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

Optimal. Leaf size=42 \[ -\frac{2 a A \cos (c+d x)}{d}-\frac{a A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a A x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0733261, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {21, 3788, 2638, 4045, 8} \[ -\frac{2 a A \cos (c+d x)}{d}-\frac{a A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a A x}{2} \]

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]

(3*a*A*x)/2 - (2*a*A*Cos[c + d*x])/d - (a*A*Cos[c + d*x]*Sin[c + d*x])/(2*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 2638

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0632426, size = 33, normalized size = 0.79 \[ -\frac{a A (-6 (c+d x)+\sin (2 (c+d x))+8 \cos (c+d x))}{4 d} \]

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]

-(a*A*(-6*(c + d*x) + 8*Cos[c + d*x] + Sin[2*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.078, size = 49, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( Aa \left ( -{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) -2\,Aa\cos \left ( dx+c \right ) +Aa \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)^2,x)

[Out]

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

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Maxima [A]  time = 0.990114, size = 63, normalized size = 1.5 \begin{align*} \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 - 8 \, A a \cos \left (d x + c\right )}{4 \, 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)^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 - 8*A*a*cos(d*x + c))/d

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Fricas [A]  time = 0.476211, size = 97, normalized size = 2.31 \begin{align*} \frac{3 \, A a d x - A a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, A a \cos \left (d x + c\right )}{2 \, 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)^2,x, algorithm="fricas")

[Out]

1/2*(3*A*a*d*x - A*a*cos(d*x + c)*sin(d*x + c) - 4*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 ^{2}{\left (c + d x \right )} \csc{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\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)**2,x)

[Out]

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

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Giac [B]  time = 1.49724, size = 107, normalized size = 2.55 \begin{align*} \frac{3 \,{\left (d x + c\right )} A a + \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, 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)^2,x, algorithm="giac")

[Out]

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