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

Optimal. Leaf size=27 \[ \frac{a A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a A \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0529919, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3962, 2592, 321, 206} \[ \frac{a A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a A \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0244733, size = 46, normalized size = 1.7 \[ -a A \left (\frac{\cos (c+d x)}{d}+\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 38, normalized size = 1.4 \begin{align*} -{\frac{Aa\cos \left ( dx+c \right ) }{d}}-{\frac{Aa\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{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]

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

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Maxima [A]  time = 1.1122, size = 54, normalized size = 2. \begin{align*} \frac{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*(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 = 0.500541, size = 132, normalized size = 4.89 \begin{align*} -\frac{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*(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 \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(sin(c + d*x)*csc(c + d*x)**2, x) + Integral(-sin(c + d*x), x))

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Giac [B]  time = 1.36561, size = 81, normalized size = 3. \begin{align*} -\frac{A a \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac{4 \, A a}{\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1}}{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="giac")

[Out]

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