3.11.95 \(\int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx\) [1095]

Optimal. Leaf size=104 \[ -\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {5 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}+\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}} \]

[Out]

-5/32*a*x*(b*x^4+a)^(3/4)/b^2+1/8*x^5*(b*x^4+a)^(3/4)/b+5/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)+5/6
4*a^2*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 = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {327, 246, 218, 212, 209} \begin {gather*} \frac {5 a^2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}+\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^4)^(1/4),x]

[Out]

(-5*a*x*(a + b*x^4)^(3/4))/(32*b^2) + (x^5*(a + b*x^4)^(3/4))/(8*b) + (5*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1
/4)])/(64*b^(9/4)) + (5*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*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 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}{\sqrt [4]{a+b x^4}} \, dx &=\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {(5 a) \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx}{8 b}\\ &=-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {\left (5 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{32 b^2}\\ &=-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{32 b^2}\\ &=-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^2}+\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^2}\\ &=-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {5 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}+\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 87, normalized size = 0.84 \begin {gather*} \frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4} \left (-5 a+4 b x^4\right )+5 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+5 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^4)^(1/4),x]

[Out]

(2*b^(1/4)*x*(a + b*x^4)^(3/4)*(-5*a + 4*b*x^4) + 5*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + 5*a^2*ArcTanh[
(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(9/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^4+a)^(1/4),x)

[Out]

int(x^8/(b*x^4+a)^(1/4),x)

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Maxima [A]
time = 0.51, size = 151, normalized size = 1.45 \begin {gather*} -\frac {5 \, a^{2} {\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 )}}{128 \, b^{2}} + \frac {\frac {9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b}{x^{3}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{x^{7}}}{32 \, {\left (b^{4} - \frac {2 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

-5/128*a^2*(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 + 1/32*(9*(b*x^4 + a)^(3/4)*a^2*b/x^3 - 5*(b*x^4 + a)^(7/4)*a^2/x^7)/(b^4 -
2*(b*x^4 + a)*b^3/x^4 + (b*x^4 + a)^2*b^2/x^8)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (80) = 160\).
time = 0.40, size = 228, normalized size = 2.19 \begin {gather*} \frac {20 \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} - b^{2} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \sqrt {\frac {a^{8} b^{5} x^{2} \sqrt {\frac {a^{8}}{b^{9}}} + \sqrt {b x^{4} + a} a^{12}}{x^{2}}}}{a^{8} x}\right ) + 5 \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {125 \, {\left (b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 5 \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) + 4 \, {\left (4 \, b x^{5} - 5 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

1/128*(20*b^2*(a^8/b^9)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*a^6*b^2*(a^8/b^9)^(1/4) - b^2*x*(a^8/b^9)^(1/4)*sqrt(
(a^8*b^5*x^2*sqrt(a^8/b^9) + sqrt(b*x^4 + a)*a^12)/x^2))/(a^8*x)) + 5*b^2*(a^8/b^9)^(1/4)*log(125*(b^7*x*(a^8/
b^9)^(3/4) + (b*x^4 + a)^(1/4)*a^6)/x) - 5*b^2*(a^8/b^9)^(1/4)*log(-125*(b^7*x*(a^8/b^9)^(3/4) - (b*x^4 + a)^(
1/4)*a^6)/x) + 4*(4*b*x^5 - 5*a*x)*(b*x^4 + a)^(3/4))/b^2

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Sympy [C] Result contains complex when optimal does not.
time = 1.49, size = 37, normalized size = 0.36 \begin {gather*} \frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \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)**(1/4),x)

[Out]

x**9*gamma(9/4)*hyper((1/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/4)*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)^(1/4),x, algorithm="giac")

[Out]

integrate(x^8/(b*x^4 + a)^(1/4), 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 )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^4)^(1/4),x)

[Out]

int(x^8/(a + b*x^4)^(1/4), x)

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