3.1131 \(\int \frac{\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=154 \[ -\frac{2 \left (a^2 b^2+a^4-2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^2 d \sqrt{a^2-b^2}}+\frac{\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a d (a+b \sin (c+d x))}+\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.303716, antiderivative size = 154, 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 = {2890, 3057, 2660, 618, 204, 3770} \[ -\frac{2 \left (a^2 b^2+a^4-2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^2 d \sqrt{a^2-b^2}}+\frac{\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a d (a+b \sin (c+d x))}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2890

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]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b*d*(n + 1)*(m + 1)), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1)*Simp[a^2*(n + 1)
*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m
+ n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[((a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n
+ 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*b*d^2*f*(n + 1)*(m + 1)), x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]

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

Mathematica [A]  time = 1.76709, size = 182, normalized size = 1.18 \[ \frac{-\frac{4 \left (a^2 b^2+a^4-2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^2 \sqrt{a^2-b^2}}+\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a^2 b (a+b \sin (c+d x))}-\frac{4 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{4 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{a^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{a^2}+\frac{2 (c+d x)}{b^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.161, size = 396, normalized size = 2.6 \begin{align*}{\frac{1}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{2}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) a}}-2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{1}{bd \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{b}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{a}{d{b}^{2}\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{1}{da\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{{b}^{2}}{d{a}^{3}\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\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-2\,{\frac{b\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)+2/d/b^2*arctan(tan(1/2*d*x+1/2*c))+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*
c)*b+a)/a*tan(1/2*d*x+1/2*c)-2/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b^2*tan(1/2*d*x+1/2*c)+
2/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b
+a)*b-2/d/b^2*a/(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/a/(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))+4/d/a^3/(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))*b^2-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^3*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)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.93085, size = 1557, normalized size = 10.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(2*a^3*b*d*x*cos(d*x + c)^2 - 2*a^3*b*d*x + 2*a^2*b^2*cos(d*x + c) - (a^2*b + 2*b^3 - (a^2*b + 2*b^3)*cos
(d*x + c)^2 + (a^3 + 2*a*b^2)*sin(d*x + c))*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)) + 2*(b^4*cos(d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(1/2*cos(d*x + c) + 1/2
) - 2*(b^4*cos(d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^4*d*x + (a^3*b - 2*a
*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^4*b^2*d*sin(d*x + c) - a^3*b^3*d), (a^3*b*d*x*
cos(d*x + c)^2 - a^3*b*d*x + a^2*b^2*cos(d*x + c) - (a^2*b + 2*b^3 - (a^2*b + 2*b^3)*cos(d*x + c)^2 + (a^3 + 2
*a*b^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (b^4*cos(
d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(1/2*cos(d*x + c) + 1/2) - (b^4*cos(d*x + c)^2 - a*b^3*sin(d*x + c)
- b^4)*log(-1/2*cos(d*x + c) + 1/2) - (a^4*d*x + (a^3*b - 2*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(
d*x + c)^2 - a^4*b^2*d*sin(d*x + c) - a^3*b^3*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.37536, size = 351, normalized size = 2.28 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}}{b^{2}} - \frac{12 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{12 \,{\left (a^{4} + a^{2} b^{2} - 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^{3} b^{2}} + \frac{4 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 14 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} a^{3} b}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)/b^2 - 12*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 3*tan(1/2*d*x + 1/2*c)/a^2 - 12*(a^4 + a^2*b^
2 - 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^3*b^2) + (4*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 4*b^3*tan(1/2*d*x
+ 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) - 14*a*b^2*tan(1/2*d*x + 1/2*c) - 3*a^2*b)/((a*tan(1/2*d*x + 1/2*c)^
3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))*a^3*b))/d