3.1056 \(\int x^5 (a+b x^4)^{5/4} \, dx\)

Optimal. Leaf size=122 \[ -\frac{10 a^{7/2} \left (\frac{b x^4}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{231 b^{3/2} \left (a+b x^4\right )^{3/4}}+\frac{5 a^2 x^2 \sqrt [4]{a+b x^4}}{231 b}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}+\frac{5}{77} a x^6 \sqrt [4]{a+b x^4} \]

[Out]

(5*a^2*x^2*(a + b*x^4)^(1/4))/(231*b) + (5*a*x^6*(a + b*x^4)^(1/4))/77 + (x^6*(a + b*x^4)^(5/4))/11 - (10*a^(7
/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(231*b^(3/2)*(a + b*x^4)^(3/4))

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Rubi [A]  time = 0.0792693, antiderivative size = 122, normalized size of antiderivative = 1., 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 = {275, 279, 321, 233, 231} \[ -\frac{10 a^{7/2} \left (\frac{b x^4}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{231 b^{3/2} \left (a+b x^4\right )^{3/4}}+\frac{5 a^2 x^2 \sqrt [4]{a+b x^4}}{231 b}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}+\frac{5}{77} a x^6 \sqrt [4]{a+b x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^4)^(5/4),x]

[Out]

(5*a^2*x^2*(a + b*x^4)^(1/4))/(231*b) + (5*a*x^6*(a + b*x^4)^(1/4))/77 + (x^6*(a + b*x^4)^(5/4))/11 - (10*a^(7
/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(231*b^(3/2)*(a + b*x^4)^(3/4))

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 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*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 )^{5/4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{5/4} \, dx,x,x^2\right )\\ &=\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}+\frac{1}{22} (5 a) \operatorname{Subst}\left (\int x^2 \sqrt [4]{a+b x^2} \, dx,x,x^2\right )\\ &=\frac{5}{77} a x^6 \sqrt [4]{a+b x^4}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}+\frac{1}{154} \left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=\frac{5 a^2 x^2 \sqrt [4]{a+b x^4}}{231 b}+\frac{5}{77} a x^6 \sqrt [4]{a+b x^4}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}-\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{231 b}\\ &=\frac{5 a^2 x^2 \sqrt [4]{a+b x^4}}{231 b}+\frac{5}{77} a x^6 \sqrt [4]{a+b x^4}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}-\frac{\left (5 a^3 \left (1+\frac{b x^4}{a}\right )^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{231 b \left (a+b x^4\right )^{3/4}}\\ &=\frac{5 a^2 x^2 \sqrt [4]{a+b x^4}}{231 b}+\frac{5}{77} a x^6 \sqrt [4]{a+b x^4}+\frac{1}{11} x^6 \left (a+b x^4\right )^{5/4}-\frac{10 a^{7/2} \left (1+\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{231 b^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0878867, size = 69, normalized size = 0.57 \[ \frac{x^2 \sqrt [4]{a+b x^4} \left (\left (a+b x^4\right )^2-\frac{a^2 \, _2F_1\left (-\frac{5}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^4}{a}\right )}{\sqrt [4]{\frac{b x^4}{a}+1}}\right )}{11 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^4)^(5/4),x]

[Out]

(x^2*(a + b*x^4)^(1/4)*((a + b*x^4)^2 - (a^2*Hypergeometric2F1[-5/4, 1/2, 3/2, -((b*x^4)/a)])/(1 + (b*x^4)/a)^
(1/4)))/(11*b)

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \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^5*(b*x^4+a)^(5/4),x)

[Out]

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

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

[Out]

integrate((b*x^4 + a)^(5/4)*x^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{9} + a x^{5}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^9 + a*x^5)*(b*x^4 + a)^(1/4), x)

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Sympy [C]  time = 3.79356, size = 29, normalized size = 0.24 \begin{align*} \frac{a^{\frac{5}{4}} x^{6}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{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)**(5/4),x)

[Out]

a**(5/4)*x**6*hyper((-5/4, 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{5}{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)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)*x^5, x)