3.1130 \(\int \frac{\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=137 \[ \frac{2 \sqrt{a^2-b^2} \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 a x}{b^3}-\frac{\cos (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.273874, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2892, 3057, 2660, 618, 204, 3770} \[ \frac{2 \sqrt{a^2-b^2} \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 a x}{b^3}-\frac{\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 2892

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[((a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*b^2*d*
f*(m + 1)), x] + (-Dist[1/(a*b^2*(m + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Sim
p[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2)*(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 + 2)*(d*Sin[e +
f*x])^(n + 1))/(b^2*d*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2
*m, 2*n] && LtQ[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 ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac{\cos (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (b^2-a b \sin (c+d x)-2 a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{a b^2}\\ &=-\frac{2 a x}{b^3}-\frac{\cos (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \csc (c+d x) \, dx}{a^2}-\frac{\left (-2 a^4+a^2 b^2+b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^2 b^3}\\ &=-\frac{2 a x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cos (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac{\left (2 \left (-2 a^4+a^2 b^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^2 b^3 d}\\ &=-\frac{2 a x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cos (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\left (4 \left (-2 a^4+a^2 b^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^2 b^3 d}\\ &=-\frac{2 a x}{b^3}+\frac{2 \left (2 a^4-a^2 b^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^2 b^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cos (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.70204, size = 161, normalized size = 1.18 \[ \frac{\frac{2 \left (-a^2 b^2+2 a^4-b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 \sqrt{a^2-b^2}}+\frac{\left (b^2-a^2\right ) \cos (c+d x)}{a b^2 (a+b \sin (c+d x))}+\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}-\frac{2 a (c+d x)}{b^3}-\frac{\cos (c+d x)}{b^2}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.151, size = 380, normalized size = 2.8 \begin{align*} -2\,{\frac{1}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{3}}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{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{\tan \left ( 1/2\,dx+c/2 \right ) 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} \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}{da \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 ) }}+4\,{\frac{{a}^{2}}{d{b}^{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 ) }-2\,{\frac{1}{bd\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}{d{a}^{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 ) }+{\frac{1}{d{a}^{2}}\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)^4*csc(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.02764, size = 1226, normalized size = 8.95 \begin{align*} \left [-\frac{4 \, a^{4} d x -{\left (2 \, a^{3} + a b^{2} +{\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\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 ) + 2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) +{\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (2 \, a^{3} b d x + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{4} d \sin \left (d x + c\right ) + a^{3} b^{3} d\right )}}, -\frac{4 \, a^{4} d x + 2 \,{\left (2 \, a^{3} + a b^{2} +{\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) +{\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (2 \, a^{3} b d x + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{4} d \sin \left (d x + c\right ) + a^{3} b^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{4}{\left (c + d x \right )} \csc{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.38625, size = 386, normalized size = 2.82 \begin{align*} -\frac{\frac{2 \,{\left (d x + c\right )} a}{b^{3}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{2 \,{\left (2 \, a^{4} - 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^{3}} + \frac{2 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, 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 )} a^{2} b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(d*x + c)*a/b^3 - log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 2*(2*a^4 - a^2*b^2 - b^4)*(pi*floor(1/2*(d*x + c)/p
i + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^2*b^3) + 2*(a^2*b*t
an(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 + 2*a^3*tan(1/2*d*x + 1/2*c)^2 - a*b^2*tan(1/2*d*x + 1/2*c)
^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c) + 2*a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^4 + 2*b
*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^2*b^2))/d