3.1306 \(\int \frac{\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=123 \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b d}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{x}{b} \]

[Out]

x/b - (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b*d) + ((3*a^2 - 2*b^2)*ArcT
anh[Cos[c + d*x]])/(2*a^3*d) + (b*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.301393, antiderivative size = 123, 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 = {2893, 3057, 2660, 618, 204, 3770} \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b d}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/b - (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b*d) + ((3*a^2 - 2*b^2)*ArcT
anh[Cos[c + d*x]])/(2*a^3*d) + (b*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

Rule 2893

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^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[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 (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}-\frac{\int \frac{\csc (c+d x) \left (3 a^2-2 b^2-a b \sin (c+d x)-2 a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac{x}{b}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}-\frac{\left (3 a^2-2 b^2\right ) \int \csc (c+d x) \, dx}{2 a^3}-\frac{\left (a^2-b^2\right )^2 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3 b}\\ &=\frac{x}{b}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 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^3 b d}\\ &=\frac{x}{b}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 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^3 b d}\\ &=\frac{x}{b}-\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^3 b d}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}\\ \end{align*}

Mathematica [A]  time = 1.53311, size = 204, normalized size = 1.66 \[ \frac{-16 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-a^2 b \csc ^2\left (\frac{1}{2} (c+d x)\right )+a^2 b \sec ^2\left (\frac{1}{2} (c+d x)\right )-12 a^2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 a^3 c+8 a^3 d x-4 a b^2 \tan \left (\frac{1}{2} (c+d x)\right )+4 a b^2 \cot \left (\frac{1}{2} (c+d x)\right )+8 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^3*c + 8*a^3*d*x - 16*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 4*a*b^2*Cot[(c
+ d*x)/2] - a^2*b*Csc[(c + d*x)/2]^2 + 12*a^2*b*Log[Cos[(c + d*x)/2]] - 8*b^3*Log[Cos[(c + d*x)/2]] - 12*a^2*b
*Log[Sin[(c + d*x)/2]] + 8*b^3*Log[Sin[(c + d*x)/2]] + a^2*b*Sec[(c + d*x)/2]^2 - 4*a*b^2*Tan[(c + d*x)/2])/(8
*a^3*b*d)

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Maple [B]  time = 0.12, size = 286, normalized size = 2.3 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{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 ) }{bd}}-2\,{\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 ) }+4\,{\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 ) }-2\,{\frac{{b}^{3}}{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}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.65531, size = 1337, normalized size = 10.87 \begin{align*} \left [\frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) - 2 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 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 ) -{\left (3 \, a^{2} b - 2 \, b^{3} -{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (3 \, a^{2} b - 2 \, b^{3} -{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b d\right )}}, \frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\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 ) -{\left (3 \, a^{2} b - 2 \, b^{3} -{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (3 \, a^{2} b - 2 \, b^{3} -{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b 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)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.24938, size = 293, normalized size = 2.38 \begin{align*} \frac{\frac{8 \,{\left (d x + c\right )}}{b} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{4 \,{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{16 \,{\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^{3} b} + \frac{18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)/b + (a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(3*a^2 - 2*b^2)*log(abs(tan
(1/2*d*x + 1/2*c)))/a^3 - 16*(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^3*b) + (18*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(
1/2*d*x + 1/2*c)^2 + 4*a*b*tan(1/2*d*x + 1/2*c) - a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d