3.363 \(\int (1+\cot ^3(x)) (a \sec ^2(x)-\sin (2 x))^2 \, dx\)

Optimal. Leaf size=88 \[ \frac {1}{3} a^2 \tan ^3(x)+a^2 \tan (x)-\frac {1}{2} a^2 \cot ^2(x)+\left (a^2+4\right ) \log (\sin (x))+4 a x+4 a \cot (x)+(4-a) a \log (\cos (x))+\frac {x}{2}+\cos ^4(x)+2 \cos ^2(x)-\sin (x) \cos ^3(x)+\frac {1}{2} \sin (x) \cos (x) \]

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Rubi [A]  time = 0.55, antiderivative size = 84, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1805, 1802, 635, 203, 260} \[ \frac {1}{3} a^2 \tan ^3(x)+a^2 \tan (x)-\frac {1}{2} a^2 \cot ^2(x)+\left (a^2+4\right ) \log (\tan (x))+\frac {1}{2} (8 a+1) x+4 a \cot (x)+4 (a+1) \log (\cos (x))+\cos ^4(x) (1-\tan (x))+\frac {1}{2} \cos ^2(x) (\tan (x)+4) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cot[x]^3)*(a*Sec[x]^2 - Sin[2*x])^2,x]

[Out]

((1 + 8*a)*x)/2 + 4*a*Cot[x] - (a^2*Cot[x]^2)/2 + 4*(1 + a)*Log[Cos[x]] + (4 + a^2)*Log[Tan[x]] + Cos[x]^4*(1
- Tan[x]) + a^2*Tan[x] + (a^2*Tan[x]^3)/3 + (Cos[x]^2*(4 + Tan[x]))/2

Rule 203

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

Rule 260

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \left (1+\cot ^3(x)\right ) \left (a \sec ^2(x)-\sin (2 x)\right )^2 \, dx &=\operatorname {Subst}\left (\int \frac {\left (1+x^3\right ) \left (a-2 x+2 a x^2+a x^4\right )^2}{x^3 \left (1+x^2\right )^3} \, dx,x,\tan (x)\right )\\ &=\cos ^4(x) (1-\tan (x))-\frac {1}{4} \operatorname {Subst}\left (\int \frac {-4 a^2+16 a x-4 \left (4+3 a^2\right ) x^2-4 \left (1-4 a+a^2\right ) x^3+4 (4-3 a) a x^4-12 a^2 x^5+4 (4-a) a x^6-12 a^2 x^7-4 a^2 x^9}{x^3 \left (1+x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\cos ^4(x) (1-\tan (x))+\frac {1}{2} \cos ^2(x) (4+\tan (x))+\frac {1}{8} \operatorname {Subst}\left (\int \frac {8 a^2-32 a x+16 \left (2+a^2\right ) x^2+4 \left (1+2 a^2\right ) x^3-8 (4-a) a x^4+16 a^2 x^5+8 a^2 x^7}{x^3 \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=\cos ^4(x) (1-\tan (x))+\frac {1}{2} \cos ^2(x) (4+\tan (x))+\frac {1}{8} \operatorname {Subst}\left (\int \left (8 a^2+\frac {8 a^2}{x^3}-\frac {32 a}{x^2}+\frac {8 \left (4+a^2\right )}{x}+8 a^2 x^2+\frac {4 (1+8 a-8 (1+a) x)}{1+x^2}\right ) \, dx,x,\tan (x)\right )\\ &=4 a \cot (x)-\frac {1}{2} a^2 \cot ^2(x)+\left (4+a^2\right ) \log (\tan (x))+\cos ^4(x) (1-\tan (x))+a^2 \tan (x)+\frac {1}{3} a^2 \tan ^3(x)+\frac {1}{2} \cos ^2(x) (4+\tan (x))+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+8 a-8 (1+a) x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=4 a \cot (x)-\frac {1}{2} a^2 \cot ^2(x)+\left (4+a^2\right ) \log (\tan (x))+\cos ^4(x) (1-\tan (x))+a^2 \tan (x)+\frac {1}{3} a^2 \tan ^3(x)+\frac {1}{2} \cos ^2(x) (4+\tan (x))-(4 (1+a)) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (x)\right )+\frac {1}{2} (1+8 a) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} (1+8 a) x+4 a \cot (x)-\frac {1}{2} a^2 \cot ^2(x)+4 (1+a) \log (\cos (x))+\left (4+a^2\right ) \log (\tan (x))+\cos ^4(x) (1-\tan (x))+a^2 \tan (x)+\frac {1}{3} a^2 \tan ^3(x)+\frac {1}{2} \cos ^2(x) (4+\tan (x))\\ \end {align*}

