3.22.11 \(\int \frac {1}{x^6 (-b+a x^5)^{3/4}} \, dx\) [2111]

Optimal. Leaf size=153 \[ \frac {\sqrt [4]{-b+a x^5}}{5 b x^5}-\frac {3 a \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}{-\sqrt {b}+\sqrt {-b+a x^5}}\right )}{10 \sqrt {2} b^{7/4}}+\frac {3 a \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^5}}\right )}{10 \sqrt {2} b^{7/4}} \]

[Out]

1/5*(a*x^5-b)^(1/4)/b/x^5-3/20*a*arctan(2^(1/2)*b^(1/4)*(a*x^5-b)^(1/4)/(-b^(1/2)+(a*x^5-b)^(1/2)))*2^(1/2)/b^
(7/4)+3/20*a*arctanh((1/2*b^(1/4)*2^(1/2)+1/2*(a*x^5-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x^5-b)^(1/4))*2^(1/2)/b^(7/4
)

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Rubi [A]
time = 0.15, antiderivative size = 228, normalized size of antiderivative = 1.49, number of steps used = 12, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {272, 44, 65, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {3 a \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^5-b}}{\sqrt [4]{b}}\right )}{10 \sqrt {2} b^{7/4}}+\frac {3 a \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{a x^5-b}}{\sqrt [4]{b}}+1\right )}{10 \sqrt {2} b^{7/4}}-\frac {3 a \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^5-b}+\sqrt {a x^5-b}+\sqrt {b}\right )}{20 \sqrt {2} b^{7/4}}+\frac {3 a \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^5-b}+\sqrt {a x^5-b}+\sqrt {b}\right )}{20 \sqrt {2} b^{7/4}}+\frac {\sqrt [4]{a x^5-b}}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(-b + a*x^5)^(3/4)),x]

[Out]

(-b + a*x^5)^(1/4)/(5*b*x^5) - (3*a*ArcTan[1 - (Sqrt[2]*(-b + a*x^5)^(1/4))/b^(1/4)])/(10*Sqrt[2]*b^(7/4)) + (
3*a*ArcTan[1 + (Sqrt[2]*(-b + a*x^5)^(1/4))/b^(1/4)])/(10*Sqrt[2]*b^(7/4)) - (3*a*Log[Sqrt[b] - Sqrt[2]*b^(1/4
)*(-b + a*x^5)^(1/4) + Sqrt[-b + a*x^5]])/(20*Sqrt[2]*b^(7/4)) + (3*a*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^
5)^(1/4) + Sqrt[-b + a*x^5]])/(20*Sqrt[2]*b^(7/4))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

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

Rule 272

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (-b+a x^5\right )^{3/4}} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {1}{x^2 (-b+a x)^{3/4}} \, dx,x,x^5\right )\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{x (-b+a x)^{3/4}} \, dx,x,x^5\right )}{20 b}\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}+\frac {3 \text {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{5 b}\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{10 b^{3/2}}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{10 b^{3/2}}\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}-\frac {(3 a) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}+2 x}{-\sqrt {b}-\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}-\frac {(3 a) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}-2 x}{-\sqrt {b}+\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{20 b^{3/2}}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{20 b^{3/2}}\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}-\frac {3 a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}+\frac {3 a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{10 \sqrt {2} b^{7/4}}-\frac {(3 a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{10 \sqrt {2} b^{7/4}}\\ &=\frac {\sqrt [4]{-b+a x^5}}{5 b x^5}-\frac {3 a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{10 \sqrt {2} b^{7/4}}+\frac {3 a \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{10 \sqrt {2} b^{7/4}}-\frac {3 a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}+\frac {3 a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{20 \sqrt {2} b^{7/4}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 147, normalized size = 0.96 \begin {gather*} \frac {4 b^{3/4} \sqrt [4]{-b+a x^5}+3 \sqrt {2} a x^5 \text {ArcTan}\left (\frac {-\sqrt {b}+\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}\right )+3 \sqrt {2} a x^5 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}{\sqrt {b}+\sqrt {-b+a x^5}}\right )}{20 b^{7/4} x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(-b + a*x^5)^(3/4)),x]

[Out]

(4*b^(3/4)*(-b + a*x^5)^(1/4) + 3*Sqrt[2]*a*x^5*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^5])/(Sqrt[2]*b^(1/4)*(-b + a*
x^5)^(1/4))] + 3*Sqrt[2]*a*x^5*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^5)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^5])])/(20
*b^(7/4)*x^5)

