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 + (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)
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Rubi [A] time = 0.0682019, antiderivative size = 66, normalized size of antiderivative = 1.,
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 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]
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]
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*}
Mathematica [A] time = 0.135017, size = 46, normalized size = 0.7 \[ \frac{b (2 (c+d x) \sin (2 (a+b x))+b x (2 c+d x))+d \cos (2 (a+b x))}{2 b^2} \]
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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Maple [A] time = 0.037, size = 119, normalized size = 1.8 \begin{align*} -cx-{\frac{d{x}^{2}}{2}}+4\,{\frac{c \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) }{b}}+4\,{\frac{d \left ( \left ( bx+a \right ) \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) -1/4\, \left ( bx+a \right ) ^{2}-1/4\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}-a \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) \right ) }{{b}^{2}}} \end{align*}
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 = 1.06442, size = 74, normalized size = 1.12 \begin{align*} \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}} \end{align*}
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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Fricas [A] time = 0.484193, size = 132, normalized size = 2. \begin{align*} \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}} \end{align*}
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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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Giac [B] time = 1.20591, size = 1242, normalized size = 18.82 \begin{align*} \text{result too large to display} \end{align*}
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)