3.4.44 \(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [344]

Optimal. Leaf size=104 \[ -\frac {a x}{b^2}+\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^2 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d} \]

[Out]

-a*x/b^2+2*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/b^2/d+b*arctanh(cos(d*x+c))/a^
2/d-cos(d*x+c)/b/d-cot(d*x+c)/a/d

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Rubi [A]
time = 0.17, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2973, 3136, 2739, 632, 210, 3855} \begin {gather*} \frac {2 \left (a^2-b^2\right )^{3/2} \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^2 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {a x}{b^2}-\frac {\cot (c+d x)}{a d}-\frac {\cos (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

-((a*x)/b^2) + (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^2*d) + (b*ArcTanh
[Cos[c + d*x]])/(a^2*d) - Cos[c + d*x]/(b*d) - Cot[c + d*x]/(a*d)

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2973

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] +
(Dist[1/(a*b*d*(n + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n
+ 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n
+ 4))*Sin[e + f*x]^2, x], x], x] - Simp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 2)/(b
*d^2*f*(m + n + 4))), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m
, 2*n]) &&  !m < -1 && LtQ[n, -1] && NeQ[m + n + 4, 0]

Rule 3136

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}-\frac {\int \frac {\csc (c+d x) \left (b^2+2 a b \sin (c+d x)+a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{a b}\\ &=-\frac {a x}{b^2}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}-\frac {b \int \csc (c+d x) \, dx}{a^2}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2 b^2}\\ &=-\frac {a x}{b^2}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b^2 d}\\ &=-\frac {a x}{b^2}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}-\frac {\left (4 \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b^2 d}\\ &=-\frac {a x}{b^2}+\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^2 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 146, normalized size = 1.40 \begin {gather*} -\frac {2 a^3 c+2 a^3 d x-4 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+2 a^2 b \cos (c+d x)+a b^2 \cot \left (\frac {1}{2} (c+d x)\right )-2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-a b^2 \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

-1/2*(2*a^3*c + 2*a^3*d*x - 4*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 2*a^2*b*Cos
[c + d*x] + a*b^2*Cot[(c + d*x)/2] - 2*b^3*Log[Cos[(c + d*x)/2]] + 2*b^3*Log[Sin[(c + d*x)/2]] - a*b^2*Tan[(c
+ d*x)/2])/(a^2*b^2*d)

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Maple [A]
time = 0.23, size = 155, normalized size = 1.49

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {\left (4 a^{4}-8 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(155\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {\left (4 a^{4}-8 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(155\)
risch	\(-\frac {a x}{b^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}\)	\(308\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/2/a*tan(1/2*d*x+1/2*c)-2/b^2*(b/(1+tan(1/2*d*x+1/2*c)^2)+a*arctan(tan(1/2*d*x+1/2*c)))+1/2*(4*a^4-8*a^2
*b^2+4*b^4)/a^2/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/2/a/tan(1/2*d*x
+1/2*c)-1/a^2*b*ln(tan(1/2*d*x+1/2*c)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.47, size = 396, normalized size = 3.81 \begin {gather*} \left [\frac {b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{3} d x + a^{2} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, a^{2} b^{2} d \sin \left (d x + c\right )}, \frac {b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{3} d x + a^{2} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, a^{2} b^{2} d \sin \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*(b^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - b^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*a*b^2*c
os(d*x + c) - (a^2 - b^2)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2
+ 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c)
- a^2 - b^2))*sin(d*x + c) - 2*(a^3*d*x + a^2*b*cos(d*x + c))*sin(d*x + c))/(a^2*b^2*d*sin(d*x + c)), 1/2*(b^3
*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - b^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*a*b^2*cos(d*x +
c) - 2*(a^2 - b^2)^(3/2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) - 2*(a^3*d*
x + a^2*b*cos(d*x + c))*sin(d*x + c))/(a^2*b^2*d*sin(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*cot(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2*cot(c + d*x)**2/(a + b*sin(c + d*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (99) = 198\).
time = 5.61, size = 221, normalized size = 2.12 \begin {gather*} -\frac {\frac {6 \, {\left (d x + c\right )} a}{b^{2}} + \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{2}} - \frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2} b}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(6*(d*x + c)*a/b^2 + 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - 12*(a^4 - 2*a^2*
b^2 + b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(s
qrt(a^2 - b^2)*a^2*b^2) - (2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*tan(1/2*d*x +
1/2*c) + 2*b^2*tan(1/2*d*x + 1/2*c) - 3*a*b)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*a^2*b))/d

