3.1026 \(\int x^5 (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}}-\frac{2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b} \]

[Out]

(-2*a^2*x^2)/(15*b*(a + b*x^4)^(1/4)) + (a*x^2*(a + b*x^4)^(3/4))/(15*b) + (x^6*(a + b*x^4)^(3/4))/9 + (2*a^(5
/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*b^(3/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0751214, antiderivative size = 125, 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 = {275, 279, 321, 229, 227, 196} \[ \frac{2 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}}-\frac{2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^4)^(3/4),x]

[Out]

(-2*a^2*x^2)/(15*b*(a + b*x^4)^(1/4)) + (a*x^2*(a + b*x^4)^(3/4))/(15*b) + (x^6*(a + b*x^4)^(3/4))/9 + (2*a^(5
/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*b^(3/2)*(a + b*x^4)^(1/4))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int x^5 \left (a+b x^4\right )^{3/4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{3/4} \, dx,x,x^2\right )\\ &=\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{15 b}\\ &=\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}-\frac{\left (a^2 \sqrt [4]{1+\frac{b x^4}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx,x,x^2\right )}{15 b \sqrt [4]{a+b x^4}}\\ &=-\frac{2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac{\left (a^2 \sqrt [4]{1+\frac{b x^4}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{15 b \sqrt [4]{a+b x^4}}\\ &=-\frac{2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac{a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac{1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac{2 a^{5/2} \sqrt [4]{1+\frac{b x^4}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0575493, size = 64, normalized size = 0.51 \[ \frac{x^2 \left (a+b x^4\right )^{3/4} \left (-\frac{a \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^4}{a}\right )}{\left (\frac{b x^4}{a}+1\right )^{3/4}}+a+b x^4\right )}{9 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^4)^(3/4),x]

[Out]

(x^2*(a + b*x^4)^(3/4)*(a + b*x^4 - (a*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^4)/a)])/(1 + (b*x^4)/a)^(3/4))
)/(9*b)

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \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^5*(b*x^4+a)^(3/4),x)

[Out]

int(x^5*(b*x^4+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{\frac{3}{4}} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)*x^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac{3}{4}} x^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^5, x)

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Sympy [C]  time = 1.9699, size = 29, normalized size = 0.23 \begin{align*} \frac{a^{\frac{3}{4}} x^{6}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**4+a)**(3/4),x)

[Out]

a**(3/4)*x**6*hyper((-3/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/6

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{\frac{3}{4}} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)*x^5, x)