\(\int \frac {x^6}{(b+a x^4)^{3/4}} \, dx\) [1065]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {x^3 \sqrt [4]{b+a x^4}}{4 a}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{7/4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{7/4}} \]

[Out]

1/4*x^3*(a*x^4+b)^(1/4)/a+3/8*b*arctan(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(7/4)-3/8*b*arctanh(a^(1/4)*x/(a*x^4+b)^(1
/4))/a^(7/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {327, 338, 304, 209, 212} \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{7/4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{7/4}}+\frac {x^3 \sqrt [4]{a x^4+b}}{4 a} \]

[In]

Int[x^6/(b + a*x^4)^(3/4),x]

[Out]

(x^3*(b + a*x^4)^(1/4))/(4*a) + (3*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(7/4)) - (3*b*ArcTanh[(a^(1/4
)*x)/(b + a*x^4)^(1/4)])/(8*a^(7/4))

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 327

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]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \sqrt [4]{b+a x^4}}{4 a}-\frac {(3 b) \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx}{4 a} \\ & = \frac {x^3 \sqrt [4]{b+a x^4}}{4 a}-\frac {(3 b) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 a} \\ & = \frac {x^3 \sqrt [4]{b+a x^4}}{4 a}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/2}}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/2}} \\ & = \frac {x^3 \sqrt [4]{b+a x^4}}{4 a}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{7/4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{7/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {2 a^{3/4} x^3 \sqrt [4]{b+a x^4}+3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )-3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{7/4}} \]

[In]

Integrate[x^6/(b + a*x^4)^(3/4),x]

[Out]

(2*a^(3/4)*x^3*(b + a*x^4)^(1/4) + 3*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)] - 3*b*ArcTanh[(a^(1/4)*x)/(b + a*
x^4)^(1/4)])/(8*a^(7/4))

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06

method result size
pseudoelliptic \(-\frac {3 \left (-\frac {4 \left (a \,x^{4}+b \right )^{\frac {1}{4}} x^{3} a^{\frac {3}{4}}}{3}+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b \right )}{16 a^{\frac {7}{4}}}\) \(85\)

[In]

int(x^6/(a*x^4+b)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-3/16/a^(7/4)*(-4/3*(a*x^4+b)^(1/4)*x^3*a^(3/4)+ln((-a^(1/4)*x-(a*x^4+b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4)))*b
+2*arctan(1/a^(1/4)/x*(a*x^4+b)^(1/4))*b)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.51 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} x^{3} - 3 \, a \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (\frac {3 \, {\left (a^{2} x \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + 3 \, a \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (a^{2} x \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + 3 i \, a \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (i \, a^{2} x \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) - 3 i \, a \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (-i \, a^{2} x \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right )}{16 \, a} \]

[In]

integrate(x^6/(a*x^4+b)^(3/4),x, algorithm="fricas")

[Out]

1/16*(4*(a*x^4 + b)^(1/4)*x^3 - 3*a*(b^4/a^7)^(1/4)*log(3*(a^2*x*(b^4/a^7)^(1/4) + (a*x^4 + b)^(1/4)*b)/x) + 3
*a*(b^4/a^7)^(1/4)*log(-3*(a^2*x*(b^4/a^7)^(1/4) - (a*x^4 + b)^(1/4)*b)/x) + 3*I*a*(b^4/a^7)^(1/4)*log(-3*(I*a
^2*x*(b^4/a^7)^(1/4) - (a*x^4 + b)^(1/4)*b)/x) - 3*I*a*(b^4/a^7)^(1/4)*log(-3*(-I*a^2*x*(b^4/a^7)^(1/4) - (a*x
^4 + b)^(1/4)*b)/x))/a

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.46 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate(x**6/(a*x**4+b)**(3/4),x)

[Out]

x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*gamma(11/4))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=-\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{16 \, a} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{4 \, {\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x} \]

[In]

integrate(x^6/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\int { \frac {x^{6}}{{\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^6/(a*x^4+b)^(3/4),x, algorithm="giac")

[Out]

integrate(x^6/(a*x^4 + b)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4}} \, dx=\int \frac {x^6}{{\left (a\,x^4+b\right )}^{3/4}} \,d x \]

[In]

int(x^6/(b + a*x^4)^(3/4),x)

[Out]

int(x^6/(b + a*x^4)^(3/4), x)