3.11.25 \(\int x^9 (a+b x^4)^{3/4} \, dx\) [1025]

Optimal. Leaf size=149 \[ \frac {4 a^3 x^2}{65 b^2 \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}-\frac {4 a^{7/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

4/65*a^3*x^2/b^2/(b*x^4+a)^(1/4)-2/65*a^2*x^2*(b*x^4+a)^(3/4)/b^2+1/39*a*x^6*(b*x^4+a)^(3/4)/b+1/13*x^10*(b*x^
4+a)^(3/4)-4/65*a^(7/2)*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^
(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4+a)^(1/4)

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Rubi [A]
time = 0.07, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {281, 285, 327, 235, 233, 202} \begin {gather*} -\frac {4 a^{7/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^4}}+\frac {4 a^3 x^2}{65 b^2 \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9*(a + b*x^4)^(3/4),x]

[Out]

(4*a^3*x^2)/(65*b^2*(a + b*x^4)^(1/4)) - (2*a^2*x^2*(a + b*x^4)^(3/4))/(65*b^2) + (a*x^6*(a + b*x^4)^(3/4))/(3
9*b) + (x^10*(a + b*x^4)^(3/4))/13 - (4*a^(7/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/
2, 2])/(65*b^(5/2)*(a + b*x^4)^(1/4))

Rule 202

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

Rule 233

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 235

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 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 285

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 327

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^9 \left (a+b x^4\right )^{3/4} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^4 \left (a+b x^2\right )^{3/4} \, dx,x,x^2\right )\\ &=\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}+\frac {1}{26} (3 a) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}-\frac {a^2 \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{13 b}\\ &=-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{65 b^2}\\ &=-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}+\frac {\left (2 a^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx,x,x^2\right )}{65 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac {4 a^3 x^2}{65 b^2 \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}-\frac {\left (2 a^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{65 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac {4 a^3 x^2}{65 b^2 \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}-\frac {4 a^{7/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^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 = 8.39, size = 95, normalized size = 0.64 \begin {gather*} \frac {x^2 \left (a+b x^4\right )^{3/4} \left (\left (1+\frac {b x^4}{a}\right )^{3/4} \left (-2 a^2+a b x^4+3 b^2 x^8\right )+2 a^2 \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^4}{a}\right )\right )}{39 b^2 \left (1+\frac {b x^4}{a}\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9*(a + b*x^4)^(3/4),x]

[Out]

(x^2*(a + b*x^4)^(3/4)*((1 + (b*x^4)/a)^(3/4)*(-2*a^2 + a*b*x^4 + 3*b^2*x^8) + 2*a^2*Hypergeometric2F1[-3/4, 1
/2, 3/2, -((b*x^4)/a)]))/(39*b^2*(1 + (b*x^4)/a)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(b*x^4+a)^(3/4),x)

[Out]

int(x^9*(b*x^4+a)^(3/4),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(x^9*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)*x^9, x)

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Fricas [F]
time = 0.08, size = 15, normalized size = 0.10 \begin {gather*} {\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{9}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^9, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.80, size = 29, normalized size = 0.19 \begin {gather*} \frac {a^{\frac {3}{4}} x^{10} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(b*x**4+a)**(3/4),x)

[Out]

a**(3/4)*x**10*hyper((-3/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/10

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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(x^9*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)*x^9, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^9\,{\left (b\,x^4+a\right )}^{3/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(a + b*x^4)^(3/4),x)

[Out]

int(x^9*(a + b*x^4)^(3/4), x)

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