3.1028 \(\int \frac {(a+b x^4)^{3/4}}{x^3} \, dx\)

Optimal. Leaf size=98 \[ \frac {3 b x^2}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 x^2}-\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

[Out]

3/2*b*x^2/(b*x^4+a)^(1/4)-1/2*(b*x^4+a)^(3/4)/x^2-3/2*(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))*a^(1/2)*
b^(1/2)/(b*x^4+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 98, 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 = {275, 277, 229, 227, 196} \[ \frac {3 b x^2}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 x^2}-\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^3,x]

[Out]

(3*b*x^2)/(2*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(2*x^2) - (3*Sqrt[a]*Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*Ellipti
cE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(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 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 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 277

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.52 \[ -\frac {\left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{2};-\frac {b x^4}{a}\right )}{2 x^2 \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^3,x]

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^3,x)

[Out]

int((a + b*x^4)^(3/4)/x^3, x)

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sympy [C]  time = 1.25, size = 32, normalized size = 0.33 \[ - \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/4)*hyper((-3/4, -1/2), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*x**2)

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