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]

-(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

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Rubi [A]  time = 0.0218054, antiderivative size = 75, normalized size of antiderivative = 1., 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 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]

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 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 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 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])

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*}

Mathematica [C]  time = 0.008634, 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]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [C]  time = 1.94056, size = 42, normalized size = 0.56 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{x^{4}}\,{d x} \end{align*}

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)