3.1088 \(\int \frac{\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.477451, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2889, 3056, 3001, 3770, 2660, 618, 204} \[ -\frac{b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac{\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 2889

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rule 3770

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

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])

Rubi steps

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

Mathematica [A]  time = 1.08966, size = 154, normalized size = 1. \[ \frac{\frac{2 b \left (2 b^2-3 a^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a \cos (c+d x) \left (b \left (a^2-2 b^2\right ) \sin (c+d x)+2 a^3-3 a b^2\right )}{(a-b) (a+b) (a+b \sin (c+d x))^2}+2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

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

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Maple [B]  time = 0.18, size = 632, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-4/d/a^2/(tan(1/2*d*x+
1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+2/d*a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2+1/d*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-6/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(
1/2*d*x+1/2*c)^2*b^4+5/d*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)-8/
d/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^3/(a^2-b^2)*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)^2*a/(a^2-b^2)-3/d/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^2
/(a^2-b^2)-3/d/a*b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/d/a^3*b^3/(a^2-b
^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+1/d/a^3*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*csc(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.69597, size = 2190, normalized size = 14.22 \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)^2*csc(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)*sin(d*x + c) - (3*a^4*b + a^2*b^3 - 2*b^5 - (3*a^2*b^3 - 2
*b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 2*a*b^4)*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*(2*a^6 - 5*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c) - 2*(a^6 -
a^4*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x +
c))*log(1/2*cos(d*x + c) + 1/2) + 2*(a^6 - a^4*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)
^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7*b^2 - 2*a^5*b^4 + a^3*b^6
)*d*cos(d*x + c)^2 - 2*(a^8*b - 2*a^6*b^3 + a^4*b^5)*d*sin(d*x + c) - (a^9 - a^7*b^2 - a^5*b^4 + a^3*b^6)*d),
-1/2*((a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)*sin(d*x + c) + (3*a^4*b + a^2*b^3 - 2*b^5 - (3*a^2*b^3 - 2*b^
5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 2*a*b^4)*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))) + (2*a^6 - 5*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c) - (a^6 - a^4*b^2 - a^2*b^4 + b^6 - (a^
4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1
/2) + (a^6 - a^4*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b
^5)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7*b^2 - 2*a^5*b^4 + a^3*b^6)*d*cos(d*x + c)^2 - 2*(a^8*b -
2*a^6*b^3 + a^4*b^5)*d*sin(d*x + c) - (a^9 - a^7*b^2 - a^5*b^4 + a^3*b^6)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )} \csc{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)/(a + b*sin(c + d*x))**3, x)

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Giac [A]  time = 1.37014, size = 374, normalized size = 2.43 \begin{align*} -\frac{\frac{{\left (3 \, a^{2} b - 2 \, b^{3}\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 )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{5} - a^{3} b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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