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

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

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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 = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {195, 240, 212, 206, 203} \begin {gather*} \frac {1}{4} x \left (a x^4+b\right )^{3/4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 \sqrt [4]{a}}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 \sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^4)^(3/4),x]

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.61 \begin {gather*} \frac {x \left (a x^4+b\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};-\frac {a x^4}{b}\right )}{\left (\frac {a x^4}{b}+1\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^4)^(3/4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.35, size = 75, normalized size = 1.00 \begin {gather*} \frac {1}{4} x \left (b+a x^4\right )^{3/4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt [4]{a}}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^4)^(3/4),x]

[Out]

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

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fricas [B]  time = 0.48, size = 192, normalized size = 2.56 \begin {gather*} \frac {1}{4} \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} x + \frac {3}{4} \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} b^{3} - \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} x \sqrt {\frac {\sqrt {\frac {b^{4}}{a}} a b^{4} x^{2} + \sqrt {a x^{4} + b} b^{6}}{x^{2}}}}{b^{4} x}\right ) + \frac {3}{16} \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} b^{3} + \left (\frac {b^{4}}{a}\right )^{\frac {3}{4}} a x\right )}}{x}\right ) - \frac {3}{16} \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} b^{3} - \left (\frac {b^{4}}{a}\right )^{\frac {3}{4}} a x\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b)^(3/4), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \left (a \,x^{4}+b \right )^{\frac {3}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 102, normalized size = 1.36 \begin {gather*} -\frac {3}{16} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}} b}{4 \, {\left (a - \frac {a x^{4} + b}{x^{4}}\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*b*(2*arctan((a*x^4 + b)^(1/4)/(a^(1/4)*x))/a^(1/4) + log(-(a^(1/4) - (a*x^4 + b)^(1/4)/x)/(a^(1/4) + (a*
x^4 + b)^(1/4)/x))/a^(1/4)) - 1/4*(a*x^4 + b)^(3/4)*b/((a - (a*x^4 + b)/x^4)*x^3)

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mupad [B]  time = 0.78, size = 37, normalized size = 0.49 \begin {gather*} \frac {x\,{\left (a\,x^4+b\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^4}{b}\right )}{{\left (\frac {a\,x^4}{b}+1\right )}^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^4)^(3/4),x)

[Out]

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

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sympy [C]  time = 1.18, size = 37, normalized size = 0.49 \begin {gather*} \frac {b^{\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 {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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