3.3 \(\int (c+d x)^3 \cos (a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=120 \[ -\frac {3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}-\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac {3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x)^3 \sin ^2(a+b x)}{2 b}+\frac {3 d^3 x}{8 b^3}-\frac {(c+d x)^3}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4404, 3311, 32, 2635, 8} \[ -\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac {3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac {3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}+\frac {(c+d x)^3 \sin ^2(a+b x)}{2 b}+\frac {3 d^3 x}{8 b^3}-\frac {(c+d x)^3}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 71, normalized size = 0.59 \[ \frac {3 d \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )-2 b (c+d x) \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )}{16 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 166, normalized size = 1.38 \[ \frac {2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} - 2 \, {\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 2 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \, {\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 3 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \, {\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x}{8 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 121, normalized size = 1.01 \[ -\frac {{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} - 3 \, b d^{3} x - 3 \, b c d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{4}} + \frac {3 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{16 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 466, normalized size = 3.88 \[ \frac {\frac {d^{3} \left (-\frac {\left (b x +a \right )^{3} \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\frac {3 \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 )}{2}+\frac {3 \left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{4}-\frac {3 \cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}-\frac {3 b x}{8}-\frac {3 a}{8}-\frac {\left (b x +a \right )^{3}}{2}\right )}{b^{3}}-\frac {3 a \,d^{3} \left (-\frac {\left (b x +a \right )^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\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 )}{b^{3}}+\frac {3 c \,d^{2} \left (-\frac {\left (b x +a \right )^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\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 )}{b^{2}}+\frac {3 a^{2} d^{3} \left (-\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}\right )}{b^{3}}-\frac {6 a c \,d^{2} \left (-\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}\right )}{b^{2}}+\frac {3 c^{2} d \left (-\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}\right )}{b}+\frac {a^{3} d^{3} \left (\cos ^{2}\left (b x +a \right )\right )}{2 b^{3}}-\frac {3 a^{2} c \,d^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2 b^{2}}+\frac {3 a \,c^{2} d \left (\cos ^{2}\left (b x +a \right )\right )}{2 b}-\frac {c^{3} \left (\cos ^{2}\left (b x +a \right )\right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.36, size = 342, normalized size = 2.85 \[ -\frac {8 \, c^{3} \cos \left (b x + a\right )^{2} - \frac {24 \, a c^{2} d \cos \left (b x + a\right )^{2}}{b} + \frac {24 \, a^{2} c d^{2} \cos \left (b x + a\right )^{2}}{b^{2}} - \frac {8 \, a^{3} d^{3} \cos \left (b x + a\right )^{2}}{b^{3}} + \frac {6 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} d}{b} - \frac {12 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a c d^{2}}{b^{2}} + \frac {6 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} d^{3}}{b^{3}} + \frac {6 \, {\left ({\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c d^{2}}{b^{2}} - \frac {6 \, {\left ({\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{3}}{b^{3}} + \frac {{\left (2 \, {\left (2 \, {\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{3}}{b^{3}}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*c^3*cos(b*x + a)^2 - 24*a*c^2*d*cos(b*x + a)^2/b + 24*a^2*c*d^2*cos(b*x + a)^2/b^2 - 8*a^3*d^3*cos(b*
x + a)^2/b^3 + 6*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*c^2*d/b - 12*(2*(b*x + a)*cos(2*b*x + 2*a)
- sin(2*b*x + 2*a))*a*c*d^2/b^2 + 6*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a^2*d^3/b^3 + 6*((2*(b*x
+ a)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*c*d^2/b^2 - 6*((2*(b*x + a)^2 - 1)*cos(2*b*x + 2
*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*a*d^3/b^3 + (2*(2*(b*x + a)^3 - 3*b*x - 3*a)*cos(2*b*x + 2*a) - 3*(2*(b*x
+ a)^2 - 1)*sin(2*b*x + 2*a))*d^3/b^3)/b

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mupad [B]  time = 0.85, size = 165, normalized size = 1.38 \[ \frac {\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {3\,c\,d^2}{4}-\frac {b^2\,c^3}{2}\right )}{2\,b^3}-\frac {3\,\sin \left (2\,a+2\,b\,x\right )\,\left (d^3-2\,b^2\,c^2\,d\right )}{16\,b^4}-\frac {d^3\,x^3\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}+\frac {3\,d^3\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{8\,b^2}+\frac {3\,x\,\cos \left (2\,a+2\,b\,x\right )\,\left (d^3-2\,b^2\,c^2\,d\right )}{8\,b^3}+\frac {3\,c\,d^2\,x\,\sin \left (2\,a+2\,b\,x\right )}{4\,b^2}-\frac {3\,c\,d^2\,x^2\,\cos \left (2\,a+2\,b\,x\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.39, size = 342, normalized size = 2.85 \[ \begin {cases} - \frac {c^{3} \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {3 c^{2} d x \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {3 c^{2} d x \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {3 c d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {3 c d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {d^{3} x^{3} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d^{3} x^{3} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {3 c^{2} d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{2}} + \frac {3 c d^{2} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{2}} + \frac {3 d^{3} x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{2}} + \frac {3 c d^{2} \cos ^{2}{\left (a + b x \right )}}{4 b^{3}} - \frac {3 d^{3} x \sin ^{2}{\left (a + b x \right )}}{8 b^{3}} + \frac {3 d^{3} x \cos ^{2}{\left (a + b x \right )}}{8 b^{3}} - \frac {3 d^{3} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b^{4}} & \text {for}\: b \neq 0 \\\left (c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4}\right ) \sin {\relax (a )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*cos(b*x+a)*sin(b*x+a),x)

[Out]

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

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