3.1.75 \(\int \frac {1}{a+b \cos ^6(x)} \, dx\) [75]

Optimal. Leaf size=171 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}+\sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}+\sqrt [3]{b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b}}} \]

[Out]

-1/3*arctan(cot(x)*(a^(1/3)+b^(1/3))^(1/2)/a^(1/6))/a^(5/6)/(a^(1/3)+b^(1/3))^(1/2)-1/3*arctan(cot(x)*(a^(1/3)
-(-1)^(1/3)*b^(1/3))^(1/2)/a^(1/6))/a^(5/6)/(a^(1/3)-(-1)^(1/3)*b^(1/3))^(1/2)-1/3*arctan(cot(x)*(a^(1/3)+(-1)
^(2/3)*b^(1/3))^(1/2)/a^(1/6))/a^(5/6)/(a^(1/3)+(-1)^(2/3)*b^(1/3))^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3290, 3260, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}+\sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}+\sqrt [3]{b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b}} \cot (x)}{\sqrt [6]{a}}\right )}{3 a^{5/6} \sqrt {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^6)^(-1),x]

[Out]

-1/3*ArcTan[(Sqrt[a^(1/3) + b^(1/3)]*Cot[x])/a^(1/6)]/(a^(5/6)*Sqrt[a^(1/3) + b^(1/3)]) - ArcTan[(Sqrt[a^(1/3)
- (-1)^(1/3)*b^(1/3)]*Cot[x])/a^(1/6)]/(3*a^(5/6)*Sqrt[a^(1/3) - (-1)^(1/3)*b^(1/3)]) - ArcTan[(Sqrt[a^(1/3)
+ (-1)^(2/3)*b^(1/3)]*Cot[x])/a^(1/6)]/(3*a^(5/6)*Sqrt[a^(1/3) + (-1)^(2/3)*b^(1/3)])

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 3260

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 3290

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{k}, Dist[2/(a*n), Sum[Int[1/(1 - Si
n[e + f*x]^2/((-1)^(4*(k/n))*Rt[-a/b, n/2])), x], {k, 1, n/2}], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.24, size = 146, normalized size = 0.85 \begin {gather*} \frac {8}{3} \text {RootSum}\left [b+6 b \text {$\#$1}+15 b \text {$\#$1}^2+64 a \text {$\#$1}^3+20 b \text {$\#$1}^3+15 b \text {$\#$1}^4+6 b \text {$\#$1}^5+b \text {$\#$1}^6\&,\frac {2 \text {ArcTan}\left (\frac {\sin (2 x)}{\cos (2 x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (2 x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b+5 b \text {$\#$1}+32 a \text {$\#$1}^2+10 b \text {$\#$1}^2+10 b \text {$\#$1}^3+5 b \text {$\#$1}^4+b \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^6)^(-1),x]

[Out]

(8*RootSum[b + 6*b*#1 + 15*b*#1^2 + 64*a*#1^3 + 20*b*#1^3 + 15*b*#1^4 + 6*b*#1^5 + b*#1^6 & , (2*ArcTan[Sin[2*
x]/(Cos[2*x] - #1)]*#1^2 - I*Log[1 - 2*Cos[2*x]*#1 + #1^2]*#1^2)/(b + 5*b*#1 + 32*a*#1^2 + 10*b*#1^2 + 10*b*#1
^3 + 5*b*#1^4 + b*#1^5) & ])/3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.43, size = 60, normalized size = 0.35

method	result	size
default	\(\frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+3 a \,\textit {\_Z}^{2}+a +b \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (x \right )-\textit {\_R} \right )}{\textit {\_R}^{5}+2 \textit {\_R}^{3}+\textit {\_R}}}{6 a}\)	\(60\)
risch	\(\munderset {\textit {\_R} =\RootOf \left (1+\left (46656 a^{6}+46656 a^{5} b \right ) \textit {\_Z}^{6}+3888 a^{4} \textit {\_Z}^{4}+108 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-\frac {15552 i a^{6}}{b}-15552 i a^{5}\right ) \textit {\_R}^{5}+\left (\frac {2592 a^{5}}{b}+2592 a^{4}\right ) \textit {\_R}^{4}+\left (-\frac {864 i a^{4}}{b}+432 i a^{3}\right ) \textit {\_R}^{3}+\left (\frac {144 a^{3}}{b}-72 a^{2}\right ) \textit {\_R}^{2}+\left (-\frac {12 i a^{2}}{b}-12 i a \right ) \textit {\_R} +\frac {2 a}{b}+1\right )\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^6),x,method=_RETURNVERBOSE)

