\(\int \frac {-b+a x^3}{x^3 (b+a x^3) \sqrt [4]{-b x+a x^4}} \, dx\) [1913]

Optimal result

Integrand size = 36, antiderivative size = 133 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{9 b x^3}+\frac {2\ 2^{3/4} a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b}+\frac {2\ 2^{3/4} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b} \]

[Out]

-4/9*(a*x^4-b*x)^(3/4)/b/x^3+2/3*2^(3/4)*a^(3/4)*arctan(2^(1/4)*a^(1/4)*(a*x^4-b*x)^(3/4)/(a*x^3-b))/b+2/3*2^(
3/4)*a^(3/4)*arctanh(2^(1/4)*a^(1/4)*(a*x^4-b*x)^(3/4)/(a*x^3-b))/b

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2081, 477, 476, 525, 524} \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\frac {4 \left (b-a x^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\frac {2 a x^3}{b-a x^3}\right )}{9 b x^2 \sqrt [4]{a x^4-b x}} \]

[In]

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

[Out]

(4*(b - a*x^3)*Hypergeometric2F1[-3/4, 1, 1/4, (-2*a*x^3)/(b - a*x^3)])/(9*b*x^2*(-(b*x) + a*x^4)^(1/4))

Rule 476

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \int \frac {\left (-b+a x^3\right )^{3/4}}{x^{13/4} \left (b+a x^3\right )} \, dx}{\sqrt [4]{-b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {\left (-b+a x^{12}\right )^{3/4}}{x^{10} \left (b+a x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {\left (-b+a x^4\right )^{3/4}}{x^4 \left (b+a x^4\right )} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{-b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \left (-b+a x^3\right )\right ) \text {Subst}\left (\int \frac {\left (1-\frac {a x^4}{b}\right )^{3/4}}{x^4 \left (b+a x^4\right )} \, dx,x,x^{3/4}\right )}{3 \left (1-\frac {a x^3}{b}\right )^{3/4} \sqrt [4]{-b x+a x^4}} \\ & = \frac {4 \left (b-a x^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\frac {2 a x^3}{b-a x^3}\right )}{9 b x^2 \sqrt [4]{-b x+a x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.59 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (1+\frac {a x^3}{b}\right )^{3/4} \left (-b x+a x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{4},\frac {1}{4},\frac {2 a x^3}{b+a x^3}\right )}{9 b x^3 \left (1-\frac {a x^3}{b}\right )^{3/4}} \]

[In]

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

[Out]

(-4*(1 + (a*x^3)/b)^(3/4)*(-(b*x) + a*x^4)^(3/4)*Hypergeometric2F1[-3/4, -3/4, 1/4, (2*a*x^3)/(b + a*x^3)])/(9
*b*x^3*(1 - (a*x^3)/b)^(3/4))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(-\frac {2 \,2^{\frac {3}{4}} \left (\arctan \left (\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) a \,x^{3}-\frac {\ln \left (\frac {x 2^{\frac {1}{4}} a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{-x 2^{\frac {1}{4}} a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}\right ) a \,x^{3}}{2}+\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {3}{4}} 2^{\frac {1}{4}} a^{\frac {1}{4}}}{3}\right )}{3 a^{\frac {1}{4}} x^{3} b}\)	\(122\)

[In]

int((a*x^3-b)/x^3/(a*x^3+b)/(a*x^4-b*x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 48.59 (sec) , antiderivative size = 581, normalized size of antiderivative = 4.37 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\frac {3 \cdot 8^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} + 8^{\frac {3}{4}} \sqrt {a x^{4} - b x} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 4 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a^{2} + 8^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} - a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b}\right ) + 3 i \cdot 8^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} + i \cdot 8^{\frac {3}{4}} \sqrt {a x^{4} - b x} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 4 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a^{2} + 8^{\frac {1}{4}} {\left (-3 i \, a^{2} b x^{3} + i \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b}\right ) - 3 i \cdot 8^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} - i \cdot 8^{\frac {3}{4}} \sqrt {a x^{4} - b x} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 4 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a^{2} + 8^{\frac {1}{4}} {\left (3 i \, a^{2} b x^{3} - i \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b}\right ) - 3 \cdot 8^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} - 8^{\frac {3}{4}} \sqrt {a x^{4} - b x} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 4 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a^{2} - 8^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} - a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b}\right ) - 8 \, {\left (a x^{4} - b x\right )}^{\frac {3}{4}}}{18 \, b x^{3}} \]

[In]

integrate((a*x^3-b)/x^3/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="fricas")

[Out]

1/18*(3*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) + 8^(3/4)*sqr
t(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) + 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^(1/4)*(3*a^2*b*x^3 - a*b^2)*(a^3/b^4)^(1/
4))/(a*x^3 + b)) + 3*I*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(-(4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^
4) + I*8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^(1/4)*(-3*I*a^2*b*x^3 +
I*a*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 3*I*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(-(4*sqrt(2)*(a*x^4 - b*x)^(1/4
)*a*b^2*x^2*sqrt(a^3/b^4) - I*8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^
(1/4)*(3*I*a^2*b*x^3 - I*a*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 3*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(2)
*(a*x^4 - b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) - 8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) + 4*(a*x^4 - b*
x)^(3/4)*a^2 - 8^(1/4)*(3*a^2*b*x^3 - a*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 8*(a*x^4 - b*x)^(3/4))/(b*x^3)

Sympy [F]

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

[In]

integrate((a*x**3-b)/x**3/(a*x**3+b)/(a*x**4-b*x)**(1/4),x)

[Out]

Integral((a*x**3 - b)/(x**3*(x*(a*x**3 - b))**(1/4)*(a*x**3 + b)), x)

Maxima [F]

\[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int { \frac {a x^{3} - b}{{\left (a x^{4} - b x\right )}^{\frac {1}{4}} {\left (a x^{3} + b\right )} x^{3}} \,d x } \]

[In]

integrate((a*x^3-b)/x^3/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (105) = 210\).

Time = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.62 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b} + \frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b} - \frac {4 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {3}{4}}}{9 \, b} \]

[In]

integrate((a*x^3-b)/x^3/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="giac")

[Out]

2/3*2^(1/4)*(-a)^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a - b/x^3)^(1/4))/(-a)^(1/4))/b + 2/3*2^(1/
4)*(-a)^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) - 2*(a - b/x^3)^(1/4))/(-a)^(1/4))/b - 1/3*2^(1/4)*(-a)^
(3/4)*log(2^(3/4)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^3))/b + 1/3*2^(1/4)*(-a)^(3/4
)*log(-2^(3/4)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^3))/b - 4/9*(a - b/x^3)^(3/4)/b

Mupad [F(-1)]

Timed out. \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int -\frac {b-a\,x^3}{x^3\,{\left (a\,x^4-b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \]

[In]

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

[Out]

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