\(\int x^4 (-b+a x^4)^{3/4} \, dx\) [1360]

Optimal result

Integrand size = 17, antiderivative size = 98 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {\left (-b+a x^4\right )^{3/4} \left (-3 b x+4 a x^5\right )}{32 a}-\frac {3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {285, 327, 246, 218, 212, 209} \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=-\frac {3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}-\frac {3 b x \left (a x^4-b\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (a x^4-b\right )^{3/4} \]

[In]

Int[x^4*(-b + a*x^4)^(3/4),x]

[Out]

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

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 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]

Rule 285

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 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]

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {2 \sqrt [4]{a} x \left (-b+a x^4\right )^{3/4} \left (-3 b+4 a x^4\right )-3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )-3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}} \]

[In]

Integrate[x^4*(-b + a*x^4)^(3/4),x]

[Out]

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

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(\frac {16 a^{\frac {5}{4}} \left (a \,x^{4}-b \right )^{\frac {3}{4}} x^{5}-12 b x \,a^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {3}{4}}+6 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{2}-3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) b^{2}}{128 a^{\frac {5}{4}}}\)	\(116\)

[In]

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

[Out]

1/128/a^(5/4)*(16*a^(5/4)*(a*x^4-b)^(3/4)*x^5-12*b*x*a^(1/4)*(a*x^4-b)^(3/4)+6*arctan(1/a^(1/4)/x*(a*x^4-b)^(1
/4))*b^2-3*ln((-a^(1/4)*x-(a*x^4-b)^(1/4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))*b^2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=-\frac {3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} + \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 3 i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} + i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) + 3 i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} - i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} - \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 4 \, {\left (4 \, a x^{5} - 3 \, b x\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}}{128 \, a} \]

[In]

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

[Out]

-1/128*(3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^6 + (b^8/a^5)^(3/4)*a^4*x)/x) - 3*I*(b^8/a^5)^(1/4)*a*
log(27*((a*x^4 - b)^(1/4)*b^6 + I*(b^8/a^5)^(3/4)*a^4*x)/x) + 3*I*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*
b^6 - I*(b^8/a^5)^(3/4)*a^4*x)/x) - 3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^6 - (b^8/a^5)^(3/4)*a^4*x)
/x) - 4*(4*a*x^5 - 3*b*x)*(a*x^4 - b)^(3/4))/a

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {b^{\frac {3}{4}} x^{5} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).

Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.65 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {3 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a} + \frac {\frac {{\left (a x^{4} - b\right )}^{\frac {3}{4}} a b^{2}}{x^{3}} + \frac {3 \, {\left (a x^{4} - b\right )}^{\frac {7}{4}} b^{2}}{x^{7}}}{32 \, {\left (a^{3} - \frac {2 \, {\left (a x^{4} - b\right )} a^{2}}{x^{4}} + \frac {{\left (a x^{4} - b\right )}^{2} a}{x^{8}}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((a*x^4 - b)^(3/4)*x^4, x)

Mupad [F(-1)]

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

[In]

int(x^4*(a*x^4 - b)^(3/4),x)

[Out]

int(x^4*(a*x^4 - b)^(3/4), x)