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

Optimal. Leaf size=29 \[ -\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} a A x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3962, 2635, 8} \[ -\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} a A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*A*x)/2 - (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 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 3962

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*} \int (a-a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^2(c+d x) \, dx &=-\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*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.86 \[ -\frac {a A (2 (c+d x)+\sin (2 (c+d x)))}{4 d} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.49, size = 26, normalized size = 0.90 \[ -\frac {A a d x + A a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.46, size = 35, normalized size = 1.21 \[ -\frac {{\left (d x + c\right )} A a + \frac {A a \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

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)^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

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maple [A]  time = 0.50, size = 40, normalized size = 1.38 \[ \frac {a A \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a A \left (d x +c \right )}{d} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 37, normalized size = 1.28 \[ \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 4 \, {\left (d x + c\right )} A a}{4 \, d} \]

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)^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

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mupad [B]  time = 0.66, size = 53, normalized size = 1.83 \[ \frac {A\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-A\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {A\,a\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 5.53, size = 54, normalized size = 1.86 \[ - A a x + A a \left (\begin {cases} \frac {x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {\sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \sin ^{2}{\relax (c )} & \text {otherwise} \end {cases}\right ) \]

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)**2,x)

[Out]

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

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