3.1162 \(\int \frac {x^{14}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=129 \[ \frac {77 a^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{7/2} \sqrt [4]{a+b x^4}}+\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}} \]

[Out]

77/120*a^2*x^3/b^3/(b*x^4+a)^(1/4)-11/60*a*x^7/b^2/(b*x^4+a)^(1/4)+1/10*x^11/b/(b*x^4+a)^(1/4)+77/40*a^(5/2)*(
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)))*Ellipti
cE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(7/2)/(b*x^4+a)^(1/4)

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Rubi [A]  time = 0.06, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {282, 281, 335, 275, 196} \[ \frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}+\frac {77 a^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{7/2} \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^14/(a + b*x^4)^(5/4),x]

[Out]

(77*a^2*x^3)/(120*b^3*(a + b*x^4)^(1/4)) - (11*a*x^7)/(60*b^2*(a + b*x^4)^(1/4)) + x^11/(10*b*(a + b*x^4)^(1/4
)) + (77*a^(5/2)*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(40*b^(7/2)*(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 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 282

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m - 3)/(b*(m - 4)*(a + b*x^4)^(1/4)), x] - Dis
t[(a*(m - 3))/(b*(m - 4)), Int[x^(m - 4)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && IGtQ[(m
- 2)/4, 0]

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 \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {(11 a) \int \frac {x^{10}}{\left (a+b x^4\right )^{5/4}} \, dx}{10 b}\\ &=-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}+\frac {\left (77 a^2\right ) \int \frac {x^6}{\left (a+b x^4\right )^{5/4}} \, dx}{60 b^2}\\ &=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {\left (77 a^3\right ) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{40 b^3}\\ &=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {\left (77 a^3 \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}{40 b^4 \sqrt [4]{a+b x^4}}\\ &=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}+\frac {\left (77 a^3 \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 )}{40 b^4 \sqrt [4]{a+b x^4}}\\ &=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}+\frac {\left (77 a^3 \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 )}{80 b^4 \sqrt [4]{a+b x^4}}\\ &=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}+\frac {77 a^{5/2} \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 )}{40 b^{7/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 80, normalized size = 0.62 \[ \frac {x^3 \left (-77 a^2 \sqrt [4]{\frac {b x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^4}{a}\right )+77 a^2-22 a b x^4+12 b^2 x^8\right )}{120 b^3 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/(a + b*x^4)^(5/4),x]

[Out]

(x^3*(77*a^2 - 22*a*b*x^4 + 12*b^2*x^8 - 77*a^2*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^
4)/a)]))/(120*b^3*(a + b*x^4)^(1/4))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{14}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^14/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^14/(b*x^4 + a)^(5/4), x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(b*x^4+a)^(5/4),x)

[Out]

int(x^14/(b*x^4+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^14/(b*x^4 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{14}}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(a + b*x^4)^(5/4),x)

[Out]

int(x^14/(a + b*x^4)^(5/4), x)

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sympy [C]  time = 2.27, size = 37, normalized size = 0.29 \[ \frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {19}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(b*x**4+a)**(5/4),x)

[Out]

x**15*gamma(15/4)*hyper((5/4, 15/4), (19/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(19/4))

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