3.11.73 \(\int x^4 (a+b x^4)^{5/4} \, dx\) [1073]

Optimal. Leaf size=123 \[ \frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {a^{5/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 \sqrt {b} \left (a+b x^4\right )^{3/4}} \]

[Out]

1/24*a^2*x*(b*x^4+a)^(1/4)/b+1/12*a*x^5*(b*x^4+a)^(1/4)+1/10*x^5*(b*x^4+a)^(5/4)+1/24*a^(5/2)*(1+a/b/x^4)^(3/4
)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*ar
ccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^4+a)^(3/4)/b^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {285, 327, 243, 342, 281, 237} \begin {gather*} \frac {a^{5/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 \sqrt {b} \left (a+b x^4\right )^{3/4}}+\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^4)^(5/4),x]

[Out]

(a^2*x*(a + b*x^4)^(1/4))/(24*b) + (a*x^5*(a + b*x^4)^(1/4))/12 + (x^5*(a + b*x^4)^(5/4))/10 + (a^(5/2)*(1 + a
/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*Sqrt[b]*(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 243

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

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]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int x^4 \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {1}{2} a \int x^4 \sqrt [4]{a+b x^4} \, dx\\ &=\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {1}{12} a^2 \int \frac {x^4}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}-\frac {a^3 \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{24 b}\\ &=\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}-\frac {\left (a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{24 b \left (a+b x^4\right )^{3/4}}\\ &=\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {\left (a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{24 b \left (a+b x^4\right )^{3/4}}\\ &=\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {\left (a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{48 b \left (a+b x^4\right )^{3/4}}\\ &=\frac {a^2 x \sqrt [4]{a+b x^4}}{24 b}+\frac {1}{12} a x^5 \sqrt [4]{a+b x^4}+\frac {1}{10} x^5 \left (a+b x^4\right )^{5/4}+\frac {a^{5/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 \sqrt {b} \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 = 8.61, size = 67, normalized size = 0.54 \begin {gather*} \frac {x \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}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}}\right )}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^4)^(5/4),x]

[Out]

(x*(a + b*x^4)^(1/4)*((a + b*x^4)^2 - (a^2*Hypergeometric2F1[-5/4, 1/4, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^(1
/4)))/(10*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^4 + a)^(5/4)*x^4, x)

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Fricas [F]
time = 0.08, size = 23, normalized size = 0.19 \begin {gather*} {\rm integral}\left ({\left (b x^{8} + a x^{4}\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^4*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^8 + a*x^4)*(b*x^4 + a)^(1/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.72, size = 39, normalized size = 0.32 \begin {gather*} \frac {a^{\frac {5}{4}} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/4)*x**5*gamma(5/4)*hyper((-5/4, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4))

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

[Out]

integrate((b*x^4 + a)^(5/4)*x^4, x)

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

[Out]

int(x^4*(a + b*x^4)^(5/4), x)

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