3.448 \(\int \frac {\tan (x)}{(a^3-b^3 \cos ^n(x))^{4/3}} \, dx\)

Optimal. Leaf size=112 \[ \frac {\log (\cos (x))}{2 a^4}-\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}-\frac {3 \log \left (a-\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^4 n}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 \cos ^n(x)}+a}{\sqrt {3} a}\right )}{a^4 n} \]

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Rubi [A]  time = 0.17, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3230, 266, 51, 55, 617, 204, 31} \[ -\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}-\frac {3 \log \left (a-\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^4 n}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 \cos ^n(x)}+a}{\sqrt {3} a}\right )}{a^4 n}+\frac {\log (\cos (x))}{2 a^4} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a^3 - b^3*Cos[x]^n)^(4/3),x]

[Out]

-((Sqrt[3]*ArcTan[(a + 2*(a^3 - b^3*Cos[x]^n)^(1/3))/(Sqrt[3]*a)])/(a^4*n)) - 3/(a^3*n*(a^3 - b^3*Cos[x]^n)^(1
/3)) + Log[Cos[x]]/(2*a^4) - (3*Log[a - (a^3 - b^3*Cos[x]^n)^(1/3)])/(2*a^4*n)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 204

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

Rule 266

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 617

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 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {\tan (x)}{\left (a^3-b^3 \cos ^n(x)\right )^{4/3}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x \left (a^3-b^3 x^n\right )^{4/3}} \, dx,x,\cos (x)\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x \left (a^3-b^3 x\right )^{4/3}} \, dx,x,\cos ^n(x)\right )}{n}\\ &=-\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{a^3-b^3 x}} \, dx,x,\cos ^n(x)\right )}{a^3 n}\\ &=-\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}+\frac {\log (\cos (x))}{2 a^4}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{a-x} \, dx,x,\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^4 n}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{a^2+a x+x^2} \, dx,x,\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^3 n}\\ &=-\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}+\frac {\log (\cos (x))}{2 a^4}-\frac {3 \log \left (a-\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^4 n}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a^3-b^3 \cos ^n(x)}}{a}\right )}{a^4 n}\\ &=-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a^3-b^3 \cos ^n(x)}}{a}}{\sqrt {3}}\right )}{a^4 n}-\frac {3}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}}+\frac {\log (\cos (x))}{2 a^4}-\frac {3 \log \left (a-\sqrt [3]{a^3-b^3 \cos ^n(x)}\right )}{2 a^4 n}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 47, normalized size = 0.42 \[ -\frac {3 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};1-\frac {b^3 \cos ^n(x)}{a^3}\right )}{a^3 n \sqrt [3]{a^3-b^3 \cos ^n(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(a^3 - b^3*Cos[x]^n)^(4/3),x]

[Out]

(-3*Hypergeometric2F1[-1/3, 1, 2/3, 1 - (b^3*Cos[x]^n)/a^3])/(a^3*n*(a^3 - b^3*Cos[x]^n)^(1/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan (x)}{\left (a^3-b^3 \cos ^n(x)\right )^{4/3}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Tan[x]/(a^3 - b^3*Cos[x]^n)^(4/3),x]

[Out]

Could not integrate

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fricas [A]  time = 1.36, size = 185, normalized size = 1.65 \[ -\frac {2 \, {\left (\sqrt {3} b^{3} \cos \relax (x)^{n} - \sqrt {3} a^{3}\right )} \arctan \left (\frac {\sqrt {3} a + 2 \, \sqrt {3} {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}}}{3 \, a}\right ) - {\left (b^{3} \cos \relax (x)^{n} - a^{3}\right )} \log \left (a^{2} + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}} a + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (b^{3} \cos \relax (x)^{n} - a^{3}\right )} \log \left (-a + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}}\right ) - 6 \, {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {2}{3}} a}{2 \, {\left (a^{4} b^{3} n \cos \relax (x)^{n} - a^{7} n\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(sqrt(3)*b^3*cos(x)^n - sqrt(3)*a^3)*arctan(1/3*(sqrt(3)*a + 2*sqrt(3)*(-b^3*cos(x)^n + a^3)^(1/3))/a)
- (b^3*cos(x)^n - a^3)*log(a^2 + (-b^3*cos(x)^n + a^3)^(1/3)*a + (-b^3*cos(x)^n + a^3)^(2/3)) + 2*(b^3*cos(x)
^n - a^3)*log(-a + (-b^3*cos(x)^n + a^3)^(1/3)) - 6*(-b^3*cos(x)^n + a^3)^(2/3)*a)/(a^4*b^3*n*cos(x)^n - a^7*n
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \relax (x)}{{\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)/(-b^3*cos(x)^n + a^3)^(4/3), x)

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maple [A]  time = 0.51, size = 127, normalized size = 1.13

method	result	size
derivativedivides	\(-\frac {\frac {\ln \left (a -\left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}\right )}{a^{4}}+\frac {-\frac {\ln \left (a^{2}+a \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}+\left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (a +2 \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )}{a^{4}}+\frac {3}{a^{3} \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}}}{n}\)	\(127\)
default	\(-\frac {\frac {\ln \left (a -\left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}\right )}{a^{4}}+\frac {-\frac {\ln \left (a^{2}+a \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}+\left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (a +2 \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )}{a^{4}}+\frac {3}{a^{3} \left (a^{3}-b^{3} \left (\cos ^{n}\relax (x )\right )\right )^{\frac {1}{3}}}}{n}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 136, normalized size = 1.21 \[ -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a + 2 \, {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{a^{4} n} + \frac {\log \left (a^{2} + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}} a + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {2}{3}}\right )}{2 \, a^{4} n} - \frac {\log \left (-a + {\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}}\right )}{a^{4} n} - \frac {3}{{\left (-b^{3} \cos \relax (x)^{n} + a^{3}\right )}^{\frac {1}{3}} a^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*arctan(1/3*sqrt(3)*(a + 2*(-b^3*cos(x)^n + a^3)^(1/3))/a)/(a^4*n) + 1/2*log(a^2 + (-b^3*cos(x)^n + a^
3)^(1/3)*a + (-b^3*cos(x)^n + a^3)^(2/3))/(a^4*n) - log(-a + (-b^3*cos(x)^n + a^3)^(1/3))/(a^4*n) - 3/((-b^3*c
os(x)^n + a^3)^(1/3)*a^3*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tan}\relax (x)}{{\left (a^3-b^3\,{\cos \relax (x)}^n\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a^3 - b^3*cos(x)^n)^(4/3),x)

[Out]

int(tan(x)/(a^3 - b^3*cos(x)^n)^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{\left (a^{3} - b^{3} \cos ^{n}{\relax (x )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)/(a**3 - b**3*cos(x)**n)**(4/3), x)

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