3.370 \(\int (c+d x)^2 \csc (a+b x) \sin (3 a+3 b x) \, dx\)
Optimal. Leaf size=112 \[ -\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{3 d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac{d^2 \sin (a+b x) \cos (a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b}-\frac{d^2 x}{2 b^2}+\frac{(c+d x)^3}{3 d} \]
[Out]
-(d^2*x)/(2*b^2) + (c + d*x)^3/(3*d) + (3*d*(c + d*x)*Cos[a + b*x]^2)/(2*b^2) - (d^2*Cos[a + b*x]*Sin[a + b*x]
)/b^3 + (2*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/b - (d*(c + d*x)*Sin[a + b*x]^2)/(2*b^2)
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Rubi [A] time = 0.13595, antiderivative size = 112, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used
= {4431, 3311, 32, 2635, 8} \[ -\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{3 d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac{d^2 \sin (a+b x) \cos (a+b x)}{b^3}+\frac{2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b}-\frac{d^2 x}{2 b^2}+\frac{(c+d x)^3}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^2*Csc[a + b*x]*Sin[3*a + 3*b*x],x]
[Out]
-(d^2*x)/(2*b^2) + (c + d*x)^3/(3*d) + (3*d*(c + d*x)*Cos[a + b*x]^2)/(2*b^2) - (d^2*Cos[a + b*x]*Sin[a + b*x]
)/b^3 + (2*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/b - (d*(c + d*x)*Sin[a + b*x]^2)/(2*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 3311
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
Rule 32
Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int (c+d x)^2 \csc (a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^2 \cos ^2(a+b x)-(c+d x)^2 \sin ^2(a+b x)\right ) \, dx\\ &=3 \int (c+d x)^2 \cos ^2(a+b x) \, dx-\int (c+d x)^2 \sin ^2(a+b x) \, dx\\ &=\frac{3 d (c+d x) \cos ^2(a+b x)}{2 b^2}+\frac{2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b}-\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}-\frac{1}{2} \int (c+d x)^2 \, dx+\frac{3}{2} \int (c+d x)^2 \, dx+\frac{d^2 \int \sin ^2(a+b x) \, dx}{2 b^2}-\frac{\left (3 d^2\right ) \int \cos ^2(a+b x) \, dx}{2 b^2}\\ &=\frac{(c+d x)^3}{3 d}+\frac{3 d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac{d^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac{2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b}-\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{d^2 \int 1 \, dx}{4 b^2}-\frac{\left (3 d^2\right ) \int 1 \, dx}{4 b^2}\\ &=-\frac{d^2 x}{2 b^2}+\frac{(c+d x)^3}{3 d}+\frac{3 d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac{d^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac{2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b}-\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.394428, size = 73, normalized size = 0.65 \[ \frac{\sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )}{2 b^3}+\frac{d (c+d x) \cos (2 (a+b x))}{b^2}+c^2 x+c d x^2+\frac{d^2 x^3}{3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^2*Csc[a + b*x]*Sin[3*a + 3*b*x],x]
[Out]
c^2*x + c*d*x^2 + (d^2*x^3)/3 + (d*(c + d*x)*Cos[2*(a + b*x)])/b^2 + ((-d^2 + 2*b^2*(c + d*x)^2)*Sin[2*(a + b*
x)])/(2*b^3)
