3.370 \(\int (c+d x)^2 \csc (a+b x) \sin (3 a+3 b x) \, dx\)

Optimal. Leaf size=112 \[ -\frac {d^2 \sin (a+b x) \cos (a+b x)}{b^3}-\frac {d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac {3 d (c+d x) \cos ^2(a+b x)}{2 b^2}+\frac {2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b}-\frac {d^2 x}{2 b^2}+\frac {(c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 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)^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*}

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Mathematica [A]  time = 0.40, size = 73, normalized size = 0.65 \[ \frac {d (c+d x) \cos (2 (a+b x))}{b^2}+\frac {\sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )}{2 b^3}+c^2 x+c d x^2+\frac {d^2 x^3}{3} \]

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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fricas [A]  time = 0.45, size = 111, normalized size = 0.99 \[ \frac {b^{3} d^{2} x^{3} + 3 \, b^{3} c d x^{2} + 6 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 3 \, {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \, {\left (b^{3} c^{2} - b d^{2}\right )} x}{3 \, b^{3}} \]

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

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giac [B]  time = 5.77, size = 1880, normalized size = 16.79 \[ \text {result too large to display} \]

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)

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maple [B]  time = 0.03, size = 294, normalized size = 2.62 \[ -c^{2} x -\frac {d^{2} x^{3}}{3}+\frac {4 c^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}-c d \,x^{2}+\frac {4 d^{2} \left (\left (b x +a \right )^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )+\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}-\frac {b x}{4}-\frac {a}{4}-\frac {\left (b x +a \right )^{3}}{3}-2 a \left (\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}\right )+a^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )\right )}{b^{3}}+\frac {8 c d \left (\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}-a \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^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))

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maxima [A]  time = 0.35, size = 108, normalized size = 0.96 \[ \frac {{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2}}{b} + \frac {{\left (b^{2} x^{2} + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} c d}{b^{2}} + \frac {{\left (2 \, b^{3} x^{3} + 6 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{6 \, b^{3}} \]

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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mupad [B]  time = 0.31, size = 121, normalized size = 1.08 \[ c^2\,x+\frac {d^2\,x^3}{3}+\frac {c^2\,\sin \left (2\,a+2\,b\,x\right )}{b}-\frac {d^2\,\sin \left (2\,a+2\,b\,x\right )}{2\,b^3}+c\,d\,x^2+\frac {d^2\,x\,\cos \left (2\,a+2\,b\,x\right )}{b^2}+\frac {d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{b}+\frac {c\,d\,\cos \left (2\,a+2\,b\,x\right )}{b^2}+\frac {2\,c\,d\,x\,\sin \left (2\,a+2\,b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(3*a + 3*b*x)*(c + d*x)^2)/sin(a + b*x),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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