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

Optimal. Leaf size=143 \[ -\frac {8 a^{7/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b} \]

[Out]

8/65*a^3*x/b^2/(b*x^2+a)^(1/4)-4/65*a^2*x*(b*x^2+a)^(3/4)/b^2+2/39*a*x^3*(b*x^2+a)^(3/4)/b+2/13*x^5*(b*x^2+a)^
(3/4)-8/65*a^(7/2)*(1+b*x^2/a)^(1/4)*(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)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {279, 321, 229, 227, 196} \[ -\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {8 a^{7/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^2)^(3/4),x]

[Out]

(8*a^3*x)/(65*b^2*(a + b*x^2)^(1/4)) - (4*a^2*x*(a + b*x^2)^(3/4))/(65*b^2) + (2*a*x^3*(a + b*x^2)^(3/4))/(39*
b) + (2*x^5*(a + b*x^2)^(3/4))/13 - (8*a^(7/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2,
2])/(65*b^(5/2)*(a + b*x^2)^(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 227

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

Rule 229

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

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 321

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]

Rubi steps

\begin {align*} \int x^4 \left (a+b x^2\right )^{3/4} \, dx &=\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}+\frac {1}{13} (3 a) \int \frac {x^4}{\sqrt [4]{a+b x^2}} \, dx\\ &=\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}-\frac {\left (2 a^2\right ) \int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx}{13 b}\\ &=-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}+\frac {\left (4 a^3\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{65 b^2}\\ &=-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}+\frac {\left (4 a^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{65 b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}-\frac {\left (4 a^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{65 b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}-\frac {8 a^{7/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 93, normalized size = 0.65 \[ \frac {2 x \left (a+b x^2\right )^{3/4} \left (\left (\frac {b x^2}{a}+1\right )^{3/4} \left (-2 a^2+a b x^2+3 b^2 x^4\right )+2 a^2 \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )\right )}{39 b^2 \left (\frac {b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^2)^(3/4),x]

[Out]

(2*x*(a + b*x^2)^(3/4)*((1 + (b*x^2)/a)^(3/4)*(-2*a^2 + a*b*x^2 + 3*b^2*x^4) + 2*a^2*Hypergeometric2F1[-3/4, 1
/2, 3/2, -((b*x^2)/a)]))/(39*b^2*(1 + (b*x^2)/a)^(3/4))

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(3/4)*x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/4)*x^4, x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2}+a \right )^{\frac {3}{4}} x^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/4)*x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^2)^(3/4),x)

[Out]

int(x^4*(a + b*x^2)^(3/4), x)

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sympy [C]  time = 1.20, size = 29, normalized size = 0.20 \[ \frac {a^{\frac {3}{4}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/4)*x**5*hyper((-3/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5

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