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

Optimal. Leaf size=133 \[ \frac {2\ 2^{3/4} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )}{3 b}+\frac {2\ 2^{3/4} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )}{3 b}-\frac {4 \left (a x^4-b x\right )^{3/4}}{9 b x^3} \]

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Rubi [C]  time = 0.41, antiderivative size = 56, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2056, 466, 465, 511, 510} \begin {gather*} \frac {4 \left (b-a x^3\right ) \, _2F_1\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}} \end {gather*}

Antiderivative was successfully verified.

[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 465

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 466

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 2056

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*} \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \, _2F_1\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*}

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Mathematica [C]  time = 0.05, size = 78, normalized size = 0.59 \begin {gather*} -\frac {4 \left (\frac {a x^3}{b}+1\right )^{3/4} \left (a x^4-b x\right )^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};\frac {2 a x^3}{a x^3+b}\right )}{9 b x^3 \left (1-\frac {a x^3}{b}\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[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))

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IntegrateAlgebraic [A]  time = 0.44, size = 133, normalized size = 1.00 \begin {gather*} -\frac {4 \left (-b x+a x^4\right )^{3/4}}{9 b x^3}+\frac {2\ 2^{3/4} a^{3/4} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 91.11, size = 488, normalized size = 3.67 \begin {gather*} -\frac {12 \cdot 8^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \arctan \left (\frac {16 \cdot 8^{\frac {1}{4}} {\left (a x^{4} - b x\right )}^{\frac {1}{4}} a^{4} b x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + 4 \cdot 8^{\frac {3}{4}} {\left (a x^{4} - b x\right )}^{\frac {3}{4}} a^{2} b^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + \sqrt {2} {\left (8 \cdot 8^{\frac {1}{4}} \sqrt {a x^{4} - b x} a^{2} b x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + 8^{\frac {3}{4}} {\left (3 \, a b^{3} x^{3} - b^{4}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}}\right )} \sqrt {\sqrt {2} a^{2} b^{2} \sqrt {\frac {a^{3}}{b^{4}}}}}{8 \, {\left (a^{5} x^{3} + a^{4} b\right )}}\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 ) + 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*(12*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*arctan(1/8*(16*8^(1/4)*(a*x^4 - b*x)^(1/4)*a^4*b*x^2*(a^3/b^4)^(1/4) +
4*8^(3/4)*(a*x^4 - b*x)^(3/4)*a^2*b^3*(a^3/b^4)^(3/4) + sqrt(2)*(8*8^(1/4)*sqrt(a*x^4 - b*x)*a^2*b*x*(a^3/b^4
)^(1/4) + 8^(3/4)*(3*a*b^3*x^3 - b^4)*(a^3/b^4)^(3/4))*sqrt(sqrt(2)*a^2*b^2*sqrt(a^3/b^4)))/(a^5*x^3 + a^4*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)) + 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)

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giac [B]  time = 0.23, size = 215, normalized size = 1.62 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{3}-b}{x^{3} \left (a \,x^{3}+b \right ) \left (a \,x^{4}-b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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