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Maple [B] time = 0.041, size = 294, normalized size = 2.6 \begin{align*} -{c}^{2}x-{\frac{{d}^{2}{x}^{3}}{3}}+4\,{\frac{{c}^{2} \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) }{b}}-cd{x}^{2}+4\,{\frac{{d}^{2} \left ( \left ( bx+a \right ) ^{2} \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) +1/2\, \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1/4\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) -1/4\,bx-a/4-1/3\, \left ( bx+a \right ) ^{3}-2\,a \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} \right ) +{a}^{2} \left ( 1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) \right ) }{{b}^{3}}}+8\,{\frac{cd \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)^2*csc(b*x+a)*sin(3*b*x+3*a),x)
[Out]
-c^2*x-1/3*d^2*x^3+4*c^2/b*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-c*d*x^2+4*d^2/b^3*((b*x+a)^2*(1/2*cos(b*x
+a)*sin(b*x+a)+1/2*b*x+1/2*a)+1/2*(b*x+a)*cos(b*x+a)^2-1/4*cos(b*x+a)*sin(b*x+a)-1/4*b*x-1/4*a-1/3*(b*x+a)^3-2
*a*((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^2*(1/2*cos(b*x+a)*sin(
b*x+a)+1/2*b*x+1/2*a))+8*c*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))
________________________________________________________________________________________
Maxima [A] time = 1.07337, size = 146, normalized size = 1.3 \begin{align*} \frac{{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2}}{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 )} c d}{b^{2}} + \frac{{\left (2 \, b^{3} x^{3} + 6 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \,{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{6 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="maxima")
[Out]
(b*x + sin(2*b*x + 2*a))*c^2/b + (b^2*x^2 + 2*b*x*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*c*d/b^2 + 1/6*(2*b^3*x^
3 + 6*b*x*cos(2*b*x + 2*a) + 3*(2*b^2*x^2 - 1)*sin(2*b*x + 2*a))*d^2/b^3
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Fricas [A] time = 0.497992, size = 240, normalized size = 2.14 \begin{align*} \frac{b^{3} d^{2} x^{3} + 3 \, b^{3} c d x^{2} + 6 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 3 \,{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \,{\left (b^{3} c^{2} - b d^{2}\right )} x}{3 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="fricas")
[Out]
1/3*(b^3*d^2*x^3 + 3*b^3*c*d*x^2 + 6*(b*d^2*x + b*c*d)*cos(b*x + a)^2 + 3*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2
*c^2 - d^2)*cos(b*x + a)*sin(b*x + a) + 3*(b^3*c^2 - b*d^2)*x)/b^3
________________________________________________________________________________________
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)**2*csc(b*x+a)*sin(3*b*x+3*a),x)
[Out]
Timed out
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Giac [B] time = 1.3095, size = 2538, normalized size = 22.66 \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)^2*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="giac")
[Out]
1/3*(b^3*d^2*x^3*tan(1/2*b*x)^4*tan(1/2*a)^4 + 3*b^3*c*d*x^2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^3*d^2*x^3*tan(1
/2*b*x)^4*tan(1/2*a)^2 + 2*b^3*d^2*x^3*tan(1/2*b*x)^2*tan(1/2*a)^4 + 3*b^3*c^2*x*tan(1/2*b*x)^4*tan(1/2*a)^4 +
6*b^3*c*d*x^2*tan(1/2*b*x)^4*tan(1/2*a)^2 - 12*b^2*d^2*x^2*tan(1/2*b*x)^4*tan(1/2*a)^3 + 6*b^3*c*d*x^2*tan(1/
2*b*x)^2*tan(1/2*a)^4 - 12*b^2*d^2*x^2*tan(1/2*b*x)^3*tan(1/2*a)^4 + b^3*d^2*x^3*tan(1/2*b*x)^4 + 4*b^3*d^2*x^
3*tan(1/2*b*x)^2*tan(1/2*a)^2 + 6*b^3*c^2*x*tan(1/2*b*x)^4*tan(1/2*a)^2 - 24*b^2*c*d*x*tan(1/2*b*x)^4*tan(1/2*
