3.1033 \(\int x^4 (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=101 \[ -\frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}+\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4} \]

[Out]

3/32*a*x*(b*x^4+a)^(3/4)/b+1/8*x^5*(b*x^4+a)^(3/4)-3/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)-3/64*a^2
*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {279, 321, 240, 212, 206, 203} \[ -\frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}+\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^4)^(3/4),x]

[Out]

(3*a*x*(a + b*x^4)^(3/4))/(32*b) + (x^5*(a + b*x^4)^(3/4))/8 - (3*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(
64*b^(5/4)) - (3*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(5/4))

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

Rubi steps

\begin {align*} \int x^4 \left (a+b x^4\right )^{3/4} \, dx &=\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}+\frac {1}{8} (3 a) \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}-\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{32 b}\\ &=\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{32 b}\\ &=\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b}\\ &=\frac {3 a x \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{8} x^5 \left (a+b x^4\right )^{3/4}-\frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.07, size = 62, normalized size = 0.61 \[ \frac {x \left (a+b x^4\right )^{3/4} \left (-\frac {a \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\left (\frac {b x^4}{a}+1\right )^{3/4}}+a+b x^4\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^4)^(3/4),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.94, size = 218, normalized size = 2.16 \[ -\frac {12 \, \left (\frac {a^{8}}{b^{5}}\right )^{\frac {1}{4}} b \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (\frac {a^{8}}{b^{5}}\right )^{\frac {1}{4}} a^{6} b - \left (\frac {a^{8}}{b^{5}}\right )^{\frac {1}{4}} b x \sqrt {\frac {\sqrt {\frac {a^{8}}{b^{5}}} a^{8} b^{3} x^{2} + \sqrt {b x^{4} + a} a^{12}}{x^{2}}}}{a^{8} x}\right ) + 3 \, \left (\frac {a^{8}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} + \left (\frac {a^{8}}{b^{5}}\right )^{\frac {3}{4}} b^{4} x\right )}}{x}\right ) - 3 \, \left (\frac {a^{8}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} - \left (\frac {a^{8}}{b^{5}}\right )^{\frac {3}{4}} b^{4} x\right )}}{x}\right ) - 4 \, {\left (4 \, b x^{5} + 3 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(12*(a^8/b^5)^(1/4)*b*arctan(-((b*x^4 + a)^(1/4)*(a^8/b^5)^(1/4)*a^6*b - (a^8/b^5)^(1/4)*b*x*sqrt((sqrt
(a^8/b^5)*a^8*b^3*x^2 + sqrt(b*x^4 + a)*a^12)/x^2))/(a^8*x)) + 3*(a^8/b^5)^(1/4)*b*log(27*((b*x^4 + a)^(1/4)*a
^6 + (a^8/b^5)^(3/4)*b^4*x)/x) - 3*(a^8/b^5)^(1/4)*b*log(27*((b*x^4 + a)^(1/4)*a^6 - (a^8/b^5)^(3/4)*b^4*x)/x)
 - 4*(4*b*x^5 + 3*a*x)*(b*x^4 + a)^(3/4))/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)*x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 3.03, size = 148, normalized size = 1.47 \[ \frac {3 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{128 \, b} + \frac {\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b}{x^{3}} + \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{x^{7}}}{32 \, {\left (b^{3} - \frac {2 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b}{x^{8}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*a^2*(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))/b + 1/32*((b*x^4 + a)^(3/4)*a^2*b/x^3 + 3*(b*x^4 + a)^(7/4)*a^2/x^7)/(b^3 - 2*(b
*x^4 + a)*b^2/x^4 + (b*x^4 + a)^2*b/x^8)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^4)^(3/4),x)

[Out]

int(x^4*(a + b*x^4)^(3/4), x)

________________________________________________________________________________________

sympy [C]  time = 4.29, size = 39, normalized size = 0.39 \[ \frac {a^{\frac {3}{4}} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/4)*x**5*gamma(5/4)*hyper((-3/4, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4))

________________________________________________________________________________________