\(\int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx\) [17]

Optimal result

Integrand size = 30, antiderivative size = 29 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=-\frac {1}{2} a A x-\frac {a A \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4047, 2715, 8} \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=-\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} a A x \]

[In]

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

[Out]

-1/2*(a*A*x) - (a*A*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_))^(n_.), x_Symbol] :> Dist[((-a)*c)^m, Int[ExpandTrig[(g*csc[e + f*x])^p*cot[e + f*x]^(2*m), (c
+ d*csc[e + f*x])^(n - m), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps \begin{align*} \text {integral}& = -\left ((a A) \int \cos ^2(c+d x) \, dx\right ) \\ & = -\frac {a A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} (a A) \int 1 \, dx \\ & = -\frac {1}{2} a A x-\frac {a A \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=-\frac {a A (2 (c+d x)+\sin (2 (c+d x)))}{4 d} \]

[In]

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

[Out]

-1/4*(a*A*(2*(c + d*x) + Sin[2*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(-\frac {A a \left (2 d x +\sin \left (2 d x +2 c \right )\right )}{4 d}\)	\(22\)
risch	\(-\frac {a A x}{2}-\frac {A a \sin \left (2 d x +2 c \right )}{4 d}\)	\(23\)
parts	\(\frac {A a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-a A x\)	\(35\)
derivativedivides	\(\frac {A a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A a \left (d x +c \right )}{d}\)	\(40\)
default	\(\frac {A a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A a \left (d x +c \right )}{d}\)	\(40\)
norman	\(\frac {\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\)	\(110\)

[In]

int((a+a*csc(d*x+c))*(A-A*csc(d*x+c))/csc(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*A*a*(2*d*x+sin(2*d*x+2*c))/d

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=-\frac {A a d x + A a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (27) = 54\).

Time = 3.38 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.93 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=\begin {cases} \frac {A a x \cot ^{2}{\left (c + d x \right )}}{2 \csc ^{2}{\left (c + d x \right )}} - A a x + \frac {A a x}{2 \csc ^{2}{\left (c + d x \right )}} - \frac {A a \cot {\left (c + d x \right )}}{2 d \csc ^{2}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \left (- A \csc {\left (c \right )} + A\right ) \left (a \csc {\left (c \right )} + a\right )}{\csc ^{2}{\left (c \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*a*x*cot(c + d*x)**2/(2*csc(c + d*x)**2) - A*a*x + A*a*x/(2*csc(c + d*x)**2) - A*a*cot(c + d*x)/(2
*d*csc(c + d*x)**2), Ne(d, 0)), (x*(-A*csc(c) + A)*(a*csc(c) + a)/csc(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 4 \, {\left (d x + c\right )} A a}{4 \, d} \]

[In]

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

[Out]

1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*A*a - 4*(d*x + c)*A*a)/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=-\frac {{\left (d x + c\right )} A a + \frac {A a \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

-1/2*((d*x + c)*A*a + A*a*tan(d*x + c)/(tan(d*x + c)^2 + 1))/d

Mupad [B] (verification not implemented)

Time = 19.43 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int (a+a \csc (c+d x)) (A-A \csc (c+d x)) \sin ^2(c+d x) \, dx=\frac {A\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-A\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {A\,a\,x}{2} \]

[In]

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

[Out]

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