a)^3 + b^3*d^2*x^3*tan(1/2*a)^4 + 6*b^3*c^2*x*tan(1/2*b*x)^2*tan(1/2*a)^4 - 24*b^2*c*d*x*tan(1/2*b*x)^3*tan(1/
2*a)^4 + 3*b*d^2*x*tan(1/2*b*x)^4*tan(1/2*a)^4 + 3*b^3*c*d*x^2*tan(1/2*b*x)^4 + 12*b^2*d^2*x^2*tan(1/2*b*x)^4*
tan(1/2*a) + 12*b^3*c*d*x^2*tan(1/2*b*x)^2*tan(1/2*a)^2 + 72*b^2*d^2*x^2*tan(1/2*b*x)^3*tan(1/2*a)^2 + 72*b^2*
d^2*x^2*tan(1/2*b*x)^2*tan(1/2*a)^3 - 12*b^2*c^2*tan(1/2*b*x)^4*tan(1/2*a)^3 + 3*b^3*c*d*x^2*tan(1/2*a)^4 + 12
*b^2*d^2*x^2*tan(1/2*b*x)*tan(1/2*a)^4 - 12*b^2*c^2*tan(1/2*b*x)^3*tan(1/2*a)^4 + 3*b*c*d*tan(1/2*b*x)^4*tan(1
/2*a)^4 + 2*b^3*d^2*x^3*tan(1/2*b*x)^2 + 3*b^3*c^2*x*tan(1/2*b*x)^4 + 24*b^2*c*d*x*tan(1/2*b*x)^4*tan(1/2*a) +
2*b^3*d^2*x^3*tan(1/2*a)^2 + 12*b^3*c^2*x*tan(1/2*b*x)^2*tan(1/2*a)^2 + 144*b^2*c*d*x*tan(1/2*b*x)^3*tan(1/2*
a)^2 - 18*b*d^2*x*tan(1/2*b*x)^4*tan(1/2*a)^2 + 144*b^2*c*d*x*tan(1/2*b*x)^2*tan(1/2*a)^3 - 48*b*d^2*x*tan(1/2
*b*x)^3*tan(1/2*a)^3 + 3*b^3*c^2*x*tan(1/2*a)^4 + 24*b^2*c*d*x*tan(1/2*b*x)*tan(1/2*a)^4 - 18*b*d^2*x*tan(1/2*
b*x)^2*tan(1/2*a)^4 + 6*b^3*c*d*x^2*tan(1/2*b*x)^2 - 12*b^2*d^2*x^2*tan(1/2*b*x)^3 - 72*b^2*d^2*x^2*tan(1/2*b*
x)^2*tan(1/2*a) + 12*b^2*c^2*tan(1/2*b*x)^4*tan(1/2*a) + 6*b^3*c*d*x^2*tan(1/2*a)^2 - 72*b^2*d^2*x^2*tan(1/2*b
*x)*tan(1/2*a)^2 + 72*b^2*c^2*tan(1/2*b*x)^3*tan(1/2*a)^2 - 18*b*c*d*tan(1/2*b*x)^4*tan(1/2*a)^2 - 12*b^2*d^2*
x^2*tan(1/2*a)^3 + 72*b^2*c^2*tan(1/2*b*x)^2*tan(1/2*a)^3 - 48*b*c*d*tan(1/2*b*x)^3*tan(1/2*a)^3 + 6*d^2*tan(1
/2*b*x)^4*tan(1/2*a)^3 + 12*b^2*c^2*tan(1/2*b*x)*tan(1/2*a)^4 - 18*b*c*d*tan(1/2*b*x)^2*tan(1/2*a)^4 + 6*d^2*t
an(1/2*b*x)^3*tan(1/2*a)^4 + b^3*d^2*x^3 + 6*b^3*c^2*x*tan(1/2*b*x)^2 - 24*b^2*c*d*x*tan(1/2*b*x)^3 + 3*b*d^2*
x*tan(1/2*b*x)^4 - 144*b^2*c*d*x*tan(1/2*b*x)^2*tan(1/2*a) + 48*b*d^2*x*tan(1/2*b*x)^3*tan(1/2*a) + 6*b^3*c^2*
x*tan(1/2*a)^2 - 144*b^2*c*d*x*tan(1/2*b*x)*tan(1/2*a)^2 + 108*b*d^2*x*tan(1/2*b*x)^2*tan(1/2*a)^2 - 24*b^2*c*
d*x*tan(1/2*a)^3 + 48*b*d^2*x*tan(1/2*b*x)*tan(1/2*a)^3 + 3*b*d^2*x*tan(1/2*a)^4 + 3*b^3*c*d*x^2 + 12*b^2*d^2*
x^2*tan(1/2*b*x) - 12*b^2*c^2*tan(1/2*b*x)^3 + 3*b*c*d*tan(1/2*b*x)^4 + 12*b^2*d^2*x^2*tan(1/2*a) - 72*b^2*c^2
*tan(1/2*b*x)^2*tan(1/2*a) + 48*b*c*d*tan(1/2*b*x)^3*tan(1/2*a) - 6*d^2*tan(1/2*b*x)^4*tan(1/2*a) - 72*b^2*c^2
*tan(1/2*b*x)*tan(1/2*a)^2 + 108*b*c*d*tan(1/2*b*x)^2*tan(1/2*a)^2 - 36*d^2*tan(1/2*b*x)^3*tan(1/2*a)^2 - 12*b
^2*c^2*tan(1/2*a)^3 + 48*b*c*d*tan(1/2*b*x)*tan(1/2*a)^3 - 36*d^2*tan(1/2*b*x)^2*tan(1/2*a)^3 + 3*b*c*d*tan(1/
2*a)^4 - 6*d^2*tan(1/2*b*x)*tan(1/2*a)^4 + 3*b^3*c^2*x + 24*b^2*c*d*x*tan(1/2*b*x) - 18*b*d^2*x*tan(1/2*b*x)^2
+ 24*b^2*c*d*x*tan(1/2*a) - 48*b*d^2*x*tan(1/2*b*x)*tan(1/2*a) - 18*b*d^2*x*tan(1/2*a)^2 + 12*b^2*c^2*tan(1/2
*b*x) - 18*b*c*d*tan(1/2*b*x)^2 + 6*d^2*tan(1/2*b*x)^3 + 12*b^2*c^2*tan(1/2*a) - 48*b*c*d*tan(1/2*b*x)*tan(1/2
*a) + 36*d^2*tan(1/2*b*x)^2*tan(1/2*a) - 18*b*c*d*tan(1/2*a)^2 + 36*d^2*tan(1/2*b*x)*tan(1/2*a)^2 + 6*d^2*tan(
1/2*a)^3 + 3*b*d^2*x + 3*b*c*d - 6*d^2*tan(1/2*b*x) - 6*d^2*tan(1/2*a))/(b^3*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b
^3*tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*b^3*tan(1/2*b*x)^2*tan(1/2*a)^4 + b^3*tan(1/2*b*x)^4 + 4*b^3*tan(1/2*b*x)^2
*tan(1/2*a)^2 + b^3*tan(1/2*a)^4 + 2*b^3*tan(1/2*b*x)^2 + 2*b^3*tan(1/2*a)^2 + b^3)