3.1.9 \(\int (c+d x)^3 \sin ^2(a+b x) \, dx\) [9]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 = {3392, 32, 3391} \begin {gather*} -\frac {3 d^3 \sin ^2(a+b x)}{8 b^4}+\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 {(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} \end {gather*}

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 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 3391

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 3392

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]

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*}

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Mathematica [A]
time = 0.27, size = 106, normalized size = 0.79 \begin {gather*} \frac {2 b^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )-3 d \left (-d^2+2 b^2 (c+d x)^2\right ) \cos (2 (a+b x))-2 b (c+d x) \left (-3 d^2+2 b^2 (c+d x)^2\right ) \sin (2 (a+b x))}{16 b^4} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(586\) vs. \(2(120)=240\).
time = 0.09, size = 587, normalized size = 4.38 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (120) = 240\).
time = 0.33, size = 442, normalized size = 3.30 \begin {gather*} \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 {gather*}

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 = 0.48, size = 189, normalized size = 1.41 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (131) = 262\).
time = 0.46, size = 456, normalized size = 3.40 \begin {gather*} \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 \cos ^{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} \cos ^{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 {gather*}

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*cos(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*cos(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 = 3.75, size = 153, normalized size = 1.14 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.85, size = 229, normalized size = 1.71 \begin {gather*} \frac {\frac {3\,d^3\,\cos \left (2\,a+2\,b\,x\right )}{2}+4\,b^4\,c^3\,x-2\,b^3\,c^3\,\sin \left (2\,a+2\,b\,x\right )+b^4\,d^3\,x^4-3\,b^2\,c^2\,d\,\cos \left (2\,a+2\,b\,x\right )+6\,b^4\,c^2\,d\,x^2+4\,b^4\,c\,d^2\,x^3-3\,b^2\,d^3\,x^2\,\cos \left (2\,a+2\,b\,x\right )-2\,b^3\,d^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )+3\,b\,c\,d^2\,\sin \left (2\,a+2\,b\,x\right )+3\,b\,d^3\,x\,\sin \left (2\,a+2\,b\,x\right )-6\,b^2\,c\,d^2\,x\,\cos \left (2\,a+2\,b\,x\right )-6\,b^3\,c^2\,d\,x\,\sin \left (2\,a+2\,b\,x\right )-6\,b^3\,c\,d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{8\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2*(c + d*x)^3,x)

[Out]

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

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