3.371 \(\int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac {d \sin ^2(a+b x)}{4 b^2}+\frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \sin (a+b x) \cos (a+b x)}{b}+c x+\frac {d x^2}{2} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4431, 3310} \[ -\frac {d \sin ^2(a+b x)}{4 b^2}+\frac {3 d \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x) \sin (a+b x) \cos (a+b x)}{b}+c x+\frac {d x^2}{2} \]

Antiderivative was successfully verified.

[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 3310

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 4431

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*} \int (c+d x) \csc (a+b x) \sin (3 a+3 b x) \, dx &=\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*}

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Mathematica [A]  time = 0.14, size = 46, normalized size = 0.70 \[ \frac {b (2 (c+d x) \sin (2 (a+b x))+b x (2 c+d x))+d \cos (2 (a+b x))}{2 b^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.44, size = 54, normalized size = 0.82 \[ \frac {b^{2} d x^{2} + 2 \, b^{2} c x + 2 \, d \cos \left (b x + a\right )^{2} + 4 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 1.99, size = 920, normalized size = 13.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*d*x^2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^2*c*x*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^2*d*x^2*tan(1/2*b*x)^
4*tan(1/2*a)^2 + 2*b^2*d*x^2*tan(1/2*b*x)^2*tan(1/2*a)^4 + 4*b^2*c*x*tan(1/2*b*x)^4*tan(1/2*a)^2 - 8*b*d*x*tan
(1/2*b*x)^4*tan(1/2*a)^3 + 4*b^2*c*x*tan(1/2*b*x)^2*tan(1/2*a)^4 - 8*b*d*x*tan(1/2*b*x)^3*tan(1/2*a)^4 + b^2*d
*x^2*tan(1/2*b*x)^4 + 4*b^2*d*x^2*tan(1/2*b*x)^2*tan(1/2*a)^2 - 8*b*c*tan(1/2*b*x)^4*tan(1/2*a)^3 + b^2*d*x^2*
tan(1/2*a)^4 - 8*b*c*tan(1/2*b*x)^3*tan(1/2*a)^4 + d*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^2*c*x*tan(1/2*b*x)^4 +
8*b*d*x*tan(1/2*b*x)^4*tan(1/2*a) + 8*b^2*c*x*tan(1/2*b*x)^2*tan(1/2*a)^2 + 48*b*d*x*tan(1/2*b*x)^3*tan(1/2*a)
^2 + 48*b*d*x*tan(1/2*b*x)^2*tan(1/2*a)^3 + 2*b^2*c*x*tan(1/2*a)^4 + 8*b*d*x*tan(1/2*b*x)*tan(1/2*a)^4 + 2*b^2
*d*x^2*tan(1/2*b*x)^2 + 8*b*c*tan(1/2*b*x)^4*tan(1/2*a) + 2*b^2*d*x^2*tan(1/2*a)^2 + 48*b*c*tan(1/2*b*x)^3*tan
(1/2*a)^2 - 6*d*tan(1/2*b*x)^4*tan(1/2*a)^2 + 48*b*c*tan(1/2*b*x)^2*tan(1/2*a)^3 - 16*d*tan(1/2*b*x)^3*tan(1/2
*a)^3 + 8*b*c*tan(1/2*b*x)*tan(1/2*a)^4 - 6*d*tan(1/2*b*x)^2*tan(1/2*a)^4 + 4*b^2*c*x*tan(1/2*b*x)^2 - 8*b*d*x
*tan(1/2*b*x)^3 - 48*b*d*x*tan(1/2*b*x)^2*tan(1/2*a) + 4*b^2*c*x*tan(1/2*a)^2 - 48*b*d*x*tan(1/2*b*x)*tan(1/2*
a)^2 - 8*b*d*x*tan(1/2*a)^3 + b^2*d*x^2 - 8*b*c*tan(1/2*b*x)^3 + d*tan(1/2*b*x)^4 - 48*b*c*tan(1/2*b*x)^2*tan(
1/2*a) + 16*d*tan(1/2*b*x)^3*tan(1/2*a) - 48*b*c*tan(1/2*b*x)*tan(1/2*a)^2 + 36*d*tan(1/2*b*x)^2*tan(1/2*a)^2
- 8*b*c*tan(1/2*a)^3 + 16*d*tan(1/2*b*x)*tan(1/2*a)^3 + d*tan(1/2*a)^4 + 2*b^2*c*x + 8*b*d*x*tan(1/2*b*x) + 8*
b*d*x*tan(1/2*a) + 8*b*c*tan(1/2*b*x) - 6*d*tan(1/2*b*x)^2 + 8*b*c*tan(1/2*a) - 16*d*tan(1/2*b*x)*tan(1/2*a) -
6*d*tan(1/2*a)^2 + d)/(b^2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^2*tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*b^2*tan(1/2*b*
x)^2*tan(1/2*a)^4 + b^2*tan(1/2*b*x)^4 + 4*b^2*tan(1/2*b*x)^2*tan(1/2*a)^2 + b^2*tan(1/2*a)^4 + 2*b^2*tan(1/2*
b*x)^2 + 2*b^2*tan(1/2*a)^2 + b^2)

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maple [A]  time = 0.03, size = 119, normalized size = 1.80 \[ -c x -\frac {d \,x^{2}}{2}+\frac {4 c \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+\frac {4 d \left (\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}-a \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 55, normalized size = 0.83 \[ \frac {{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c}{b} + \frac {{\left (b^{2} x^{2} + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} d}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.21, size = 53, normalized size = 0.80 \[ c\,x+\frac {d\,x^2}{2}+\frac {\frac {d\,\cos \left (2\,a+2\,b\,x\right )}{2}+b\,\left (c\,\sin \left (2\,a+2\,b\,x\right )+d\,x\,\sin \left (2\,a+2\,b\,x\right )\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \sin {\left (3 a + 3 b x \right )} \csc {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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