3.2 \(\int (c+d x)^4 \cos (a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=156 \[ \frac {3 d^4 \sin ^2(a+b x)}{4 b^5}-\frac {3 d^3 (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^4}-\frac {3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac {d (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac {3 c d^3 x}{2 b^3}+\frac {3 d^4 x^2}{4 b^3}-\frac {(c+d x)^4}{4 b} \]

[Out]

3/2*c*d^3*x/b^3+3/4*d^4*x^2/b^3-1/4*(d*x+c)^4/b-3/2*d^3*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b^4+d*(d*x+c)^3*cos(b*x+
a)*sin(b*x+a)/b^2+3/4*d^4*sin(b*x+a)^2/b^5-3/2*d^2*(d*x+c)^2*sin(b*x+a)^2/b^3+1/2*(d*x+c)^4*sin(b*x+a)^2/b

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Rubi [A]  time = 0.11, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4404, 3311, 32, 3310} \[ -\frac {3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}-\frac {3 d^3 (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac {d (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b^2}+\frac {3 d^4 \sin ^2(a+b x)}{4 b^5}+\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac {3 c d^3 x}{2 b^3}+\frac {3 d^4 x^2}{4 b^3}-\frac {(c+d x)^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(3*c*d^3*x)/(2*b^3) + (3*d^4*x^2)/(4*b^3) - (c + d*x)^4/(4*b) - (3*d^3*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(2
*b^4) + (d*(c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/b^2 + (3*d^4*Sin[a + b*x]^2)/(4*b^5) - (3*d^2*(c + d*x)^2*Si
n[a + b*x]^2)/(2*b^3) + ((c + d*x)^4*Sin[a + b*x]^2)/(2*b)

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]

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)^4 \cos (a+b x) \sin (a+b x) \, dx &=\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}-\frac {(2 d) \int (c+d x)^3 \sin ^2(a+b x) \, dx}{b}\\ &=\frac {d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}-\frac {3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}-\frac {d \int (c+d x)^3 \, dx}{b}+\frac {\left (3 d^3\right ) \int (c+d x) \sin ^2(a+b x) \, dx}{b^3}\\ &=-\frac {(c+d x)^4}{4 b}-\frac {3 d^3 (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac {d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}+\frac {3 d^4 \sin ^2(a+b x)}{4 b^5}-\frac {3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac {\left (3 d^3\right ) \int (c+d x) \, dx}{2 b^3}\\ &=\frac {3 c d^3 x}{2 b^3}+\frac {3 d^4 x^2}{4 b^3}-\frac {(c+d x)^4}{4 b}-\frac {3 d^3 (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac {d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}+\frac {3 d^4 \sin ^2(a+b x)}{4 b^5}-\frac {3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac {(c+d x)^4 \sin ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 86, normalized size = 0.55 \[ \frac {4 b d (c+d x) \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )-2 \cos (2 (a+b x)) \left (2 b^4 (c+d x)^4-6 b^2 d^2 (c+d x)^2+3 d^4\right )}{16 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^4*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(-2*(3*d^4 - 6*b^2*d^2*(c + d*x)^2 + 2*b^4*(c + d*x)^4)*Cos[2*(a + b*x)] + 4*b*d*(c + d*x)*(-3*d^2 + 2*b^2*(c
+ d*x)^2)*Sin[2*(a + b*x)])/(16*b^5)

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fricas [A]  time = 0.66, size = 255, normalized size = 1.63 \[ \frac {b^{4} d^{4} x^{4} + 4 \, b^{4} c d^{3} x^{3} + 3 \, {\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} - {\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 2 \, b^{4} c^{4} - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \, {\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} + 4 \, {\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d - 3 \, b c d^{3} + 3 \, {\left (2 \, b^{3} c^{2} d^{2} - b d^{4}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x}{4 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

1/4*(b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 3*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 - (2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 2*b^
4*c^4 - 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 + 4*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)*cos(b*x + a
)^2 + 2*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 - b*d^4)*x)*cos(b*x + a)
*sin(b*x + a) + 2*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)/b^5

