3.16 \(\int (c+d x)^2 \cos (a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=103 \[ \frac{4 d (c+d x) \cos (a+b x)}{9 b^2}+\frac{2 d (c+d x) \sin ^2(a+b x) \cos (a+b x)}{9 b^2}-\frac{2 d^2 \sin ^3(a+b x)}{27 b^3}-\frac{4 d^2 \sin (a+b x)}{9 b^3}+\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b} \]

[Out]

(4*d*(c + d*x)*Cos[a + b*x])/(9*b^2) - (4*d^2*Sin[a + b*x])/(9*b^3) + (2*d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x]
^2)/(9*b^2) - (2*d^2*Sin[a + b*x]^3)/(27*b^3) + ((c + d*x)^2*Sin[a + b*x]^3)/(3*b)

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Rubi [A]  time = 0.0772006, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4404, 3310, 3296, 2637} \[ \frac{4 d (c+d x) \cos (a+b x)}{9 b^2}+\frac{2 d (c+d x) \sin ^2(a+b x) \cos (a+b x)}{9 b^2}-\frac{2 d^2 \sin ^3(a+b x)}{27 b^3}-\frac{4 d^2 \sin (a+b x)}{9 b^3}+\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*d*(c + d*x)*Cos[a + b*x])/(9*b^2) - (4*d^2*Sin[a + b*x])/(9*b^3) + (2*d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x]
^2)/(9*b^2) - (2*d^2*Sin[a + b*x]^3)/(27*b^3) + ((c + d*x)^2*Sin[a + b*x]^3)/(3*b)

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -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]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (c+d x)^2 \cos (a+b x) \sin ^2(a+b x) \, dx &=\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b}-\frac{(2 d) \int (c+d x) \sin ^3(a+b x) \, dx}{3 b}\\ &=\frac{2 d (c+d x) \cos (a+b x) \sin ^2(a+b x)}{9 b^2}-\frac{2 d^2 \sin ^3(a+b x)}{27 b^3}+\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b}-\frac{(4 d) \int (c+d x) \sin (a+b x) \, dx}{9 b}\\ &=\frac{4 d (c+d x) \cos (a+b x)}{9 b^2}+\frac{2 d (c+d x) \cos (a+b x) \sin ^2(a+b x)}{9 b^2}-\frac{2 d^2 \sin ^3(a+b x)}{27 b^3}+\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b}-\frac{\left (4 d^2\right ) \int \cos (a+b x) \, dx}{9 b^2}\\ &=\frac{4 d (c+d x) \cos (a+b x)}{9 b^2}-\frac{4 d^2 \sin (a+b x)}{9 b^3}+\frac{2 d (c+d x) \cos (a+b x) \sin ^2(a+b x)}{9 b^2}-\frac{2 d^2 \sin ^3(a+b x)}{27 b^3}+\frac{(c+d x)^2 \sin ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.6145, size = 93, normalized size = 0.9 \[ \frac{-2 \sin (a+b x) \left (\cos (2 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )-9 b^2 (c+d x)^2+26 d^2\right )+54 b d (c+d x) \cos (a+b x)-6 b d (c+d x) \cos (3 (a+b x))}{108 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(54*b*d*(c + d*x)*Cos[a + b*x] - 6*b*d*(c + d*x)*Cos[3*(a + b*x)] - 2*(26*d^2 - 9*b^2*(c + d*x)^2 + (-2*d^2 +
9*b^2*(c + d*x)^2)*Cos[2*(a + b*x)])*Sin[a + b*x])/(108*b^3)

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Maple [B]  time = 0.016, size = 204, normalized size = 2. \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2}}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{3}}+{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) \cos \left ( bx+a \right ) }{9}}-{\frac{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{27}}-{\frac{4\,\sin \left ( bx+a \right ) }{9}} \right ) }-2\,{\frac{a{d}^{2} \left ( 1/3\, \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{3}+1/9\, \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) \cos \left ( bx+a \right ) \right ) }{{b}^{2}}}+2\,{\frac{cd \left ( 1/3\, \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{3}+1/9\, \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) \cos \left ( bx+a \right ) \right ) }{b}}+{\frac{{a}^{2}{d}^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{3\,{b}^{2}}}-{\frac{2\,acd \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{3\,b}}+{\frac{{c}^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b^2*d^2*(1/3*(b*x+a)^2*sin(b*x+a)^3+2/9*(b*x+a)*(2+sin(b*x+a)^2)*cos(b*x+a)-2/27*sin(b*x+a)^3-4/9*sin(b
*x+a))-2/b^2*a*d^2*(1/3*(b*x+a)*sin(b*x+a)^3+1/9*(2+sin(b*x+a)^2)*cos(b*x+a))+2/b*c*d*(1/3*(b*x+a)*sin(b*x+a)^
3+1/9*(2+sin(b*x+a)^2)*cos(b*x+a))+1/3/b^2*a^2*d^2*sin(b*x+a)^3-2/3/b*a*c*d*sin(b*x+a)^3+1/3*c^2*sin(b*x+a)^3)

