3.1138 \(\int \frac{\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=175 \[ -\frac{\left (-a^2 b^2+2 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d \sqrt{a^2-b^2}}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.287894, antiderivative size = 175, 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 = {2891, 3057, 2660, 618, 204, 3770} \[ -\frac{\left (-a^2 b^2+2 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d \sqrt{a^2-b^2}}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2891

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^2*b^2*(m + 1)*(m + 2)), Int[(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^n*Simp[
a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 2)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m +
n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] + Simp[((a^2*(n - m + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(a +
b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^(n + 1))/(a^2*b^2*d*f*(m + 1)*(m + 2)), 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] && (LtQ[m, -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 ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\int \frac{\csc (c+d x) \left (-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 b^2}\\ &=\frac{x}{b^3}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac{\int \csc (c+d x) \, dx}{a^3}+\frac{\left (-2 a^4+a^2 b^2-2 b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 a^3 b^3}\\ &=\frac{x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (2 a^4-a^2 b^2+2 b^4\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^3 d}\\ &=\frac{x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac{\left (2 \left (2 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^3 d}\\ &=\frac{x}{b^3}-\frac{\left (2 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^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.74973, size = 176, normalized size = 1.01 \[ \frac{-\frac{2 \left (-a^2 b^2+2 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 \sqrt{a^2-b^2}}+\frac{\cos (c+d x) \left (b \left (3 a^2+2 b^2\right ) \sin (c+d x)+2 a^3+3 a b^2\right )}{a^2 b^2 (a+b \sin (c+d x))^2}+2 \left (\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{c+d x}{b^3}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.184, size = 600, normalized size = 3.4 \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)^4*csc(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

2/d/b^3*arctan(tan(1/2*d*x+1/2*c))+1/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c
)^3+4/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3+2/d*a/b^2/(tan(1/2*d*x+
1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2+7/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b
+a)^2/a*tan(1/2*d*x+1/2*c)^2+6/d/a^3*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c
)^2+7/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+8/d/a^2*b/(tan(1/2*d*x+1/2*c)
^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+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)^
2*a+3/d/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2-2/d*a/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/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/(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^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)^4*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 = 4.54033, size = 2211, normalized size = 12.63 \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)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.32965, size = 371, normalized size = 2.12 \begin{align*} \frac{\frac{d x + c}{b^{3}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{{\left (2 \, 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^{3}} + \frac{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} + 7 \, 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} + 7 \, 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 \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} a^{3} 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))^3,x, algorithm="giac")

[Out]

((d*x + c)/b^3 + log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (2*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)))/(sqrt(a^2 - b^2)*a^3*b^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 + 7*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 + 7*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*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^3*b^2))/d