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giac [A]  time = 0.20, size = 181, normalized size = 1.16 \[ -\frac {{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 12 \, b^{4} c^{2} d^{2} x^{2} + 8 \, b^{4} c^{3} d x + 2 \, b^{4} c^{4} - 6 \, b^{2} d^{4} x^{2} - 12 \, b^{2} c d^{3} x - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{5}} + \frac {{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 6 \, b^{3} c^{2} d^{2} x + 2 \, b^{3} c^{3} d - 3 \, b d^{4} x - 3 \, b c d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-1/8*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 12*b^4*c^2*d^2*x^2 + 8*b^4*c^3*d*x + 2*b^4*c^4 - 6*b^2*d^4*x^2 - 12*b^
2*c*d^3*x - 6*b^2*c^2*d^2 + 3*d^4)*cos(2*b*x + 2*a)/b^5 + 1/4*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 6*b^3*c^2*d^2
*x + 2*b^3*c^3*d - 3*b*d^4*x - 3*b*c*d^3)*sin(2*b*x + 2*a)/b^5

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maple [B]  time = 0.06, size = 853, normalized size = 5.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x)

[Out]

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

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maxima [B]  time = 0.39, size = 586, normalized size = 3.76 \[ -\frac {4 \, c^{4} \cos \left (b x + a\right )^{2} - \frac {16 \, a c^{3} d \cos \left (b x + a\right )^{2}}{b} + \frac {24 \, a^{2} c^{2} d^{2} \cos \left (b x + a\right )^{2}}{b^{2}} - \frac {16 \, a^{3} c d^{3} \cos \left (b x + a\right )^{2}}{b^{3}} + \frac {4 \, a^{4} d^{4} \cos \left (b x + a\right )^{2}}{b^{4}} + \frac {4 \, {\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^{3} 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^{2} d^{2}}{b^{2}} + \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^{2} c d^{3}}{b^{3}} - \frac {4 \, {\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^{3} d^{4}}{b^{4}} + \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^{2} d^{2}}{b^{2}} - \frac {12 \, {\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 c 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 )} a^{2} d^{4}}{b^{4}} + \frac {2 \, {\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 )} c d^{3}}{b^{3}} - \frac {2 \, {\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 )} a d^{4}}{b^{4}} + \frac {{\left ({\left (2 \, {\left (b x + a\right )}^{4} - 6 \, {\left (b x + a\right )}^{2} + 3\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^{4}}{b^{4}}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/8*(4*c^4*cos(b*x + a)^2 - 16*a*c^3*d*cos(b*x + a)^2/b + 24*a^2*c^2*d^2*cos(b*x + a)^2/b^2 - 16*a^3*c*d^3*co
s(b*x + a)^2/b^3 + 4*a^4*d^4*cos(b*x + a)^2/b^4 + 4*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*c^3*d/b
- 12*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a*c^2*d^2/b^2 + 12*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(
2*b*x + 2*a))*a^2*c*d^3/b^3 - 4*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a^3*d^4/b^4 + 6*((2*(b*x + a
)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*c^2*d^2/b^2 - 12*((2*(b*x + a)^2 - 1)*cos(2*b*x + 2*
a) - 2*(b*x + a)*sin(2*b*x + 2*a))*a*c*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))*a^2*d^4/b^4 + 2*(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))*c*d^3/b^3 - 2*(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))*a*d^4/b^4 + ((2*(b*x + a)^4 - 6*(b*x + a)^2 + 3)*cos(2*b*x + 2*a) - 2*(2*(b*x + a)^3 - 3*b*x - 3*a)*sin(2
*b*x + 2*a))*d^4/b^4)/b

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mupad [B]  time = 0.49, size = 245, normalized size = 1.57 \[ \frac {3\,x^2\,\cos \left (2\,a+2\,b\,x\right )\,\left (d^4-2\,b^2\,c^2\,d^2\right )}{4\,b^3}-\frac {\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {b^4\,c^4}{2}-\frac {3\,b^2\,c^2\,d^2}{2}+\frac {3\,d^4}{4}\right )}{2\,b^5}-\frac {3\,x\,\sin \left (2\,a+2\,b\,x\right )\,\left (d^4-2\,b^2\,c^2\,d^2\right )}{4\,b^4}-\frac {d^4\,x^4\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}-\frac {\sin \left (2\,a+2\,b\,x\right )\,\left (3\,c\,d^3-2\,b^2\,c^3\,d\right )}{4\,b^4}+\frac {x\,\cos \left (2\,a+2\,b\,x\right )\,\left (3\,c\,d^3-2\,b^2\,c^3\,d\right )}{2\,b^3}+\frac {d^4\,x^3\,\sin \left (2\,a+2\,b\,x\right )}{2\,b^2}-\frac {c\,d^3\,x^3\,\cos \left (2\,a+2\,b\,x\right )}{b}+\frac {3\,c\,d^3\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*sin(a + b*x)*(c + d*x)^4,x)

