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

Optimal. Leaf size=134 \[ \frac{3 d^2 (c+d x) \sin (a+b x) \cos (a+b x)}{4 b^3}+\frac{3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}-\frac{3 d^3 \sin ^2(a+b x)}{8 b^4}-\frac{(c+d x)^3 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{3 c d^2 x}{4 b^2}-\frac{3 d^3 x^2}{8 b^2}+\frac{(c+d x)^4}{8 d} \]

[Out]

(-3*c*d^2*x)/(4*b^2) - (3*d^3*x^2)/(8*b^2) + (c + d*x)^4/(8*d) + (3*d^2*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(
4*b^3) - ((c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/(2*b) - (3*d^3*Sin[a + b*x]^2)/(8*b^4) + (3*d*(c + d*x)^2*Sin
[a + b*x]^2)/(4*b^2)

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Rubi [A]  time = 0.0742275, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3311, 32, 3310} \[ \frac{3 d^2 (c+d x) \sin (a+b x) \cos (a+b x)}{4 b^3}+\frac{3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}-\frac{3 d^3 \sin ^2(a+b x)}{8 b^4}-\frac{(c+d x)^3 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{3 c d^2 x}{4 b^2}-\frac{3 d^3 x^2}{8 b^2}+\frac{(c+d x)^4}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sin[a + b*x]^2,x]

[Out]

(-3*c*d^2*x)/(4*b^2) - (3*d^3*x^2)/(8*b^2) + (c + d*x)^4/(8*d) + (3*d^2*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(
4*b^3) - ((c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/(2*b) - (3*d^3*Sin[a + b*x]^2)/(8*b^4) + (3*d*(c + d*x)^2*Sin
[a + b*x]^2)/(4*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 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)^3 \sin ^2(a+b x) \, dx &=-\frac{(c+d x)^3 \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}+\frac{1}{2} \int (c+d x)^3 \, dx-\frac{\left (3 d^2\right ) \int (c+d x) \sin ^2(a+b x) \, dx}{2 b^2}\\ &=\frac{(c+d x)^4}{8 d}+\frac{3 d^2 (c+d x) \cos (a+b x) \sin (a+b x)}{4 b^3}-\frac{(c+d x)^3 \cos (a+b x) \sin (a+b x)}{2 b}-\frac{3 d^3 \sin ^2(a+b x)}{8 b^4}+\frac{3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}-\frac{\left (3 d^2\right ) \int (c+d x) \, dx}{4 b^2}\\ &=-\frac{3 c d^2 x}{4 b^2}-\frac{3 d^3 x^2}{8 b^2}+\frac{(c+d x)^4}{8 d}+\frac{3 d^2 (c+d x) \cos (a+b x) \sin (a+b x)}{4 b^3}-\frac{(c+d x)^3 \cos (a+b x) \sin (a+b x)}{2 b}-\frac{3 d^3 \sin ^2(a+b x)}{8 b^4}+\frac{3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.419724, size = 106, normalized size = 0.79 \[ \frac{-2 b (c+d x) \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )-3 d \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )+2 b^4 x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )}{16 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Sin[a + b*x]^2,x]

[Out]

(2*b^4*x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3) - 3*d*(-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] - 2*b*(
c + d*x)*(-3*d^2 + 2*b^2*(c + d*x)^2)*Sin[2*(a + b*x)])/(16*b^4)

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Maple [B]  time = 0.007, size = 587, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06971, size = 597, normalized size = 4.46 \begin{align*} \frac{4 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} c^{3} - \frac{12 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a c^{2} d}{b} + \frac{12 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} c d^{2}}{b^{2}} - \frac{4 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{3} d^{3}}{b^{3}} + \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^{2} d}{b} - \frac{12 \,{\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 c 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 )} a^{2} d^{3}}{b^{3}} + \frac{2 \,{\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 )} c d^{2}}{b^{2}} - \frac{2 \,{\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 )} a d^{3}}{b^{3}} + \frac{{\left (2 \,{\left (b x + a\right )}^{4} - 3 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \,{\left (2 \,{\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{3}}{b^{3}}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/16*(4*(2*b*x + 2*a - sin(2*b*x + 2*a))*c^3 - 12*(2*b*x + 2*a - sin(2*b*x + 2*a))*a*c^2*d/b + 12*(2*b*x + 2*a
- sin(2*b*x + 2*a))*a^2*c*d^2/b^2 - 4*(2*b*x + 2*a - sin(2*b*x + 2*a))*a^3*d^3/b^3 + 6*(2*(b*x + a)^2 - 2*(b*
x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a))*c^2*d/b - 12*(2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2
*b*x + 2*a))*a*c*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))*a^2*d^3/b^3 + 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))*c*d^2/b^2 - 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))*a*d^3/b^3 + (2*(b*x + a)^4 -
3*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 2*(2*(b*x + a)^3 - 3*b*x - 3*a)*sin(2*b*x + 2*a))*d^3/b^3)/b

