3.232 \(\int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\)

Optimal. Leaf size=86 \[ -\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{5 a^3 A \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d} \]

[Out]

(5*a^3*A*ArcTanh[Cos[c + d*x]])/(8*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (3*a^3*A*Cot[c + d*x]*Csc[c + d*x])/(
8*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.147784, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {2966, 3770, 3767, 8, 3768} \[ -\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{5 a^3 A \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(5*a^3*A*ArcTanh[Cos[c + d*x]])/(8*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (3*a^3*A*Cot[c + d*x]*Csc[c + d*x])/(
8*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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 3768

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

Rubi steps

\begin{align*} \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\int \left (-a^3 A \csc (c+d x)-2 a^3 A \csc ^2(c+d x)+2 a^3 A \csc ^4(c+d x)+a^3 A \csc ^5(c+d x)\right ) \, dx\\ &=-\left (\left (a^3 A\right ) \int \csc (c+d x) \, dx\right )+\left (a^3 A\right ) \int \csc ^5(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc ^2(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^4(c+d x) \, dx\\ &=\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{4} \left (3 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\frac{\left (2 a^3 A\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 A\right ) \int \csc (c+d x) \, dx\\ &=\frac{5 a^3 A \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [B]  time = 0.0689217, size = 210, normalized size = 2.44 \[ a^3 A \left (-\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{3 d}+\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{3 d}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{3 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{3 \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}-\frac{5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{12 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{12 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*(Cot[(c + d*x)/2]/(3*d) - (3*Csc[(c + d*x)/2]^2)/(32*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(12*d) -
Csc[(c + d*x)/2]^4/(64*d) + (5*Log[Cos[(c + d*x)/2]])/(8*d) - (5*Log[Sin[(c + d*x)/2]])/(8*d) + (3*Sec[(c + d
*x)/2]^2)/(32*d) + Sec[(c + d*x)/2]^4/(64*d) - Tan[(c + d*x)/2]/(3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/
(12*d))

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Maple [A]  time = 0.058, size = 109, normalized size = 1.3 \begin{align*} -{\frac{5\,{a}^{3}A\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{3}A\cot \left ( dx+c \right ) }{3\,d}}-{\frac{2\,{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{3}A\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

-5/8/d*a^3*A*ln(csc(d*x+c)-cot(d*x+c))+2/3*a^3*A*cot(d*x+c)/d-2/3/d*a^3*A*cot(d*x+c)*csc(d*x+c)^2-1/4*a^3*A*co
t(d*x+c)*csc(d*x+c)^3/d-3/8*a^3*A*cot(d*x+c)*csc(d*x+c)/d

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Maxima [A]  time = 0.973548, size = 196, normalized size = 2.28 \begin{align*} \frac{3 \, A a^{3}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{3}{\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{96 \, A a^{3}}{\tan \left (d x + c\right )} - \frac{32 \,{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} A a^{3}}{\tan \left (d x + c\right )^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*A*a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x +
c) + 1) + 3*log(cos(d*x + c) - 1)) + 24*A*a^3*(log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 96*A*a^3/tan(
d*x + c) - 32*(3*tan(d*x + c)^2 + 1)*A*a^3/tan(d*x + c)^3)/d

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Fricas [B]  time = 1.98415, size = 431, normalized size = 5.01 \begin{align*} -\frac{32 \, A a^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 18 \, A a^{3} \cos \left (d x + c\right )^{3} + 30 \, A a^{3} \cos \left (d x + c\right ) - 15 \,{\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 15 \,{\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{48 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(32*A*a^3*cos(d*x + c)^3*sin(d*x + c) - 18*A*a^3*cos(d*x + c)^3 + 30*A*a^3*cos(d*x + c) - 15*(A*a^3*cos(
d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 + A*a^3)*log(1/2*cos(d*x + c) + 1/2) + 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*
cos(d*x + c)^2 + A*a^3)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + 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(csc(d*x+c)**5*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.24556, size = 235, normalized size = 2.73 \begin{align*} \frac{3 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 120 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 48 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{250 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 48 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(3*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 16*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 24*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 12
0*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 48*A*a^3*tan(1/2*d*x + 1/2*c) + (250*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 4
8*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 24*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 16*A*a^3*tan(1/2*d*x + 1/2*c) - 3*A*a^3)/ta
n(1/2*d*x + 1/2*c)^4)/d