[Out]

(3*x^2*cos(2*a + 2*b*x)*(d^4 - 2*b^2*c^2*d^2))/(4*b^3) - (cos(2*a + 2*b*x)*((3*d^4)/4 + (b^4*c^4)/2 - (3*b^2*c
^2*d^2)/2))/(2*b^5) - (3*x*sin(2*a + 2*b*x)*(d^4 - 2*b^2*c^2*d^2))/(4*b^4) - (d^4*x^4*cos(2*a + 2*b*x))/(4*b)
- (sin(2*a + 2*b*x)*(3*c*d^3 - 2*b^2*c^3*d))/(4*b^4) + (x*cos(2*a + 2*b*x)*(3*c*d^3 - 2*b^2*c^3*d))/(2*b^3) +
(d^4*x^3*sin(2*a + 2*b*x))/(2*b^2) - (c*d^3*x^3*cos(2*a + 2*b*x))/b + (3*c*d^3*x^2*sin(2*a + 2*b*x))/(2*b^2)

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sympy [A]  time = 4.25, size = 502, normalized size = 3.22 \[ \begin {cases} - \frac {c^{4} \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {c^{3} d x \sin ^{2}{\left (a + b x \right )}}{b} - \frac {c^{3} d x \cos ^{2}{\left (a + b x \right )}}{b} + \frac {3 c^{2} d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{2 b} - \frac {3 c^{2} d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {c d^{3} x^{3} \sin ^{2}{\left (a + b x \right )}}{b} - \frac {c d^{3} x^{3} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {d^{4} x^{4} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d^{4} x^{4} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {c^{3} d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {3 c^{2} d^{2} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {3 c d^{3} x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {d^{4} x^{3} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {3 c^{2} d^{2} \cos ^{2}{\left (a + b x \right )}}{2 b^{3}} - \frac {3 c d^{3} x \sin ^{2}{\left (a + b x \right )}}{2 b^{3}} + \frac {3 c d^{3} x \cos ^{2}{\left (a + b x \right )}}{2 b^{3}} - \frac {3 d^{4} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b^{3}} + \frac {3 d^{4} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b^{3}} - \frac {3 c d^{3} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{4}} - \frac {3 d^{4} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{4}} - \frac {3 d^{4} \cos ^{2}{\left (a + b x \right )}}{4 b^{5}} & \text {for}\: b \neq 0 \\\left (c^{4} x + 2 c^{3} d x^{2} + 2 c^{2} d^{2} x^{3} + c d^{3} x^{4} + \frac {d^{4} x^{5}}{5}\right ) \sin {\relax (a )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*cos(b*x+a)*sin(b*x+a),x)

[Out]

Piecewise((-c**4*cos(a + b*x)**2/(2*b) + c**3*d*x*sin(a + b*x)**2/b - c**3*d*x*cos(a + b*x)**2/b + 3*c**2*d**2
*x**2*sin(a + b*x)**2/(2*b) - 3*c**2*d**2*x**2*cos(a + b*x)**2/(2*b) + c*d**3*x**3*sin(a + b*x)**2/b - c*d**3*
x**3*cos(a + b*x)**2/b + d**4*x**4*sin(a + b*x)**2/(4*b) - d**4*x**4*cos(a + b*x)**2/(4*b) + c**3*d*sin(a + b*
x)*cos(a + b*x)/b**2 + 3*c**2*d**2*x*sin(a + b*x)*cos(a + b*x)/b**2 + 3*c*d**3*x**2*sin(a + b*x)*cos(a + b*x)/
b**2 + d**4*x**3*sin(a + b*x)*cos(a + b*x)/b**2 + 3*c**2*d**2*cos(a + b*x)**2/(2*b**3) - 3*c*d**3*x*sin(a + b*
x)**2/(2*b**3) + 3*c*d**3*x*cos(a + b*x)**2/(2*b**3) - 3*d**4*x**2*sin(a + b*x)**2/(4*b**3) + 3*d**4*x**2*cos(
a + b*x)**2/(4*b**3) - 3*c*d**3*sin(a + b*x)*cos(a + b*x)/(2*b**4) - 3*d**4*x*sin(a + b*x)*cos(a + b*x)/(2*b**
4) - 3*d**4*cos(a + b*x)**2/(4*b**5), Ne(b, 0)), ((c**4*x + 2*c**3*d*x**2 + 2*c**2*d**2*x**3 + c*d**3*x**4 + d
**4*x**5/5)*sin(a)*cos(a), True))

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