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{6} \left (a \,x^{5}-b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(a*x^5-b)^(3/4),x)

[Out]

int(1/x^6/(a*x^5-b)^(3/4),x)

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Maxima [A]
time = 0.47, size = 202, normalized size = 1.32 \begin {gather*} \frac {{\left (a x^{5} - b\right )}^{\frac {1}{4}} a}{5 \, {\left ({\left (a x^{5} - b\right )} b + b^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a \log \left (\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a \log \left (-\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{40 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(a*x^5-b)^(3/4),x, algorithm="maxima")

[Out]

1/5*(a*x^5 - b)^(1/4)*a/((a*x^5 - b)*b + b^2) + 3/40*(2*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x
^5 - b)^(1/4))/b^(1/4))/b^(3/4) + 2*sqrt(2)*a*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^5 - b)^(1/4))/b^(1
/4))/b^(3/4) + sqrt(2)*a*log(sqrt(2)*(a*x^5 - b)^(1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(3/4) - sqrt(2)*
a*log(-sqrt(2)*(a*x^5 - b)^(1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(3/4))/b

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Fricas [A]
time = 0.44, size = 212, normalized size = 1.39 \begin {gather*} \frac {12 \, b x^{5} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{5} - b\right )}^{\frac {1}{4}} a b^{5} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}} - \sqrt {b^{4} \sqrt {-\frac {a^{4}}{b^{7}}} + \sqrt {a x^{5} - b} a^{2}} b^{5} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}}}{a^{4}}\right ) + 3 \, b x^{5} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a\right ) - 3 \, b x^{5} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a\right ) + 4 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}}{20 \, b x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(a*x^5-b)^(3/4),x, algorithm="fricas")

[Out]

1/20*(12*b*x^5*(-a^4/b^7)^(1/4)*arctan(-((a*x^5 - b)^(1/4)*a*b^5*(-a^4/b^7)^(3/4) - sqrt(b^4*sqrt(-a^4/b^7) +
sqrt(a*x^5 - b)*a^2)*b^5*(-a^4/b^7)^(3/4))/a^4) + 3*b*x^5*(-a^4/b^7)^(1/4)*log(3*b^2*(-a^4/b^7)^(1/4) + 3*(a*x
^5 - b)^(1/4)*a) - 3*b*x^5*(-a^4/b^7)^(1/4)*log(-3*b^2*(-a^4/b^7)^(1/4) + 3*(a*x^5 - b)^(1/4)*a) + 4*(a*x^5 -
b)^(1/4))/(b*x^5)

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Sympy [C] Result contains complex when optimal does not.
time = 0.86, size = 42, normalized size = 0.27 \begin {gather*} - \frac {\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 {b e^{2 i \pi }}{a x^{5}}} \right )}}{5 a^{\frac {3}{4}} x^{\frac {35}{4}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(a*x**5-b)**(3/4),x)

[Out]

-gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*exp_polar(2*I*pi)/(a*x**5))/(5*a**(3/4)*x**(35/4)*gamma(11/4))

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Giac [A]
time = 0.40, size = 199, normalized size = 1.30 \begin {gather*} \frac {\frac {6 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {6 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {3 \, \sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} - \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} + \frac {8 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a}{b x^{5}}}{40 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(a*x^5-b)^(3/4),x, algorithm="giac")

[Out]

1/40*(6*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^5 - b)^(1/4))/b^(1/4))/b^(7/4) + 6*sqrt(2)*a^
2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^5 - b)^(1/4))/b^(1/4))/b^(7/4) + 3*sqrt(2)*a^2*log(sqrt(2)*(a*
x^5 - b)^(1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(7/4) - 3*sqrt(2)*a^2*log(-sqrt(2)*(a*x^5 - b)^(1/4)*b^(
1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(7/4) + 8*(a*x^5 - b)^(1/4)*a/(b*x^5))/a

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Mupad [B]
time = 1.31, size = 72, normalized size = 0.47 \begin {gather*} \frac {{\left (a\,x^5-b\right )}^{1/4}}{5\,b\,x^5}+\frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^5-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{10\,{\left (-b\right )}^{7/4}}+\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^5-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{10\,{\left (-b\right )}^{7/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a*x^5 - b)^(3/4)),x)

[Out]

(a*x^5 - b)^(1/4)/(5*b*x^5) + (3*a*atan((a*x^5 - b)^(1/4)/(-b)^(1/4)))/(10*(-b)^(7/4)) + (3*a*atanh((a*x^5 - b
)^(1/4)/(-b)^(1/4)))/(10*(-b)^(7/4))

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