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]

3/2*a*A*x-2*a*A*cos(d*x+c)/d-1/2*a*A*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, 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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[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 ^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*}

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Mathematica [A]  time = 0.07, 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]

-1/4*(a*A*(-6*(c + d*x) + 8*Cos[c + d*x] + Sin[2*(c + d*x)]))/d

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fricas [A]  time = 0.49, size = 38, normalized size = 0.90 \[ \frac {3 \, A a d x - A a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, A a \cos \left (d x + c\right )}{2 \, d} \]

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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giac [B]  time = 0.35, size = 79, normalized size = 1.88 \[ \frac {3 \, {\left (d x + c\right )} A a + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.91, size = 49, normalized size = 1.17 \[ \frac {a A \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-2 A \cos \left (d x +c \right ) a +a A \left (d x +c \right )}{d} \]

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

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maxima [A]  time = 0.32, size = 47, normalized size = 1.12 \[ \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 4 \, {\left (d x + c\right )} A a - 8 \, A a \cos \left (d x + c\right )}{4 \, d} \]

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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mupad [B]  time = 0.39, size = 116, normalized size = 2.76 \[ \frac {3\,A\,a\,x}{2}-\frac {-A\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a\,\left (3\,c+3\,d\,x\right )-\frac {A\,a\,\left (6\,c+6\,d\,x-8\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+A\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {A\,a\,\left (3\,c+3\,d\,x\right )}{2}-\frac {A\,a\,\left (3\,c+3\,d\,x-8\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

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]

(3*A*a*x)/2 - (tan(c/2 + (d*x)/2)^2*(A*a*(3*c + 3*d*x) - (A*a*(6*c + 6*d*x - 8))/2) - A*a*tan(c/2 + (d*x)/2)^3
+ (A*a*(3*c + 3*d*x))/2 - (A*a*(3*c + 3*d*x - 8))/2 + A*a*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^2
)

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

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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