3.1126 \(\int \frac{x^6}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=80 \[ \frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}+\frac{x^3 \sqrt [4]{a+b x^4}}{4 b} \]

[Out]

(x^3*(a + b*x^4)^(1/4))/(4*b) + (3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(7/4)) - (3*a*ArcTanh[(b^(1/4
)*x)/(a + b*x^4)^(1/4)])/(8*b^(7/4))

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Rubi [A]  time = 0.0251478, antiderivative size = 80, normalized size of antiderivative = 1., 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, 331, 298, 203, 206} \[ \frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}+\frac{x^3 \sqrt [4]{a+b x^4}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^4)^(3/4),x]

[Out]

(x^3*(a + b*x^4)^(1/4))/(4*b) + (3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(7/4)) - (3*a*ArcTanh[(b^(1/4
)*x)/(a + b*x^4)^(1/4)])/(8*b^(7/4))

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 298

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

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])

Rubi steps

\begin{align*} \int \frac{x^6}{\left (a+b x^4\right )^{3/4}} \, dx &=\frac{x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac{(3 a) \int \frac{x^2}{\left (a+b x^4\right )^{3/4}} \, dx}{4 b}\\ &=\frac{x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{x^2}{1-b x^4} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )}{4 b}\\ &=\frac{x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac{(3 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^{3/2}}+\frac{(3 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^{3/2}}\\ &=\frac{x^3 \sqrt [4]{a+b x^4}}{4 b}+\frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}\\ \end{align*}

Mathematica [A]  time = 0.0146289, size = 75, normalized size = 0.94 \[ \frac{2 b^{3/4} x^3 \sqrt [4]{a+b x^4}+3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^4)^(3/4),x]

[Out]

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4) + 3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - 3*a*ArcTanh[(b^(1/4)*x)/(a + b*
x^4)^(1/4)])/(8*b^(7/4))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{{x}^{6} \left ( b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^4+a)^(3/4),x)

[Out]

int(x^6/(b*x^4+a)^(3/4),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^6/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.65984, size = 473, normalized size = 5.91 \begin{align*} \frac{4 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{3} + 12 \, b \left (\frac{a^{4}}{b^{7}}\right )^{\frac{1}{4}} \arctan \left (\frac{b^{5} x \sqrt{\frac{b^{4} x^{2} \sqrt{\frac{a^{4}}{b^{7}}} + \sqrt{b x^{4} + a} a^{2}}{x^{2}}} \left (\frac{a^{4}}{b^{7}}\right )^{\frac{3}{4}} -{\left (b x^{4} + a\right )}^{\frac{1}{4}} a b^{5} \left (\frac{a^{4}}{b^{7}}\right )^{\frac{3}{4}}}{a^{4} x}\right ) - 3 \, b \left (\frac{a^{4}}{b^{7}}\right )^{\frac{1}{4}} \log \left (\frac{3 \,{\left (b^{2} x \left (\frac{a^{4}}{b^{7}}\right )^{\frac{1}{4}} +{\left (b x^{4} + a\right )}^{\frac{1}{4}} a\right )}}{x}\right ) + 3 \, b \left (\frac{a^{4}}{b^{7}}\right )^{\frac{1}{4}} \log \left (-\frac{3 \,{\left (b^{2} x \left (\frac{a^{4}}{b^{7}}\right )^{\frac{1}{4}} -{\left (b x^{4} + a\right )}^{\frac{1}{4}} a\right )}}{x}\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/16*(4*(b*x^4 + a)^(1/4)*x^3 + 12*b*(a^4/b^7)^(1/4)*arctan((b^5*x*sqrt((b^4*x^2*sqrt(a^4/b^7) + sqrt(b*x^4 +
a)*a^2)/x^2)*(a^4/b^7)^(3/4) - (b*x^4 + a)^(1/4)*a*b^5*(a^4/b^7)^(3/4))/(a^4*x)) - 3*b*(a^4/b^7)^(1/4)*log(3*(
b^2*x*(a^4/b^7)^(1/4) + (b*x^4 + a)^(1/4)*a)/x) + 3*b*(a^4/b^7)^(1/4)*log(-3*(b^2*x*(a^4/b^7)^(1/4) - (b*x^4 +
a)^(1/4)*a)/x))/b

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Sympy [C]  time = 1.92499, size = 37, normalized size = 0.46 \begin{align*} \frac{x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{4}} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**4+a)**(3/4),x)

[Out]

x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/4)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^4 + a)^(3/4), x)