3.241 \(\int \frac{\csc ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=113 \[ -\frac{A \cot (c+d x)}{a^3 d}+\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{104 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)}+\frac{31 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)^2}-\frac{2 A \cot (c+d x)}{5 a^3 d (\csc (c+d x)+1)^3} \]

[Out]

(4*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (A*Cot[c + d*x])/(a^3*d) - (2*A*Cot[c + d*x])/(5*a^3*d*(1 + Csc[c + d*x]
)^3) + (31*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])^2) - (104*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])
)

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Rubi [A]  time = 0.398275, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {2950, 2709, 3770, 3767, 8, 3777, 3922, 3919, 3794} \[ -\frac{A \cot (c+d x)}{a^3 d}+\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{104 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)}+\frac{31 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)^2}-\frac{2 A \cot (c+d x)}{5 a^3 d (\csc (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(4*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (A*Cot[c + d*x])/(a^3*d) - (2*A*Cot[c + d*x])/(5*a^3*d*(1 + Csc[c + d*x]
)^3) + (31*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])^2) - (104*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])
)

Rule 2950

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3922

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[((b
*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=(a A) \int \frac{\cot ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\\ &=\frac{A \int \left (\frac{9}{a^2}-\frac{4 \csc (c+d x)}{a^2}+\frac{\csc ^2(c+d x)}{a^2}-\frac{2}{a^2 (1+\csc (c+d x))^3}+\frac{9}{a^2 (1+\csc (c+d x))^2}-\frac{16}{a^2 (1+\csc (c+d x))}\right ) \, dx}{a}\\ &=\frac{9 A x}{a^3}+\frac{A \int \csc ^2(c+d x) \, dx}{a^3}-\frac{(2 A) \int \frac{1}{(1+\csc (c+d x))^3} \, dx}{a^3}-\frac{(4 A) \int \csc (c+d x) \, dx}{a^3}+\frac{(9 A) \int \frac{1}{(1+\csc (c+d x))^2} \, dx}{a^3}-\frac{(16 A) \int \frac{1}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac{9 A x}{a^3}+\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac{3 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))^2}-\frac{16 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}+\frac{(2 A) \int \frac{-5+2 \csc (c+d x)}{(1+\csc (c+d x))^2} \, dx}{5 a^3}-\frac{(3 A) \int \frac{-3+\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^3}+\frac{(16 A) \int -1 \, dx}{a^3}-\frac{A \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{2 A x}{a^3}+\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{A \cot (c+d x)}{a^3 d}-\frac{2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac{31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac{16 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac{(2 A) \int \frac{15-7 \csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^3}-\frac{(12 A) \int \frac{\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{A \cot (c+d x)}{a^3 d}-\frac{2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac{31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac{4 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}+\frac{(44 A) \int \frac{\csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^3}\\ &=\frac{4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{A \cot (c+d x)}{a^3 d}-\frac{2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac{31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac{104 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.09343, size = 167, normalized size = 1.48 \[ -\frac{A \left (-15 \tan \left (\frac{1}{2} (c+d x)\right )+15 \cot \left (\frac{1}{2} (c+d x)\right )+120 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-120 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (-354 \sin (c+d x)+79 \cos (2 (c+d x))-287)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}+\frac{38}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{12}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}\right )}{30 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

-(A*(15*Cot[(c + d*x)/2] - 120*Log[Cos[(c + d*x)/2]] + 120*Log[Sin[(c + d*x)/2]] + 12/(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^4 + 38/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*Sin[(c + d*x)/2]*(-287 + 79*Cos[2*(c + d*x)]
- 354*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 - 15*Tan[(c + d*x)/2]))/(30*a^3*d)

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Maple [A]  time = 0.169, size = 169, normalized size = 1.5 \begin{align*}{\frac{A}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{16\,A}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+8\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-{\frac{44\,A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+14\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-18\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-4\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A/a^3*tan(1/2*d*x+1/2*c)-16/5/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^5+8/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^4-44/3/d
*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3+14/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2-18/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)-1/2/d*A/
a^3/tan(1/2*d*x+1/2*c)-4/d*A/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.02189, size = 701, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/30*(3*A*((121*sin(d*x + c)/(cos(d*x + c) + 1) + 410*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 610*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 + 425*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 125*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5
)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a
^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 5*sin(d*x + c)/(a^3*(c
os(d*x + c) + 1))) + 2*A*(2*(115*sin(d*x + c)/(cos(d*x + c) + 1) + 185*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1
35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 32)/(a^3 + 5*a^3*sin(d*x + c
)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*log(sin(d*x + c)/
(cos(d*x + c) + 1))/a^3))/d

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Fricas [B]  time = 2.10721, size = 1065, normalized size = 9.42 \begin{align*} \frac{94 \, A \cos \left (d x + c\right )^{4} + 222 \, A \cos \left (d x + c\right )^{3} - 115 \, A \cos \left (d x + c\right )^{2} - 237 \, A \cos \left (d x + c\right ) + 30 \,{\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) -{\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) + 4 \, A\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 30 \,{\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) -{\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) + 4 \, A\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (94 \, A \cos \left (d x + c\right )^{3} - 128 \, A \cos \left (d x + c\right )^{2} - 243 \, A \cos \left (d x + c\right ) - 6 \, A\right )} \sin \left (d x + c\right ) + 6 \, A}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 5 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + 4 \, a^{3} d -{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(94*A*cos(d*x + c)^4 + 222*A*cos(d*x + c)^3 - 115*A*cos(d*x + c)^2 - 237*A*cos(d*x + c) + 30*(A*cos(d*x +
c)^4 - 2*A*cos(d*x + c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) - (A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 -
2*A*cos(d*x + c) - 4*A)*sin(d*x + c) + 4*A)*log(1/2*cos(d*x + c) + 1/2) - 30*(A*cos(d*x + c)^4 - 2*A*cos(d*x +
c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) - (A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - 4*
A)*sin(d*x + c) + 4*A)*log(-1/2*cos(d*x + c) + 1/2) + (94*A*cos(d*x + c)^3 - 128*A*cos(d*x + c)^2 - 243*A*cos(
d*x + c) - 6*A)*sin(d*x + c) + 6*A)/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 5*a^3*d*cos(d*x + c)^2 +
2*a^3*d*cos(d*x + c) + 4*a^3*d - (a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3
*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{A \left (\int - \frac{\csc ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx + \int \frac{\sin{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

-A*(Integral(-csc(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x) + Integral(sin(c
+ d*x)*csc(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x))/a**3

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Giac [A]  time = 1.16707, size = 197, normalized size = 1.74 \begin{align*} -\frac{\frac{120 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{15 \,{\left (8 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{4 \,{\left (135 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 435 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 605 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 385 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 104 \, A\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(120*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 15*A*tan(1/2*d*x + 1/2*c)/a^3 - 15*(8*A*tan(1/2*d*x + 1/2*c)
- A)/(a^3*tan(1/2*d*x + 1/2*c)) + 4*(135*A*tan(1/2*d*x + 1/2*c)^4 + 435*A*tan(1/2*d*x + 1/2*c)^3 + 605*A*tan(
1/2*d*x + 1/2*c)^2 + 385*A*tan(1/2*d*x + 1/2*c) + 104*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d