[Out]

1/6/a*sum((_R^4+2*_R^2+1)/(_R^5+2*_R^3+_R)*ln(tan(x)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+3*_Z^2*a+a+b))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^6),x, algorithm="maxima")

[Out]

integrate(1/(b*cos(x)^6 + a), x)

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Fricas [C] Result contains complex when optimal does not.
time = 1.96, size = 15483, normalized size = 90.54 \begin {gather*} \text {too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^6),x, algorithm="fricas")

[Out]

-1/72*sqrt(1/2)*sqrt((-I*sqrt(3) + 1)*(1/(a^4 + a^3*b) - 1/(a^2 + a*b)^2)/(-1/93312/(a^6 + a^5*b) + 1/31104/((
a^4 + a^3*b)*(a^2 + a*b)) - 1/46656/(a^2 + a*b)^3 + 1/93312*b/((a + b)^2*a^5))^(1/3) - 1296*(I*sqrt(3) + 1)*(-
1/93312/(a^6 + a^5*b) + 1/31104/((a^4 + a^3*b)*(a^2 + a*b)) - 1/46656/(a^2 + a*b)^3 + 1/93312*b/((a + b)^2*a^5
))^(1/3) - 72/(a^2 + a*b))*log(-1/5184*(a^5 + a^4*b - 2*(a^5 + a^4*b)*cos(x)^2)*((-I*sqrt(3) + 1)*(1/(a^4 + a^
3*b) - 1/(a^2 + a*b)^2)/(-1/93312/(a^6 + a^5*b) + 1/31104/((a^4 + a^3*b)*(a^2 + a*b)) - 1/46656/(a^2 + a*b)^3
+ 1/93312*b/((a + b)^2*a^5))^(1/3) - 1296*(I*sqrt(3) + 1)*(-1/93312/(a^6 + a^5*b) + 1/31104/((a^4 + a^3*b)*(a^
2 + a*b)) - 1/46656/(a^2 + a*b)^3 + 1/93312*b/((a + b)^2*a^5))^(1/3) - 72/(a^2 + a*b))^2 + (2*a + b)*cos(x)^2
+ 1/15552*sqrt(1/2)*((a^6 + a^5*b)*((-I*sqrt(3) + 1)*(1/(a^4 + a^3*b) - 1/(a^2 + a*b)^2)/(-1/93312/(a^6 + a^5*
b) + 1/31104/((a^4 + a^3*b)*(a^2 + a*b)) - 1/46656/(a^2 + a*b)^3 + 1/93312*b/((a + b)^2*a^5))^(1/3) - 1296*(I*
sqrt(3) +  ...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a + b \cos ^{6}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)**6),x)

[Out]

Integral(1/(a + b*cos(x)**6), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^6),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 3.08, size = 184, normalized size = 1.08 \begin {gather*} \sum _{k=1}^6\ln \left ({\mathrm {root}\left (46656\,a^5\,b\,d^6+46656\,a^6\,d^6+3888\,a^4\,d^4+108\,a^2\,d^2+1,d,k\right )}^2\,a^3\,b^3\,\left ({\mathrm {root}\left (46656\,a^5\,b\,d^6+46656\,a^6\,d^6+3888\,a^4\,d^4+108\,a^2\,d^2+1,d,k\right )}^2\,a^2\,36+1\right )\,\left (\mathrm {root}\left (46656\,a^5\,b\,d^6+46656\,a^6\,d^6+3888\,a^4\,d^4+108\,a^2\,d^2+1,d,k\right )\,a\,\mathrm {tan}\left (x\right )\,6-1\right )\,36\right )\,\mathrm {root}\left (46656\,a^5\,b\,d^6+46656\,a^6\,d^6+3888\,a^4\,d^4+108\,a^2\,d^2+1,d,k\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)^6),x)

[Out]

symsum(log(36*root(46656*a^5*b*d^6 + 46656*a^6*d^6 + 3888*a^4*d^4 + 108*a^2*d^2 + 1, d, k)^2*a^3*b^3*(36*root(
46656*a^5*b*d^6 + 46656*a^6*d^6 + 3888*a^4*d^4 + 108*a^2*d^2 + 1, d, k)^2*a^2 + 1)*(6*root(46656*a^5*b*d^6 + 4
6656*a^6*d^6 + 3888*a^4*d^4 + 108*a^2*d^2 + 1, d, k)*a*tan(x) - 1))*root(46656*a^5*b*d^6 + 46656*a^6*d^6 + 388
8*a^4*d^4 + 108*a^2*d^2 + 1, d, k), k, 1, 6)

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