3.11.74 \(\int (a+b x^4)^{5/4} \, dx\) [1074]

Optimal. Leaf size=97 \[ \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {5 a^{3/2} \sqrt {b} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 \left (a+b x^4\right )^{3/4}} \]

[Out]

5/12*a*x*(b*x^4+a)^(1/4)+1/6*x*(b*x^4+a)^(5/4)-5/12*a^(3/2)*(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*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2)
)*b^(1/2)/(b*x^4+a)^(3/4)

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Rubi [A]
time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {201, 243, 342, 281, 237} \begin {gather*} -\frac {5 a^{3/2} \sqrt {b} 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 )}{12 \left (a+b x^4\right )^{3/4}}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {5}{12} a x \sqrt [4]{a+b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4),x]

[Out]

(5*a*x*(a + b*x^4)^(1/4))/12 + (x*(a + b*x^4)^(5/4))/6 - (5*a^(3/2)*Sqrt[b]*(1 + a/(b*x^4))^(3/4)*x^3*Elliptic
F[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*(a + b*x^4)^(3/4))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {1}{6} (5 a) \int \sqrt [4]{a+b x^4} \, dx\\ &=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {1}{12} \left (5 a^2\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {\left (5 a^2 \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}{12 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^2 \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 )}{12 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^2 \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 )}{24 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {5 a^{3/2} \sqrt {b} \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 )}{12 \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 = 8.38, size = 47, normalized size = 0.48 \begin {gather*} \frac {a x \sqrt [4]{a+b x^4} \, _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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (b \,x^{4}+a \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^4 + a)^(5/4), x)

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Fricas [F]
time = 0.07, size = 11, normalized size = 0.11 \begin {gather*} {\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(5/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.54, size = 37, normalized size = 0.38 \begin {gather*} \frac {a^{\frac {5}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/4)*x*gamma(1/4)*hyper((-5/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4))

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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((b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4), x)

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Mupad [B]
time = 1.07, size = 37, normalized size = 0.38 \begin {gather*} \frac {x\,{\left (b\,x^4+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4),x)

[Out]

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

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