3.4 \(\int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx\)

Optimal. Leaf size=33 \[ -\frac{a A \cos (c+d x)}{d}-\frac{a A \tanh ^{-1}(\cos (c+d x))}{d}+2 a A x \]

[Out]

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

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Rubi [A]  time = 0.0594924, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {21, 3788, 8, 4045, 3770} \[ -\frac{a A \cos (c+d x)}{d}-\frac{a A \tanh ^{-1}(\cos (c+d x))}{d}+2 a A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

Rubi steps

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

Mathematica [B]  time = 0.0225814, size = 72, normalized size = 2.18 \[ \frac{a A \sin (c) \sin (d x)}{d}-\frac{a A \cos (c) \cos (d x)}{d}+\frac{a A \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}-\frac{a A \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}+2 a A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.079, size = 50, normalized size = 1.5 \begin{align*} 2\,aAx-{\frac{Aa\cos \left ( dx+c \right ) }{d}}+{\frac{Aa\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Aac}{d}} \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),x)

[Out]

2*a*A*x-a*A*cos(d*x+c)/d+1/d*A*a*ln(csc(d*x+c)-cot(d*x+c))+2/d*A*a*c

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Maxima [A]  time = 0.988842, size = 68, normalized size = 2.06 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a - A a{\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.507614, size = 147, normalized size = 4.45 \begin{align*} \frac{4 \, A a d x - 2 \, A a \cos \left (d x + c\right ) - A a \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + A a \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.35554, size = 63, normalized size = 1.91 \begin{align*} \frac{2 \,{\left (d x + c\right )} A a + A a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \, A a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{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),x, algorithm="giac")

[Out]

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