3.1034 \(\int (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=75 \[ \frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}} \]

[Out]

(x*(a + b*x^4)^(3/4))/4 + (3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(1/4)) + (3*a*ArcTanh[(b^(1/4)*x)/(
a + b*x^4)^(1/4)])/(8*b^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0188698, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {195, 240, 212, 206, 203} \[ \frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4),x]

[Out]

(x*(a + b*x^4)^(3/4))/4 + (3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(1/4)) + (3*a*ArcTanh[(b^(1/4)*x)/(
a + b*x^4)^(1/4)])/(8*b^(1/4))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \left (a+b x^4\right )^{3/4} \, dx &=\frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{1}{4} (3 a) \int \frac{1}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{1}{4} (3 a) \operatorname{Subst}\left (\int \frac{1}{1-b x^4} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{1}{8} (3 a) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )+\frac{1}{8} (3 a) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{1}{4} x \left (a+b x^4\right )^{3/4}+\frac{3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}\\ \end{align*}

Mathematica [C]  time = 0.0056816, size = 46, normalized size = 0.61 \[ \frac{x \left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )}{\left (\frac{b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4),x]

[Out]

(x*(a + b*x^4)^(3/4)*Hypergeometric2F1[-3/4, 1/4, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^(3/4)

________________________________________________________________________________________

Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int \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)

[Out]

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

________________________________________________________________________________________

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, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.86778, size = 437, normalized size = 5.83 \begin{align*} \frac{1}{4} \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} x + \frac{3}{4} \, \left (\frac{a^{4}}{b}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (\frac{a^{4}}{b}\right )^{\frac{1}{4}} a^{3} - \left (\frac{a^{4}}{b}\right )^{\frac{1}{4}} x \sqrt{\frac{\sqrt{\frac{a^{4}}{b}} a^{4} b x^{2} + \sqrt{b x^{4} + a} a^{6}}{x^{2}}}}{a^{4} x}\right ) + \frac{3}{16} \, \left (\frac{a^{4}}{b}\right )^{\frac{1}{4}} \log \left (\frac{27 \,{\left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3} + \left (\frac{a^{4}}{b}\right )^{\frac{3}{4}} b x\right )}}{x}\right ) - \frac{3}{16} \, \left (\frac{a^{4}}{b}\right )^{\frac{1}{4}} \log \left (\frac{27 \,{\left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3} - \left (\frac{a^{4}}{b}\right )^{\frac{3}{4}} b x\right )}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^4 + a)^(3/4)*x + 3/4*(a^4/b)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*(a^4/b)^(1/4)*a^3 - (a^4/b)^(1/4)*x*sqr
t((sqrt(a^4/b)*a^4*b*x^2 + sqrt(b*x^4 + a)*a^6)/x^2))/(a^4*x)) + 3/16*(a^4/b)^(1/4)*log(27*((b*x^4 + a)^(1/4)*
a^3 + (a^4/b)^(3/4)*b*x)/x) - 3/16*(a^4/b)^(1/4)*log(27*((b*x^4 + a)^(1/4)*a^3 - (a^4/b)^(3/4)*b*x)/x)

________________________________________________________________________________________

Sympy [C]  time = 1.4305, size = 37, normalized size = 0.49 \begin{align*} \frac{a^{\frac{3}{4}} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/4)*x*gamma(1/4)*hyper((-3/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4))

________________________________________________________________________________________

Giac [B]  time = 1.14747, size = 302, normalized size = 4.03 \begin{align*} \frac{1}{32} \,{\left (\frac{6 \, \sqrt{2} \left (-b\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b} + \frac{6 \, \sqrt{2} \left (-b\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b} - \frac{3 \, \sqrt{2} \left (-b\right )^{\frac{3}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b} + \frac{3 \, \sqrt{2} \left (-b\right )^{\frac{3}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b} + \frac{8 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} x}{a}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(6*sqrt(2)*(-b)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b + 6*s
qrt(2)*(-b)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b - 3*sqrt(2)*(
-b)^(3/4)*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b + 3*sqrt(2)*(-b)^(3/4
)*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b + 8*(b*x^4 + a)^(3/4)*x/a)*a