3.1155 \(\int \frac{x^8}{(a+b x^4)^{5/4}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0311875, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {288, 321, 240, 212, 206, 203} \[ \frac{5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac{5 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac{x^5}{b \sqrt [4]{a+b x^4}} \]

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 288

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 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]

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 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 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 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])

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) \operatorname{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) \operatorname{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) \operatorname{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*}

Mathematica [C]  time = 0.0139504, size = 60, normalized size = 0.62 \[ \frac{x^5-x^5 \sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{5}{4},\frac{5}{4};\frac{9}{4};-\frac{b x^4}{a}\right )}{4 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^4)^(5/4),x]

[Out]

(x^5 - x^5*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((b*x^4)/a)])/(4*b*(a + b*x^4)^(1/4))

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{{x}^{8} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.65184, size = 597, normalized size = 6.15 \begin{align*} -\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{align*}

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]  time = 2.0083, size = 37, normalized size = 0.38 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

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)