\(\int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx\) [371]

Optimal result

Integrand size = 21, antiderivative size = 66 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=c x+\frac {d x^2}{2}+\frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \cos (a+b x) \sin (a+b x)}{b}-\frac {d \sin ^2(a+b x)}{4 b^2} \]

[Out]

c*x+1/2*d*x^2+3/4*d*cos(b*x+a)^2/b^2+2*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b-1/4*d*sin(b*x+a)^2/b^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4516, 3391} \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=-\frac {d \sin ^2(a+b x)}{4 b^2}+\frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \sin (a+b x) \cos (a+b x)}{b}+c x+\frac {d x^2}{2} \]

[In]

Int[(c + d*x)*Csc[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

c*x + (d*x^2)/2 + (3*d*Cos[a + b*x]^2)/(4*b^2) + (2*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/b - (d*Sin[a + b*x]^2
)/(4*b^2)

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 4516

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

Rubi steps \begin{align*} \text {integral}& = \int \left (3 (c+d x) \cos ^2(a+b x)-(c+d x) \sin ^2(a+b x)\right ) \, dx \\ & = 3 \int (c+d x) \cos ^2(a+b x) \, dx-\int (c+d x) \sin ^2(a+b x) \, dx \\ & = \frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \cos (a+b x) \sin (a+b x)}{b}-\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {1}{2} \int (c+d x) \, dx+\frac {3}{2} \int (c+d x) \, dx \\ & = c x+\frac {d x^2}{2}+\frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \cos (a+b x) \sin (a+b x)}{b}-\frac {d \sin ^2(a+b x)}{4 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {d \cos (2 (a+b x))+b (b x (2 c+d x)+2 (c+d x) \sin (2 (a+b x)))}{2 b^2} \]

[In]

Integrate[(c + d*x)*Csc[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

(d*Cos[2*(a + b*x)] + b*(b*x*(2*c + d*x) + 2*(c + d*x)*Sin[2*(a + b*x)]))/(2*b^2)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {d \,x^{2}}{2}+c x +\frac {d \cos \left (2 x b +2 a \right )}{2 b^{2}}+\frac {\left (d x +c \right ) \sin \left (2 x b +2 a \right )}{b}\)	\(44\)
default	\(-c x -\frac {d \,x^{2}}{2}+\frac {4 c \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )}{b}+\frac {4 d \left (\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}-a \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )\right )}{b^{2}}\)	\(119\)

[In]

int((d*x+c)*csc(b*x+a)*sin(3*b*x+3*a),x,method=_RETURNVERBOSE)

[Out]

1/2*d*x^2+c*x+1/2*d/b^2*cos(2*b*x+2*a)+1/b*(d*x+c)*sin(2*b*x+2*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {b^{2} d x^{2} + 2 \, b^{2} c x + 2 \, d \cos \left (b x + a\right )^{2} + 4 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b^{2}} \]

[In]

integrate((d*x+c)*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="fricas")

[Out]

1/2*(b^2*d*x^2 + 2*b^2*c*x + 2*d*cos(b*x + a)^2 + 4*(b*d*x + b*c)*cos(b*x + a)*sin(b*x + a))/b^2

Sympy [F]

\[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=\int \left (c + d x\right ) \sin {\left (3 a + 3 b x \right )} \csc {\left (a + b x \right )}\, dx \]

[In]

integrate((d*x+c)*csc(b*x+a)*sin(3*b*x+3*a),x)

[Out]

Integral((c + d*x)*sin(3*a + 3*b*x)*csc(a + b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c}{b} + \frac {{\left (b^{2} x^{2} + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} d}{2 \, b^{2}} \]

[In]

integrate((d*x+c)*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="maxima")

[Out]

(b*x + sin(2*b*x + 2*a))*c/b + 1/2*(b^2*x^2 + 2*b*x*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*d/b^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.61 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {4 \, b c \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b x + a\right )} d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 4 \, a d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (b x + a\right )} b c + {\left (b x + a\right )}^{2} d - 2 \, {\left (b x + a\right )} a d + d \cos \left (b x + a\right )^{2} - d \sin \left (b x + a\right )^{2}}{2 \, b^{2}} \]

[In]

integrate((d*x+c)*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="giac")

[Out]

1/2*(4*b*c*cos(b*x + a)*sin(b*x + a) + 4*(b*x + a)*d*cos(b*x + a)*sin(b*x + a) - 4*a*d*cos(b*x + a)*sin(b*x +
a) + 2*(b*x + a)*b*c + (b*x + a)^2*d - 2*(b*x + a)*a*d + d*cos(b*x + a)^2 - d*sin(b*x + a)^2)/b^2

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx=c\,x+\frac {d\,x^2}{2}+\frac {\frac {d\,\cos \left (2\,a+2\,b\,x\right )}{2}+b\,\left (c\,\sin \left (2\,a+2\,b\,x\right )+d\,x\,\sin \left (2\,a+2\,b\,x\right )\right )}{b^2} \]

[In]

int((sin(3*a + 3*b*x)*(c + d*x))/sin(a + b*x),x)

[Out]

c*x + (d*x^2)/2 + ((d*cos(2*a + 2*b*x))/2 + b*(c*sin(2*a + 2*b*x) + d*x*sin(2*a + 2*b*x)))/b^2