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

Optimal. Leaf size=95 \[ \frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{d^2 \sin (a+b x) \cos (a+b x)}{4 b^3}-\frac{(c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{d^2 x}{4 b^2}+\frac{(c+d x)^3}{6 d} \]

[Out]

-(d^2*x)/(4*b^2) + (c + d*x)^3/(6*d) + (d^2*Cos[a + b*x]*Sin[a + b*x])/(4*b^3) - ((c + d*x)^2*Cos[a + b*x]*Sin
[a + b*x])/(2*b) + (d*(c + d*x)*Sin[a + b*x]^2)/(2*b^2)

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Rubi [A]  time = 0.0538203, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 32, 2635, 8} \[ \frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{d^2 \sin (a+b x) \cos (a+b x)}{4 b^3}-\frac{(c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{d^2 x}{4 b^2}+\frac{(c+d x)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sin[a + b*x]^2,x]

[Out]

-(d^2*x)/(4*b^2) + (c + d*x)^3/(6*d) + (d^2*Cos[a + b*x]*Sin[a + b*x])/(4*b^3) - ((c + d*x)^2*Cos[a + b*x]*Sin
[a + b*x])/(2*b) + (d*(c + d*x)*Sin[a + b*x]^2)/(2*b^2)

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 \sin ^2(a+b x) \, dx &=-\frac{(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}+\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}+\frac{1}{2} \int (c+d x)^2 \, dx-\frac{d^2 \int \sin ^2(a+b x) \, dx}{2 b^2}\\ &=\frac{(c+d x)^3}{6 d}+\frac{d^2 \cos (a+b x) \sin (a+b x)}{4 b^3}-\frac{(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}+\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}-\frac{d^2 \int 1 \, dx}{4 b^2}\\ &=-\frac{d^2 x}{4 b^2}+\frac{(c+d x)^3}{6 d}+\frac{d^2 \cos (a+b x) \sin (a+b x)}{4 b^3}-\frac{(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}+\frac{d (c+d x) \sin ^2(a+b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.301784, size = 77, normalized size = 0.81 \[ \frac{-3 \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )-6 b d (c+d x) \cos (2 (a+b x))+4 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^3*x*(3*c^2 + 3*c*d*x + d^2*x^2) - 6*b*d*(c + d*x)*Cos[2*(a + b*x)] - 3*(-d^2 + 2*b^2*(c + d*x)^2)*Sin[2*(
a + b*x)])/(24*b^3)

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Maple [B]  time = 0.007, size = 289, normalized size = 3. \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2}}{{b}^{2}} \left ( \left ( bx+a \right ) ^{2} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) -{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{4}}+{\frac{bx}{4}}+{\frac{a}{4}}-{\frac{ \left ( bx+a \right ) ^{3}}{3}} \right ) }-2\,{\frac{a{d}^{2} \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 ) }{{b}^{2}}}+2\,{\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} \right ) }{b}}+{\frac{{a}^{2}{d}^{2}}{{b}^{2}} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) }-2\,{\frac{acd \left ( -1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) }{b}}+{c}^{2} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02537, size = 313, normalized size = 3.29 \begin{align*} \frac{6 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} - \frac{12 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a c d}{b} + \frac{6 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} d^{2}}{b^{2}} + \frac{6 \,{\left (2 \,{\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} c d}{b} - \frac{6 \,{\left (2 \,{\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} + \frac{{\left (4 \,{\left (b x + a\right )}^{3} - 6 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*(2*b*x + 2*a - sin(2*b*x + 2*a))*c^2 - 12*(2*b*x + 2*a - sin(2*b*x + 2*a))*a*c*d/b + 6*(2*b*x + 2*a -
sin(2*b*x + 2*a))*a^2*d^2/b^2 + 6*(2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a))*c*d/b - 6*
(2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a))*a*d^2/b^2 + (4*(b*x + a)^3 - 6*(b*x + a)*cos
(2*b*x + 2*a) - 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a))*d^2/b^2)/b

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Fricas [A]  time = 1.7763, size = 247, normalized size = 2.6 \begin{align*} \frac{2 \, b^{3} d^{2} x^{3} + 6 \, 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 (2 \, b^{3} c^{2} + b d^{2}\right )} x}{12 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b^3*d^2*x^3 + 6*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*(2*b^3*c^2 + b*d^2)*x)/b^3

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Sympy [A]  time = 1.46087, size = 264, normalized size = 2.78 \begin{align*} \begin{cases} \frac{c^{2} x \sin ^{2}{\left (a + b x \right )}}{2} + \frac{c^{2} x \cos ^{2}{\left (a + b x \right )}}{2} + \frac{c d x^{2} \sin ^{2}{\left (a + b x \right )}}{2} + \frac{c d x^{2} \cos ^{2}{\left (a + b x \right )}}{2} + \frac{d^{2} x^{3} \sin ^{2}{\left (a + b x \right )}}{6} + \frac{d^{2} x^{3} \cos ^{2}{\left (a + b x \right )}}{6} - \frac{c^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} - \frac{c d x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b} - \frac{d^{2} x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} + \frac{c d \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac{d^{2} x \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac{d^{2} x \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{d^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 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 )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**2*x*sin(a + b*x)**2/2 + c**2*x*cos(a + b*x)**2/2 + c*d*x**2*sin(a + b*x)**2/2 + c*d*x**2*cos(a +
b*x)**2/2 + d**2*x**3*sin(a + b*x)**2/6 + d**2*x**3*cos(a + b*x)**2/6 - c**2*sin(a + b*x)*cos(a + b*x)/(2*b)
- c*d*x*sin(a + b*x)*cos(a + b*x)/b - d**2*x**2*sin(a + b*x)*cos(a + b*x)/(2*b) + c*d*sin(a + b*x)**2/(2*b**2)
+ d**2*x*sin(a + b*x)**2/(4*b**2) - d**2*x*cos(a + b*x)**2/(4*b**2) + d**2*sin(a + b*x)*cos(a + b*x)/(4*b**3)
, Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)**2, True))

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Giac [A]  time = 1.12363, size = 127, normalized size = 1.34 \begin{align*} \frac{1}{6} \, d^{2} x^{3} + \frac{1}{2} \, c d x^{2} + \frac{1}{2} \, c^{2} x - \frac{{\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{3}} - \frac{{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*d^2*x^3 + 1/2*c*d*x^2 + 1/2*c^2*x - 1/4*(b*d^2*x + b*c*d)*cos(2*b*x + 2*a)/b^3 - 1/8*(2*b^2*d^2*x^2 + 4*b^
2*c*d*x + 2*b^2*c^2 - d^2)*sin(2*b*x + 2*a)/b^3