3.1035 \(\int \frac {(a+b x^4)^{3/4}}{x^4} \, dx\)

Optimal. Leaf size=75 \[ \frac {1}{2} b^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} b^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 75, 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 = {277, 240, 212, 206, 203} \[ \frac {1}{2} b^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} b^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^4,x]

[Out]

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

Rule 203

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.68 \[ -\frac {\left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};-\frac {b x^4}{a}\right )}{3 x^3 \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^4,x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.03, size = 85, normalized size = 1.13 \[ -\frac {1}{4} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^4,x)

[Out]

int((a + b*x^4)^(3/4)/x^4, x)

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sympy [C]  time = 3.34, size = 42, normalized size = 0.56 \[ \frac {a^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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