3.1136 \(\int \frac {1}{x^4 (a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3} \]

[Out]

-1/3*(b*x^4+a)^(1/4)/a/x^3+2/3*b^(3/2)*(1+a/b/x^4)^(3/4)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/co
s(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^4+a)^(
3/4)

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Rubi [A]  time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {325, 237, 335, 275, 231} \[ \frac {2 b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^4)^(3/4)),x]

[Out]

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

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 237

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.60 \[ -\frac {\left (\frac {b x^4}{a}+1\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {3}{4};\frac {1}{4};-\frac {b x^4}{a}\right )}{3 x^3 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^4)^(3/4)),x]

[Out]

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

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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b x^{8} + a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/(b*x^8 + a*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^4), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^4+a)^(3/4),x)

[Out]

int(1/x^4/(b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^4)^(3/4)),x)

[Out]

int(1/(x^4*(a + b*x^4)^(3/4)), x)

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sympy [C]  time = 1.56, size = 41, normalized size = 0.48 \[ \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**4+a)**(3/4),x)

[Out]

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

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