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

Optimal. Leaf size=54 \[ \frac{a A \cos ^3(c+d x)}{3 d}-\frac{2 a A \cos (c+d x)}{d}-\frac{a A \sin (c+d x) \cos (c+d x)}{d}+a A x \]

[Out]

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

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Rubi [A]  time = 0.0940265, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {21, 3788, 2635, 8, 4044, 3013} \[ \frac{a A \cos ^3(c+d x)}{3 d}-\frac{2 a A \cos (c+d x)}{d}-\frac{a A \sin (c+d x) \cos (c+d x)}{d}+a A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 2635

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

Rule 8

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

Mathematica [A]  time = 0.0841558, size = 43, normalized size = 0.8 \[ \frac{a A (-6 \sin (2 (c+d x))-21 \cos (c+d x)+\cos (3 (c+d x))+12 c+12 d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*(12*c + 12*d*x - 21*Cos[c + d*x] + Cos[3*(c + d*x)] - 6*Sin[2*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.089, size = 62, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{\frac{Aa \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}+2\,Aa \left ( -1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) -Aa\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/3*A*a*(2+sin(d*x+c)^2)*cos(d*x+c)+2*A*a*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-A*a*cos(d*x+c))

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Maxima [A]  time = 0.96236, size = 81, normalized size = 1.5 \begin{align*} \frac{2 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a + 3 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 6 \, A a \cos \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*A*a - 6*A*a*cos(d*x + c))/d

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Fricas [A]  time = 0.504407, size = 128, normalized size = 2.37 \begin{align*} \frac{A a \cos \left (d x + c\right )^{3} + 3 \, A a d x - 3 \, A a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, A a \cos \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*a*cos(d*x + c)^3 + 3*A*a*d*x - 3*A*a*cos(d*x + c)*sin(d*x + c) - 6*A*a*cos(d*x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39078, size = 128, normalized size = 2.37 \begin{align*} \frac{3 \,{\left (d x + c\right )} A a + \frac{2 \,{\left (3 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, A a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*A*a + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 - 3*A*a*tan(1/2*d*x + 1/2*c)^4 - 12*A*a*tan(1/2*d*x + 1
/2*c)^2 - 3*A*a*tan(1/2*d*x + 1/2*c) - 5*A*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d