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Mupad [B]
time = 6.11, size = 1167, normalized size = 11.22 \begin {gather*} -\frac {\mathrm {cot}\left (c+d\,x\right )}{a\,d}-\frac {b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^2\,d}-\frac {\sin \left (2\,c+2\,d\,x\right )}{2\,b\,d\,\sin \left (c+d\,x\right )}-\frac {2\,a\,\mathrm {atan}\left (\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}+\frac {\mathrm {atan}\left (\frac {16\,b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}-4\,a^{12}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-4\,a^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}-12\,a^3\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}+a^5\,b^7\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+4\,a^7\,b^5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-6\,a^9\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-29\,a^2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}+18\,a^4\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}+a^2\,b^{10}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-4\,a^4\,b^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+22\,a^6\,b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-32\,a^8\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+18\,a^{10}\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+8\,a\,b^5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}+5\,a^5\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6\right )}^{3/2}+2\,a^{11}\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{-3{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^{14}\,b+9{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^{13}\,b^2+36{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^{12}\,b^3-51{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^{11}\,b^4-151{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^{10}\,b^5+126{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^9\,b^6+323{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^8\,b^7-167{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^7\,b^8-390{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6\,b^9+123{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5\,b^{10}+269{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^{11}-48{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^{12}-100{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^{13}+8{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^{14}+16{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{15}}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\,2{}\mathrm {i}}{a^2\,b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*cot(c + d*x)^2)/(a + b*sin(c + d*x)),x)

[Out]

(atan((16*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 4*a^12*sin(c/2 + (d*x)/2)*(b^6 -
a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 4*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 12*a
^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a^5*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 -
3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 4*a^7*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 6*a^9*
b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 29*a^2*b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 -
3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 18*a^4*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a^2*b
^10*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 4*a^4*b^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3
*a^2*b^4 + 3*a^4*b^2)^(1/2) + 22*a^6*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 32*a^8
*b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 18*a^10*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6
- 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 8*a*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 5*a^5*
b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 2*a^11*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^
2*b^4 + 3*a^4*b^2)^(1/2))/(b^15*sin(c/2 + (d*x)/2)*16i + a*b^14*cos(c/2 + (d*x)/2)*8i - a^14*b*sin(c/2 + (d*x)
/2)*3i - a^3*b^12*cos(c/2 + (d*x)/2)*48i + a^5*b^10*cos(c/2 + (d*x)/2)*123i - a^7*b^8*cos(c/2 + (d*x)/2)*167i
+ a^9*b^6*cos(c/2 + (d*x)/2)*126i - a^11*b^4*cos(c/2 + (d*x)/2)*51i + a^13*b^2*cos(c/2 + (d*x)/2)*9i - a^2*b^1
3*sin(c/2 + (d*x)/2)*100i + a^4*b^11*sin(c/2 + (d*x)/2)*269i - a^6*b^9*sin(c/2 + (d*x)/2)*390i + a^8*b^7*sin(c
/2 + (d*x)/2)*323i - a^10*b^5*sin(c/2 + (d*x)/2)*151i + a^12*b^3*sin(c/2 + (d*x)/2)*36i))*(-(a + b)^3*(a - b)^
3)^(1/2)*2i)/(a^2*b^2*d) - (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^2*d) - sin(2*c + 2*d*x)/(2*b*d*si
n(c + d*x)) - (2*a*atan((a^3*cos(c/2 + (d*x)/2) + b^3*sin(c/2 + (d*x)/2))/(b^3*cos(c/2 + (d*x)/2) - a^3*sin(c/
2 + (d*x)/2))))/(b^2*d) - cot(c + d*x)/(a*d)

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