3.11.56 \(\int x^5 (a+b x^4)^{5/4} \, dx\) [1056]

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

[Out]

5/231*a^2*x^2*(b*x^4+a)^(1/4)/b+5/77*a*x^6*(b*x^4+a)^(1/4)+1/11*x^6*(b*x^4+a)^(5/4)-10/231*a^(7/2)*(1+b*x^4/a)
^(3/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*a
rctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^4+a)^(3/4)

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Rubi [A]
time = 0.06, antiderivative size = 122, 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 = {281, 285, 327, 239, 237} \begin {gather*} -\frac {10 a^{7/2} \left (\frac {b x^4}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\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} \end {gather*}

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 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

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 281

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 285

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 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 x^5 \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{2} \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.59, size = 69, normalized size = 0.57 \begin {gather*} \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]{1+\frac {b x^4}{a}}}\right )}{11 b} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int x^{5} \left (b \,x^{4}+a \right )^{\frac {5}{4}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.08, size = 23, normalized size = 0.19 \begin {gather*} {\rm integral}\left ({\left (b x^{9} + a x^{5}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}, x\right ) \end {gather*}

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] Result contains complex when optimal does not.
time = 0.80, size = 29, normalized size = 0.24 \begin {gather*} \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 {gather*}

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 = 0.00, 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^5*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)*x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,{\left (b\,x^4+a\right )}^{5/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^4)^(5/4),x)

[Out]

int(x^5*(a + b*x^4)^(5/4), x)

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