3.11.34 \(\int (a+b x^4)^{3/4} \, dx\) [1034]

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]

1/4*x*(b*x^4+a)^(3/4)+3/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)+3/8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))
/b^(1/4)

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Rubi [A]
time = 0.01, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {201, 246, 218, 212, 209} \begin {gather*} \frac {3 a \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac {1}{4} x \left (a+b x^4\right )^{3/4}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}} \end {gather*}

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 201

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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]

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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.22, size = 75, normalized size = 1.00 \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (b \,x^{4}+a \right )^{\frac {3}{4}}\, dx\]

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)

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Maxima [A]
time = 0.50, size = 102, normalized size = 1.36 \begin {gather*} -\frac {3}{16} \, a {\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}} a}{4 \, {\left (b - \frac {b x^{4} + a}{x^{4}}\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*a*(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)) - 1/4*(b*x^4 + a)^(3/4)*a/((b - (b*x^4 + a)/x^4)*x^3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (55) = 110\).
time = 0.39, size = 192, normalized size = 2.56 \begin {gather*} \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 {gather*}

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)

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Sympy [C] Result contains complex when optimal does not.
time = 0.58, size = 37, normalized size = 0.49 \begin {gather*} \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 {gather*}

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))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4), x)

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Mupad [B]
time = 1.06, size = 37, normalized size = 0.49 \begin {gather*} \frac {x\,{\left (b\,x^4+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4),x)

[Out]

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

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