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Mathematica [A]  time = 2.02, size = 127, normalized size = 1.44 \[ -\frac {2 \sin (x) \cos ^3(x) \left (\sin (2 x)-a \sec ^2(x)\right )^2 \left (-8 a^2 (\cos (2 x)+2) \sec ^2(x)-3 \cot (x) \left (-4 a^2 \csc ^2(x)+8 a^2 \log (\sin (x))-8 a^2 \log (\cos (x))+32 a x+32 a \log (\cos (x))+4 x-\sin (4 x)+12 \cos (2 x)+\cos (4 x)+32 \log (\sin (x))\right )-96 a \cot ^2(x)\right )}{3 (-4 a+2 \sin (2 x)+\sin (4 x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cot[x]^3)*(a*Sec[x]^2 - Sin[2*x])^2,x]

[Out]

(-2*Cos[x]^3*Sin[x]*(-(a*Sec[x]^2) + Sin[2*x])^2*(-96*a*Cot[x]^2 - 8*a^2*(2 + Cos[2*x])*Sec[x]^2 - 3*Cot[x]*(4
*x + 32*a*x + 12*Cos[2*x] + Cos[4*x] - 4*a^2*Csc[x]^2 + 32*a*Log[Cos[x]] - 8*a^2*Log[Cos[x]] + 32*Log[Sin[x]]
+ 8*a^2*Log[Sin[x]] - Sin[4*x])))/(3*(-4*a + 2*Sin[2*x] + Sin[4*x])^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (1+\cot ^3(x)\right ) \left (a \sec ^2(x)-\sin (2 x)\right )^2 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1 + Cot[x]^3)*(a*Sec[x]^2 - Sin[2*x])^2,x]

[Out]

Could not integrate

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fricas [B]  time = 0.95, size = 178, normalized size = 2.02 \[ \frac {24 \, \cos \relax (x)^{9} + 24 \, \cos \relax (x)^{7} + 3 \, {\left (4 \, {\left (8 \, a + 1\right )} x - 27\right )} \cos \relax (x)^{5} + 3 \, {\left (4 \, a^{2} - 4 \, {\left (8 \, a + 1\right )} x + 11\right )} \cos \relax (x)^{3} - 12 \, {\left ({\left (a^{2} - 4 \, a\right )} \cos \relax (x)^{5} - {\left (a^{2} - 4 \, a\right )} \cos \relax (x)^{3}\right )} \log \left (\cos \relax (x)^{2}\right ) + 12 \, {\left ({\left (a^{2} + 4\right )} \cos \relax (x)^{5} - {\left (a^{2} + 4\right )} \cos \relax (x)^{3}\right )} \log \left (-\frac {1}{4} \, \cos \relax (x)^{2} + \frac {1}{4}\right ) - 4 \, {\left (6 \, \cos \relax (x)^{8} - 9 \, \cos \relax (x)^{6} - {\left (4 \, a^{2} - 24 \, a - 3\right )} \cos \relax (x)^{4} + 2 \, a^{2} \cos \relax (x)^{2} + 2 \, a^{2}\right )} \sin \relax (x)}{24 \, {\left (\cos \relax (x)^{5} - \cos \relax (x)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cot(x)^3)*(a*sec(x)^2-sin(2*x))^2,x, algorithm="fricas")

[Out]

1/24*(24*cos(x)^9 + 24*cos(x)^7 + 3*(4*(8*a + 1)*x - 27)*cos(x)^5 + 3*(4*a^2 - 4*(8*a + 1)*x + 11)*cos(x)^3 -
12*((a^2 - 4*a)*cos(x)^5 - (a^2 - 4*a)*cos(x)^3)*log(cos(x)^2) + 12*((a^2 + 4)*cos(x)^5 - (a^2 + 4)*cos(x)^3)*
log(-1/4*cos(x)^2 + 1/4) - 4*(6*cos(x)^8 - 9*cos(x)^6 - (4*a^2 - 24*a - 3)*cos(x)^4 + 2*a^2*cos(x)^2 + 2*a^2)*
sin(x))/(cos(x)^5 - cos(x)^3)

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giac [A]  time = 0.64, size = 149, normalized size = 1.69 \[ \frac {1}{3} \, a^{2} \tan \relax (x)^{3} + a^{2} \tan \relax (x) + \frac {1}{2} \, {\left (8 \, a + 1\right )} x - 2 \, {\left (a + 1\right )} \log \left (\tan \relax (x)^{2} + 1\right ) + {\left (a^{2} + 4\right )} \log \left ({\left | \tan \relax (x) \right |}\right ) - \frac {a^{2} \tan \relax (x)^{6} - 4 \, a \tan \relax (x)^{6} + 3 \, a^{2} \tan \relax (x)^{4} - 8 \, a \tan \relax (x)^{5} - 8 \, a \tan \relax (x)^{4} - \tan \relax (x)^{5} + 3 \, a^{2} \tan \relax (x)^{2} - 16 \, a \tan \relax (x)^{3} - 4 \, \tan \relax (x)^{4} - 4 \, a \tan \relax (x)^{2} + \tan \relax (x)^{3} + a^{2} - 8 \, a \tan \relax (x) - 6 \, \tan \relax (x)^{2}}{2 \, {\left (\tan \relax (x)^{3} + \tan \relax (x)\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cot(x)^3)*(a*sec(x)^2-sin(2*x))^2,x, algorithm="giac")

