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

Optimal. Leaf size=124 \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^3 d}+\frac{x \left (2 a^2-3 b^2\right )}{2 b^3}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.279724, antiderivative size = 124, 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 = {2895, 3057, 2660, 618, 204, 3770} \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^3 d}+\frac{x \left (2 a^2-3 b^2\right )}{2 b^3}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && 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)} \, dx &=\frac{a \cos (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\int \frac{\csc (c+d x) \left (-2 b^2-a b \sin (c+d x)-\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 b^2}\\ &=\frac{\left (2 a^2-3 b^2\right ) x}{2 b^3}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \csc (c+d x) \, dx}{a}-\frac{\left (a^2-b^2\right )^2 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a b^3}\\ &=\frac{\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b 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 b^3 d}\\ &=\frac{\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b 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 b^3 d}\\ &=\frac{\left (2 a^2-3 b^2\right ) x}{2 b^3}-\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 b^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{a \cos (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.281121, size = 143, normalized size = 1.15 \[ -\frac{8 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-4 a^2 b \cos (c+d x)-4 a^3 c-4 a^3 d x+a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x-4 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 a b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-4*a^3*c + 6*a*b^2*c - 4*a^3*d*x + 6*a*b^2*d*x + 8*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^
2 - b^2]] - 4*a^2*b*Cos[c + d*x] + 4*b^3*Log[Cos[(c + d*x)/2]] - 4*b^3*Log[Sin[(c + d*x)/2]] + a*b^2*Sin[2*(c
+ d*x)])/(4*a*b^3*d)

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Maple [B]  time = 0.083, size = 334, normalized size = 2.7 \begin{align*}{\frac{1}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{bd}}-2\,{\frac{{a}^{3}}{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 ) }+4\,{\frac{a}{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}{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 ) }+{\frac{1}{da}\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)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.0426, size = 853, normalized size = 6.88 \begin{align*} \left [-\frac{a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - b^{3} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x -{\left (-a^{2} + b^{2}\right )}^{\frac{3}{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 \, a b^{3} d}, -\frac{a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - b^{3} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 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 )}{2 \, a b^{3} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 2.08771, size = 247, normalized size = 1.99 \begin{align*} \frac{\frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{{\left (2 \, a^{2} - 3 \, b^{2}\right )}{\left (d x + c\right )}}{b^{3}} - \frac{4 \,{\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 b^{3}} + \frac{2 \,{\left (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} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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