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Maxima [B]  time = 1.10982, size = 324, normalized size = 3.15 \begin{align*} \frac{36 \, c^{2} \sin \left (b x + a\right )^{3} - \frac{72 \, a c d \sin \left (b x + a\right )^{3}}{b} + \frac{36 \, a^{2} d^{2} \sin \left (b x + a\right )^{3}}{b^{2}} - \frac{6 \,{\left (3 \,{\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 9 \,{\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) - 9 \, \cos \left (b x + a\right )\right )} c d}{b} + \frac{6 \,{\left (3 \,{\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 9 \,{\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) - 9 \, \cos \left (b x + a\right )\right )} a d^{2}}{b^{2}} - \frac{{\left (6 \,{\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 54 \,{\left (b x + a\right )} \cos \left (b x + a\right ) +{\left (9 \,{\left (b x + a\right )}^{2} - 2\right )} \sin \left (3 \, b x + 3 \, a\right ) - 27 \,{\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{108 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/108*(36*c^2*sin(b*x + a)^3 - 72*a*c*d*sin(b*x + a)^3/b + 36*a^2*d^2*sin(b*x + a)^3/b^2 - 6*(3*(b*x + a)*sin(
3*b*x + 3*a) - 9*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) - 9*cos(b*x + a))*c*d/b + 6*(3*(b*x + a)*sin(3*b*x
+ 3*a) - 9*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) - 9*cos(b*x + a))*a*d^2/b^2 - (6*(b*x + a)*cos(3*b*x + 3*
a) - 54*(b*x + a)*cos(b*x + a) + (9*(b*x + a)^2 - 2)*sin(3*b*x + 3*a) - 27*((b*x + a)^2 - 2)*sin(b*x + a))*d^2
/b^2)/b

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Fricas [A]  time = 0.479784, size = 296, normalized size = 2.87 \begin{align*} -\frac{6 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 18 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) -{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} -{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 14 \, d^{2}\right )} \sin \left (b x + a\right )}{27 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/27*(6*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 18*(b*d^2*x + b*c*d)*cos(b*x + a) - (9*b^2*d^2*x^2 + 18*b^2*c*d*x
+ 9*b^2*c^2 - (9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 - 2*d^2)*cos(b*x + a)^2 - 14*d^2)*sin(b*x + a))/b^3

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Sympy [A]  time = 2.62016, size = 216, normalized size = 2.1 \begin{align*} \begin{cases} \frac{c^{2} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{2 c d x \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{d^{2} x^{2} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{2 c d \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{3 b^{2}} + \frac{4 c d \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{2 d^{2} x \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{3 b^{2}} + \frac{4 d^{2} x \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} - \frac{14 d^{2} \sin ^{3}{\left (a + b x \right )}}{27 b^{3}} - \frac{4 d^{2} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{9 b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sin ^{2}{\left (a \right )} \cos{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cos(b*x+a)*sin(b*x+a)**2,x)

[Out]

Piecewise((c**2*sin(a + b*x)**3/(3*b) + 2*c*d*x*sin(a + b*x)**3/(3*b) + d**2*x**2*sin(a + b*x)**3/(3*b) + 2*c*
d*sin(a + b*x)**2*cos(a + b*x)/(3*b**2) + 4*c*d*cos(a + b*x)**3/(9*b**2) + 2*d**2*x*sin(a + b*x)**2*cos(a + b*
x)/(3*b**2) + 4*d**2*x*cos(a + b*x)**3/(9*b**2) - 14*d**2*sin(a + b*x)**3/(27*b**3) - 4*d**2*sin(a + b*x)*cos(
a + b*x)**2/(9*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)**2*cos(a), True))

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Giac [A]  time = 1.1409, size = 185, normalized size = 1.8 \begin{align*} -\frac{{\left (b d^{2} x + b c d\right )} \cos \left (3 \, b x + 3 \, a\right )}{18 \, b^{3}} + \frac{{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )}{2 \, b^{3}} - \frac{{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (3 \, b x + 3 \, a\right )}{108 \, b^{3}} + \frac{{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/18*(b*d^2*x + b*c*d)*cos(3*b*x + 3*a)/b^3 + 1/2*(b*d^2*x + b*c*d)*cos(b*x + a)/b^3 - 1/108*(9*b^2*d^2*x^2 +
18*b^2*c*d*x + 9*b^2*c^2 - 2*d^2)*sin(3*b*x + 3*a)/b^3 + 1/4*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*si
n(b*x + a)/b^3