3.12.56 \(\int \frac {x^4}{(a+b x^4)^{5/4}} \, dx\) [1156]

Optimal. Leaf size=74 \[ -\frac {x}{b \sqrt [4]{a+b x^4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}} \]

[Out]

-x/b/(b*x^4+a)^(1/4)+1/2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)+1/2*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 = 74, 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 = {294, 246, 218, 212, 209} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}}-\frac {x}{b \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^4)^(5/4),x]

[Out]

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

Rule 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]
time = 0.26, size = 67, normalized size = 0.91 \begin {gather*} \frac {-\frac {2 \sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}+\tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^4)^(5/4),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^4+a)^(5/4),x)

[Out]

int(x^4/(b*x^4+a)^(5/4),x)

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Maxima [A]
time = 0.50, size = 88, normalized size = 1.19 \begin {gather*} -\frac {\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}}}}{4 \, b} - \frac {x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

-1/4*(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 - x/((b*x^4 + a)^(1/4)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (56) = 112\).
time = 0.39, size = 193, normalized size = 2.61 \begin {gather*} \frac {4 \, {\left (b^{2} x^{4} + a b\right )} \frac {1}{b^{5}}^{\frac {1}{4}} \arctan \left (\frac {b \frac {1}{b^{5}}^{\frac {1}{4}} x \sqrt {\frac {b^{3} \sqrt {\frac {1}{b^{5}}} x^{2} + \sqrt {b x^{4} + a}}{x^{2}}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} b \frac {1}{b^{5}}^{\frac {1}{4}}}{x}\right ) + {\left (b^{2} x^{4} + a b\right )} \frac {1}{b^{5}}^{\frac {1}{4}} \log \left (\frac {b^{4} \frac {1}{b^{5}}^{\frac {3}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) - {\left (b^{2} x^{4} + a b\right )} \frac {1}{b^{5}}^{\frac {1}{4}} \log \left (-\frac {b^{4} \frac {1}{b^{5}}^{\frac {3}{4}} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} x}{4 \, {\left (b^{2} x^{4} + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.51, size = 37, normalized size = 0.50 \begin {gather*} \frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**4+a)**(5/4),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^4/(b*x^4 + a)^(5/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^4)^(5/4),x)

[Out]

int(x^4/(a + b*x^4)^(5/4), x)

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