

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

Rule 321

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 73, normalized size = 0.94 \[ \frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4}-a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^4)^(1/4),x]

[Out]

(2*b^(1/4)*x*(a + b*x^4)^(3/4) - a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(
1/4)])/(8*b^(5/4))

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fricas [B]  time = 1.06, size = 206, normalized size = 2.64 \[ -\frac {4 \, b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} - b x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \sqrt {\frac {a^{4} b^{3} x^{2} \sqrt {\frac {a^{4}}{b^{5}}} + \sqrt {b x^{4} + a} a^{6}}{x^{2}}}}{a^{4} x}\right ) + b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} x}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^4 + a)^(1/4), x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^4+a)^(1/4),x)

[Out]

int(x^4/(b*x^4+a)^(1/4),x)

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maxima [A]  time = 2.93, size = 108, normalized size = 1.38 \[ \frac {a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{16 \, b} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} + a\right )} b}{x^{4}}\right )} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x
^4 + a)^(1/4)/x))/b^(1/4))/b - 1/4*(b*x^4 + a)^(3/4)*a/((b^2 - (b*x^4 + a)*b/x^4)*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^4)^(1/4),x)

[Out]

int(x^4/(a + b*x^4)^(1/4), x)

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sympy [C]  time = 1.55, size = 37, normalized size = 0.47 \[ \frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**4+a)**(1/4),x)

[Out]

x**5*gamma(5/4)*hyper((1/4, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/4)*gamma(9/4))

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