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

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Rubi [A]  time = 0.06, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 8

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

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 3770

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

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*}

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Mathematica [B]  time = 0.02, 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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fricas [A]  time = 0.46, size = 51, normalized size = 1.55 \[ \frac {4 \, A a d x - 2 \, A a \cos \left (d x + c\right ) - A a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + A a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \]

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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giac [A]  time = 0.39, size = 47, normalized size = 1.42 \[ \frac {2 \, {\left (d x + c\right )} A a + A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

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

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maple [A]  time = 1.26, size = 50, normalized size = 1.52 \[ 2 a A x -\frac {a A \cos \left (d x +c \right )}{d}+\frac {a A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {2 A a c}{d} \]

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),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 = 0.32, size = 50, normalized size = 1.52 \[ \frac {4 \, {\left (d x + c\right )} A a - A a {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, A a \cos \left (d x + c\right )}{2 \, d} \]

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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mupad [B]  time = 0.30, size = 129, normalized size = 3.91 \[ \frac {A\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,A\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {4\,A\,a\,\mathrm {atan}\left (\frac {16\,A^2\,a^2}{8\,A^2\,a^2-16\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,A^2\,a^2-16\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a*log(tan(c/2 + (d*x)/2)))/d - (2*A*a)/(d*(tan(c/2 + (d*x)/2)^2 + 1)) + (4*A*a*atan((16*A^2*a^2)/(8*A^2*a^2
- 16*A^2*a^2*tan(c/2 + (d*x)/2)) + (8*A^2*a^2*tan(c/2 + (d*x)/2))/(8*A^2*a^2 - 16*A^2*a^2*tan(c/2 + (d*x)/2))
))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ A a \left (\int 2 \sin {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )}\, dx\right ) \]

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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