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Fricas [A]  time = 1.70047, size = 394, normalized size = 2.94 \begin{align*} \frac{b^{4} d^{3} x^{4} + 4 \, b^{4} c d^{2} x^{3} + 3 \,{\left (2 \, b^{4} c^{2} d + b^{2} d^{3}\right )} x^{2} - 3 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2} - 2 \,{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 2 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \,{\left (2 \, b^{4} c^{3} + 3 \, b^{2} c d^{2}\right )} x}{8 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.23952, size = 456, normalized size = 3.4 \begin{align*} \begin{cases} \frac{c^{3} x \sin ^{2}{\left (a + b x \right )}}{2} + \frac{c^{3} x \cos ^{2}{\left (a + b x \right )}}{2} + \frac{3 c^{2} d x^{2} \sin ^{2}{\left (a + b x \right )}}{4} + \frac{3 c^{2} d x^{2} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{c d^{2} x^{3} \sin ^{2}{\left (a + b x \right )}}{2} + \frac{c d^{2} x^{3} \cos ^{2}{\left (a + b x \right )}}{2} + \frac{d^{3} x^{4} \sin ^{2}{\left (a + b x \right )}}{8} + \frac{d^{3} x^{4} \cos ^{2}{\left (a + b x \right )}}{8} - \frac{c^{3} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} - \frac{3 c^{2} d x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} - \frac{3 c d^{2} x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} - \frac{d^{3} x^{3} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} + \frac{3 c^{2} d \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{3 c d^{2} x \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac{3 c d^{2} x \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{3 d^{3} x^{2} \sin ^{2}{\left (a + b x \right )}}{8 b^{2}} - \frac{3 d^{3} x^{2} \cos ^{2}{\left (a + b x \right )}}{8 b^{2}} + \frac{3 c d^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{3}} + \frac{3 d^{3} x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{3}} - \frac{3 d^{3} \sin ^{2}{\left (a + b x \right )}}{8 b^{4}} & \text{for}\: b \neq 0 \\\left (c^{3} x + \frac{3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac{d^{3} x^{4}}{4}\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)**3*sin(b*x+a)**2,x)

[Out]

Piecewise((c**3*x*sin(a + b*x)**2/2 + c**3*x*cos(a + b*x)**2/2 + 3*c**2*d*x**2*sin(a + b*x)**2/4 + 3*c**2*d*x*
*2*cos(a + b*x)**2/4 + c*d**2*x**3*sin(a + b*x)**2/2 + c*d**2*x**3*cos(a + b*x)**2/2 + d**3*x**4*sin(a + b*x)*
*2/8 + d**3*x**4*cos(a + b*x)**2/8 - c**3*sin(a + b*x)*cos(a + b*x)/(2*b) - 3*c**2*d*x*sin(a + b*x)*cos(a + b*
x)/(2*b) - 3*c*d**2*x**2*sin(a + b*x)*cos(a + b*x)/(2*b) - d**3*x**3*sin(a + b*x)*cos(a + b*x)/(2*b) + 3*c**2*
d*sin(a + b*x)**2/(4*b**2) + 3*c*d**2*x*sin(a + b*x)**2/(4*b**2) - 3*c*d**2*x*cos(a + b*x)**2/(4*b**2) + 3*d**
3*x**2*sin(a + b*x)**2/(8*b**2) - 3*d**3*x**2*cos(a + b*x)**2/(8*b**2) + 3*c*d**2*sin(a + b*x)*cos(a + b*x)/(4
*b**3) + 3*d**3*x*sin(a + b*x)*cos(a + b*x)/(4*b**3) - 3*d**3*sin(a + b*x)**2/(8*b**4), Ne(b, 0)), ((c**3*x +
3*c**2*d*x**2/2 + c*d**2*x**3 + d**3*x**4/4)*sin(a)**2, True))

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Giac [A]  time = 1.13585, size = 207, normalized size = 1.54 \begin{align*} \frac{1}{8} \, d^{3} x^{4} + \frac{1}{2} \, c d^{2} x^{3} + \frac{3}{4} \, c^{2} d x^{2} + \frac{1}{2} \, c^{3} x - \frac{3 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (2 \, b x + 2 \, a\right )}{16 \, b^{4}} - \frac{{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} - 3 \, b d^{3} x - 3 \, b c d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/8*d^3*x^4 + 1/2*c*d^2*x^3 + 3/4*c^2*d*x^2 + 1/2*c^3*x - 3/16*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d -
d^3)*cos(2*b*x + 2*a)/b^4 - 1/8*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 6*b^3*c^2*d*x + 2*b^3*c^3 - 3*b*d^3*x - 3*b
*c*d^2)*sin(2*b*x + 2*a)/b^4