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

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

[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)

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Rubi [A]  time = 0.270288, antiderivative size = 104, normalized size of antiderivative = 1., 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 = {2894, 3057, 2660, 618, 204, 3770} \[ \frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^2 d}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{a x}{b^2}-\frac{\cot (c+d x)}{a d}-\frac{\cos (c+d x)}{b d} \]

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 2894

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 3057

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 2660

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 618

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 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 3770

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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.802352, size = 146, normalized size = 1.4 \[ -\frac{-4 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )+2 a^2 b \cos (c+d x)+2 a^3 c+2 a^3 d x-a b^2 \tan \left (\frac{1}{2} (c+d x)\right )+a b^2 \cot \left (\frac{1}{2} (c+d x)\right )+2 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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])/(2*a^2*b^2*d)

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Maple [B]  time = 0.076, size = 249, normalized size = 2.4 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{1}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{b}^{2}d}}+2\,{\frac{{a}^{2}}{{b}^{2}d\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-4\,{\frac{1}{d\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{b}^{2}}{{a}^{2}d\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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)

[Out]

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

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

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

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Fricas [A]  time = 2.95285, size = 981, normalized size = 9.43 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  time = 1.81875, size = 298, normalized size = 2.87 \begin{align*} -\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{align*}

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