[Out]

1/3*a^2*tan(x)^3 + a^2*tan(x) + 1/2*(8*a + 1)*x - 2*(a + 1)*log(tan(x)^2 + 1) + (a^2 + 4)*log(abs(tan(x))) - 1
/2*(a^2*tan(x)^6 - 4*a*tan(x)^6 + 3*a^2*tan(x)^4 - 8*a*tan(x)^5 - 8*a*tan(x)^4 - tan(x)^5 + 3*a^2*tan(x)^2 - 1
6*a*tan(x)^3 - 4*tan(x)^4 - 4*a*tan(x)^2 + tan(x)^3 + a^2 - 8*a*tan(x) - 6*tan(x)^2)/(tan(x)^3 + tan(x))^2

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maple [B]  time = 0.56, size = 210, normalized size = 2.39

method	result	size
default	\(\frac {2 \left (\cos ^{8}\relax (x )\right )}{\sin \relax (x )^{2}}+2 \left (\cos ^{6}\relax (x )\right )+\cos ^{4}\relax (x )+2 \left (\cos ^{2}\relax (x )\right )+4 \ln \left (\sin \relax (x )\right )-\frac {4 \left (\cos ^{7}\relax (x )\right )}{\sin \relax (x )}-4 \left (\cos ^{5}\relax (x )+\frac {5 \left (\cos ^{3}\relax (x )\right )}{4}+\frac {15 \cos \relax (x )}{8}\right ) \sin \relax (x )+\frac {x}{2}-\frac {2 \left (\cos ^{6}\relax (x )\right )}{\sin \relax (x )^{2}}+\frac {8 \left (\cos ^{5}\relax (x )\right )}{\sin \relax (x )}+8 \left (\cos ^{3}\relax (x )+\frac {3 \cos \relax (x )}{2}\right ) \sin \relax (x )-4 a \left (-\frac {\left (\cot ^{2}\relax (x )\right )}{2}-\ln \left (\sin \relax (x )\right )\right )-4 a \left (-\cot \relax (x )-x \right )-4 \cot \relax (x )-\frac {4 a}{\sin \relax (x )^{2}}+a^{2} \left (-\frac {1}{2 \sin \relax (x )^{2}}+\ln \left (\tan \relax (x )\right )\right )-a^{2} \left (\frac {1}{\sin \relax (x ) \cos \relax (x )}-2 \cot \relax (x )\right )-4 a \left (-\frac {1}{2 \sin \relax (x )^{2}}+\ln \left (\tan \relax (x )\right )\right )+a^{2} \left (\frac {1}{3 \sin \relax (x ) \cos \relax (x )^{3}}+\frac {4}{3 \sin \relax (x ) \cos \relax (x )}-\frac {8 \cot \relax (x )}{3}\right )\)	\(210\)
risch	\(\frac {x}{2}-4 i x a +4 a x +\frac {i {\mathrm e}^{4 i x}}{16}+\frac {{\mathrm e}^{4 i x}}{16}-\frac {i {\mathrm e}^{-4 i x}}{16}+\frac {3 \,{\mathrm e}^{2 i x}}{4}+\frac {3 \,{\mathrm e}^{-2 i x}}{4}+\frac {{\mathrm e}^{-4 i x}}{16}-4 i x +\frac {2 a \left (12 i {\mathrm e}^{8 i x}+3 a \,{\mathrm e}^{8 i x}+6 i a \,{\mathrm e}^{6 i x}+24 i {\mathrm e}^{6 i x}+9 a \,{\mathrm e}^{6 i x}-10 i a \,{\mathrm e}^{4 i x}+9 a \,{\mathrm e}^{4 i x}+2 i a \,{\mathrm e}^{2 i x}-24 i {\mathrm e}^{2 i x}+3 a \,{\mathrm e}^{2 i x}+2 i a -12 i\right )}{3 \left ({\mathrm e}^{2 i x}-1\right )^{2} \left (1+{\mathrm e}^{2 i x}\right )^{3}}+\ln \left ({\mathrm e}^{2 i x}-1\right ) a^{2}+4 \ln \left ({\mathrm e}^{2 i x}-1\right )-\ln \left (1+{\mathrm e}^{2 i x}\right ) a^{2}+4 \ln \left (1+{\mathrm e}^{2 i x}\right ) a\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+cot(x)^3)*(a*sec(x)^2-sin(2*x))^2,x,method=_RETURNVERBOSE)

