3.1074 \(\int (a+b x^4)^{5/4} \, dx\)

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

[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))

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Rubi [A]  time = 0.0383726, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {195, 237, 335, 275, 231} \[ -\frac{5 a^{3/2} \sqrt{b} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{12 \left (a+b x^4\right )^{3/4}}+\frac{1}{6} x \left (a+b x^4\right )^{5/4}+\frac{5}{12} a x \sqrt [4]{a+b x^4} \]

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 195

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_)^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 335

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]

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 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 \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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0050164, size = 47, normalized size = 0.48 \[ \frac{a x \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{5}{4},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )}{\sqrt [4]{\frac{b x^4}{a}+1}} \]

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 = 0.036, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

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

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

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]  time = 1.29669, size = 37, normalized size = 0.38 \begin{align*} \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{align*}

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

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)