3.9.11 \(\int (a+b x^2)^{5/4} \, dx\) [811]

Optimal. Leaf size=92 \[ \frac {10}{21} a x \sqrt [4]{a+b x^2}+\frac {2}{7} x \left (a+b x^2\right )^{5/4}+\frac {10 a^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 \sqrt {b} \left (a+b x^2\right )^{3/4}} \]

[Out]

10/21*a*x*(b*x^2+a)^(1/4)+2/7*x*(b*x^2+a)^(5/4)+10/21*a^(5/2)*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1
/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^2
+a)^(3/4)/b^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {201, 239, 237} \begin {gather*} \frac {10 a^{5/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {10}{21} a x \sqrt [4]{a+b x^2}+\frac {2}{7} x \left (a+b x^2\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/4),x]

[Out]

(10*a*x*(a + b*x^2)^(1/4))/21 + (2*x*(a + b*x^2)^(5/4))/7 + (10*a^(5/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan
[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*Sqrt[b]*(a + b*x^2)^(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 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]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^{5/4} \, dx &=\frac {2}{7} x \left (a+b x^2\right )^{5/4}+\frac {1}{7} (5 a) \int \sqrt [4]{a+b x^2} \, dx\\ &=\frac {10}{21} a x \sqrt [4]{a+b x^2}+\frac {2}{7} x \left (a+b x^2\right )^{5/4}+\frac {1}{21} \left (5 a^2\right ) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac {10}{21} a x \sqrt [4]{a+b x^2}+\frac {2}{7} x \left (a+b x^2\right )^{5/4}+\frac {\left (5 a^2 \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{21 \left (a+b x^2\right )^{3/4}}\\ &=\frac {10}{21} a x \sqrt [4]{a+b x^2}+\frac {2}{7} x \left (a+b x^2\right )^{5/4}+\frac {10 a^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 \sqrt {b} \left (a+b x^2\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 = 8.53, size = 47, normalized size = 0.51 \begin {gather*} \frac {a x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {5}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [4]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/4),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b \,x^{2}+a \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/4),x)

[Out]

int((b*x^2+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^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(5/4), x)

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Fricas [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^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(5/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.62, size = 26, normalized size = 0.28 \begin {gather*} a^{\frac {5}{4}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/4),x)

[Out]

a**(5/4)*x*hyper((-5/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)

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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^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(5/4), x)

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Mupad [B]
time = 4.86, size = 37, normalized size = 0.40 \begin {gather*} \frac {x\,{\left (b\,x^2+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/4),x)

[Out]

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

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