[Out]

2/sin(x)^2*cos(x)^8+2*cos(x)^6+cos(x)^4+2*cos(x)^2+4*ln(sin(x))-4/sin(x)*cos(x)^7-4*(cos(x)^5+5/4*cos(x)^3+15/
8*cos(x))*sin(x)+1/2*x-2/sin(x)^2*cos(x)^6+8/sin(x)*cos(x)^5+8*(cos(x)^3+3/2*cos(x))*sin(x)-4*a*(-1/2*cot(x)^2
-ln(sin(x)))-4*a*(-cot(x)-x)-4*cot(x)-4*a/sin(x)^2+a^2*(-1/2/sin(x)^2+ln(tan(x)))-a^2*(1/sin(x)/cos(x)-2*cot(x
))-4*a*(-1/2/sin(x)^2+ln(tan(x)))+a^2*(1/3/sin(x)/cos(x)^3+4/3/sin(x)/cos(x)-8/3*cot(x))

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maxima [A]  time = 1.35, size = 115, normalized size = 1.31 \[ \frac {1}{3} \, {\left (\tan \relax (x)^{3} + 3 \, \tan \relax (x)\right )} a^{2} - \frac {1}{2} \, a^{2} {\left (\frac {1}{\sin \relax (x)^{2}} + \log \left (\sin \relax (x)^{2} - 1\right ) - \log \left (\sin \relax (x)^{2}\right )\right )} + 4 \, a {\left (x + \frac {1}{\tan \relax (x)}\right )} + 2 \, a \log \left (-\sin \relax (x)^{2} + 1\right ) + \frac {1}{2} \, x + \frac {1}{8} \, \cos \left (4 \, x\right ) + \frac {3}{2} \, \cos \left (2 \, x\right ) + 2 \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + 2 \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - \frac {1}{8} \, \sin \left (4 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cot(x)^3)*(a*sec(x)^2-sin(2*x))^2,x, algorithm="maxima")

[Out]

1/3*(tan(x)^3 + 3*tan(x))*a^2 - 1/2*a^2*(1/sin(x)^2 + log(sin(x)^2 - 1) - log(sin(x)^2)) + 4*a*(x + 1/tan(x))
+ 2*a*log(-sin(x)^2 + 1) + 1/2*x + 1/8*cos(4*x) + 3/2*cos(2*x) + 2*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + 2
*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 1/8*sin(4*x)

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mupad [B]  time = 0.38, size = 133, normalized size = 1.51 \[ a^2\,\mathrm {tan}\relax (x)-\frac {{\mathrm {tan}\relax (x)}^4\,\left (\frac {a^2}{2}-2\right )-4\,a\,\mathrm {tan}\relax (x)+\frac {a^2}{2}-{\mathrm {tan}\relax (x)}^5\,\left (4\,a+\frac {1}{2}\right )-{\mathrm {tan}\relax (x)}^3\,\left (8\,a-\frac {1}{2}\right )+{\mathrm {tan}\relax (x)}^2\,\left (a^2-3\right )}{{\mathrm {tan}\relax (x)}^6+2\,{\mathrm {tan}\relax (x)}^4+{\mathrm {tan}\relax (x)}^2}-\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,\left (a\,\left (2+2{}\mathrm {i}\right )+2+\frac {1}{4}{}\mathrm {i}\right )-\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,\left (a\,\left (2-2{}\mathrm {i}\right )+2-\frac {1}{4}{}\mathrm {i}\right )+\frac {a^2\,{\mathrm {tan}\relax (x)}^3}{3}+\ln \left (\mathrm {tan}\relax (x)\right )\,\left (a^2+4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(x)^3 + 1)*(sin(2*x) - a/cos(x)^2)^2,x)

[Out]

a^2*tan(x) - (tan(x)^4*(a^2/2 - 2) - 4*a*tan(x) + a^2/2 - tan(x)^5*(4*a + 1/2) - tan(x)^3*(8*a - 1/2) + tan(x)
^2*(a^2 - 3))/(tan(x)^2 + 2*tan(x)^4 + tan(x)^6) - log(tan(x) - 1i)*(a*(2 + 2i) + (2 + 1i/4)) - log(tan(x) + 1
i)*(a*(2 - 2i) + (2 - 1i/4)) + (a^2*tan(x)^3)/3 + log(tan(x))*(a^2 + 4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cot(x)**3)*(a*sec(x)**2-sin(2*x))**2,x)

[Out]

Timed out

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