3.12.55 \(\int \frac {x^8}{(a+b x^4)^{5/4}} \, dx\) [1155]

Optimal. Leaf size=97 \[ -\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {5 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \]

[Out]

-x^5/b/(b*x^4+a)^(1/4)+5/4*x*(b*x^4+a)^(3/4)/b^2-5/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)-5/8*a*arctanh
(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)

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Rubi [A]
time = 0.02, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {294, 327, 246, 218, 212, 209} \begin {gather*} -\frac {5 a \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {x^5}{b \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^4)^(5/4),x]

[Out]

-(x^5/(b*(a + b*x^4)^(1/4))) + (5*x*(a + b*x^4)^(3/4))/(4*b^2) - (5*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(
8*b^(9/4)) - (5*a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(9/4))

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]

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]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx &=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx}{b}\\ &=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {(5 a) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{4 b^2}\\ &=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {(5 a) \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 b^2}\\ &=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {(5 a) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 b^2}-\frac {(5 a) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 b^2}\\ &=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {5 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 82, normalized size = 0.85 \begin {gather*} \frac {\frac {2 \sqrt [4]{b} x \left (5 a+b x^4\right )}{\sqrt [4]{a+b x^4}}-5 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-5 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^4)^(5/4),x]

[Out]

((2*b^(1/4)*x*(5*a + b*x^4))/(a + b*x^4)^(1/4) - 5*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - 5*a*ArcTanh[(b^(1
/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(9/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^4+a)^(5/4),x)

[Out]

int(x^8/(b*x^4+a)^(5/4),x)

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Maxima [A]
time = 0.50, size = 130, normalized size = 1.34 \begin {gather*} \frac {4 \, a b - \frac {5 \, {\left (b x^{4} + a\right )} a}{x^{4}}}{4 \, {\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}}{x^{5}}\right )}} + \frac {5 \, 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^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (75) = 150\).
time = 0.38, size = 269, normalized size = 2.77 \begin {gather*} -\frac {20 \, {\left (b^{3} x^{4} + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} b^{2} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} - b^{2} x \sqrt {\frac {a^{4} b^{5} x^{2} \sqrt {\frac {a^{4}}{b^{9}}} + \sqrt {b x^{4} + a} a^{6}}{x^{2}}} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}}}{a^{4} x}\right ) + 5 \, {\left (b^{3} x^{4} + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {125 \, {\left (b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) - 5 \, {\left (b^{3} x^{4} + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) - 4 \, {\left (b x^{5} + 5 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, {\left (b^{3} x^{4} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

-1/16*(20*(b^3*x^4 + a*b^2)*(a^4/b^9)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*a^3*b^2*(a^4/b^9)^(1/4) - b^2*x*sqrt((a
^4*b^5*x^2*sqrt(a^4/b^9) + sqrt(b*x^4 + a)*a^6)/x^2)*(a^4/b^9)^(1/4))/(a^4*x)) + 5*(b^3*x^4 + a*b^2)*(a^4/b^9)
^(1/4)*log(125*(b^7*x*(a^4/b^9)^(3/4) + (b*x^4 + a)^(1/4)*a^3)/x) - 5*(b^3*x^4 + a*b^2)*(a^4/b^9)^(1/4)*log(-1
25*(b^7*x*(a^4/b^9)^(3/4) - (b*x^4 + a)^(1/4)*a^3)/x) - 4*(b*x^5 + 5*a*x)*(b*x^4 + a)^(3/4))/(b^3*x^4 + a*b^2)

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Sympy [C] Result contains complex when optimal does not.
time = 0.85, size = 37, normalized size = 0.38 \begin {gather*} \frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**4+a)**(5/4),x)

[Out]

x**9*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(13/4))

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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^8/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^8/(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^8}{{\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^8/(a + b*x^4)^(5/4),x)

[Out]

int(x^8/(a + b*x^4)^(5/4), x)

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