3.1042 \(\int x^2 (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=99 \[ \frac {a^{3/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {a x^3}{4 \sqrt [4]{a+b x^4}} \]

[Out]

1/4*a*x^3/(b*x^4+a)^(1/4)+1/6*x^3*(b*x^4+a)^(3/4)+1/4*a^(3/2)*(1+a/b/x^4)^(1/4)*x*(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)))*EllipticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2)
)/(b*x^4+a)^(1/4)/b^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {279, 310, 281, 335, 275, 196} \[ \frac {a^{3/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {a x^3}{4 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^4)^(3/4),x]

[Out]

(a*x^3)/(4*(a + b*x^4)^(1/4)) + (x^3*(a + b*x^4)^(3/4))/6 + (a^(3/2)*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[
(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*Sqrt[b]*(a + b*x^4)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

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 279

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 281

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[(x*(1 + a/(b*x^4))^(1/4))/(b*(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 310

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4)^(1/4)), x] - Dist[a/2, Int[x^2/(a
+ b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

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]

Rubi steps

\begin {align*} \int x^2 \left (a+b x^4\right )^{3/4} \, dx &=\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {1}{2} a \int \frac {x^2}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}-\frac {1}{4} a^2 \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx\\ &=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}-\frac {\left (a^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{4 b \sqrt [4]{a+b x^4}}\\ &=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {\left (a^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{4 b \sqrt [4]{a+b x^4}}\\ &=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {\left (a^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{8 b \sqrt [4]{a+b x^4}}\\ &=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {a^{3/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {b} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.52 \[ \frac {x^3 \left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{3 \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^4)^(3/4),x]

[Out]

(x^3*(a + b*x^4)^(3/4)*Hypergeometric2F1[-3/4, 3/4, 7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^(3/4))

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)*x^2, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^4+a)^(3/4),x)

[Out]

int(x^2*(b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)*x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (b\,x^4+a\right )}^{3/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^4)^(3/4),x)

[Out]

int(x^2*(a + b*x^4)^(3/4), x)

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sympy [C]  time = 2.50, size = 39, normalized size = 0.39 \[ \frac {a^{\frac {3}{4}} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**4+a)**(3/4),x)

[Out]

a**(3/4)*x**3*gamma(3/4)*hyper((-3/4, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4))

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