\(\int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx\) [25]

Optimal result

Integrand size = 22, antiderivative size = 134 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {3 c d x}{16 b}-\frac {3 d^2 x^2}{32 b}+\frac {3 d (c+d x) \cos (a+b x) \sin (a+b x)}{16 b^2}-\frac {3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac {d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac {d^2 \sin ^4(a+b x)}{32 b^3}+\frac {(c+d x)^2 \sin ^4(a+b x)}{4 b} \]

[Out]

-3/16*c*d*x/b-3/32*d^2*x^2/b+3/16*d*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b^2-3/32*d^2*sin(b*x+a)^2/b^3+1/8*d*(d*x+c)*
cos(b*x+a)*sin(b*x+a)^3/b^2-1/32*d^2*sin(b*x+a)^4/b^3+1/4*(d*x+c)^2*sin(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4489, 3391} \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {d^2 \sin ^4(a+b x)}{32 b^3}-\frac {3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac {d (c+d x) \sin ^3(a+b x) \cos (a+b x)}{8 b^2}+\frac {3 d (c+d x) \sin (a+b x) \cos (a+b x)}{16 b^2}+\frac {(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac {3 c d x}{16 b}-\frac {3 d^2 x^2}{32 b} \]

[In]

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

[Out]

(-3*c*d*x)/(16*b) - (3*d^2*x^2)/(32*b) + (3*d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(16*b^2) - (3*d^2*Sin[a + b
*x]^2)/(32*b^3) + (d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^3)/(8*b^2) - (d^2*Sin[a + b*x]^4)/(32*b^3) + ((c + d*
x)^2*Sin[a + b*x]^4)/(4*b)

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 4489

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*} \text {integral}& = \frac {(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac {d \int (c+d x) \sin ^4(a+b x) \, dx}{2 b} \\ & = \frac {d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac {d^2 \sin ^4(a+b x)}{32 b^3}+\frac {(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac {(3 d) \int (c+d x) \sin ^2(a+b x) \, dx}{8 b} \\ & = \frac {3 d (c+d x) \cos (a+b x) \sin (a+b x)}{16 b^2}-\frac {3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac {d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac {d^2 \sin ^4(a+b x)}{32 b^3}+\frac {(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac {(3 d) \int (c+d x) \, dx}{16 b} \\ & = -\frac {3 c d x}{16 b}-\frac {3 d^2 x^2}{32 b}+\frac {3 d (c+d x) \cos (a+b x) \sin (a+b x)}{16 b^2}-\frac {3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac {d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac {d^2 \sin ^4(a+b x)}{32 b^3}+\frac {(c+d x)^2 \sin ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.68 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {-16 \left (-d^2+2 b^2 (c+d x)^2\right ) \cos (2 (a+b x))+\left (-d^2+8 b^2 (c+d x)^2\right ) \cos (4 (a+b x))-4 b d (c+d x) (-8 \sin (2 (a+b x))+\sin (4 (a+b x)))}{256 b^3} \]

[In]

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

[Out]

(-16*(-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] + (-d^2 + 8*b^2*(c + d*x)^2)*Cos[4*(a + b*x)] - 4*b*d*(c + d*
x)*(-8*Sin[2*(a + b*x)] + Sin[4*(a + b*x)]))/(256*b^3)

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\frac {16 \left (-2 \left (d x +c \right )^{2} b^{2}+d^{2}\right ) \cos \left (2 x b +2 a \right )+\left (8 \left (d x +c \right )^{2} b^{2}-d^{2}\right ) \cos \left (4 x b +4 a \right )+32 b d \left (d x +c \right ) \sin \left (2 x b +2 a \right )-4 b d \left (d x +c \right ) \sin \left (4 x b +4 a \right )+24 b^{2} c^{2}-15 d^{2}}{256 b^{3}}\)	\(111\)
risch	\(\frac {\left (8 x^{2} d^{2} b^{2}+16 b^{2} c d x +8 b^{2} c^{2}-d^{2}\right ) \cos \left (4 x b +4 a \right )}{256 b^{3}}-\frac {d \left (d x +c \right ) \sin \left (4 x b +4 a \right )}{64 b^{2}}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}-d^{2}\right ) \cos \left (2 x b +2 a \right )}{16 b^{3}}+\frac {d \left (d x +c \right ) \sin \left (2 x b +2 a \right )}{8 b^{2}}\)	\(136\)
derivativedivides	\(\frac {\frac {a^{2} d^{2} \sin \left (x b +a \right )^{4}}{4 b^{2}}-\frac {a c d \sin \left (x b +a \right )^{4}}{2 b}-\frac {2 a \,d^{2} \left (\frac {\left (x b +a \right ) \sin \left (x b +a \right )^{4}}{4}+\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{16}-\frac {3 x b}{32}-\frac {3 a}{32}\right )}{b^{2}}+\frac {c^{2} \sin \left (x b +a \right )^{4}}{4}+\frac {2 c d \left (\frac {\left (x b +a \right ) \sin \left (x b +a \right )^{4}}{4}+\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{16}-\frac {3 x b}{32}-\frac {3 a}{32}\right )}{b}+\frac {d^{2} \left (\frac {\left (x b +a \right )^{2} \sin \left (x b +a \right )^{4}}{4}-\frac {\left (x b +a \right ) \left (-\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{4}+\frac {3 x b}{8}+\frac {3 a}{8}\right )}{2}+\frac {3 \left (x b +a \right )^{2}}{32}-\frac {\left (2 \cos \left (x b +a \right )^{2}-5\right )^{2}}{128}\right )}{b^{2}}}{b}\)	\(256\)
default	\(\frac {\frac {a^{2} d^{2} \sin \left (x b +a \right )^{4}}{4 b^{2}}-\frac {a c d \sin \left (x b +a \right )^{4}}{2 b}-\frac {2 a \,d^{2} \left (\frac {\left (x b +a \right ) \sin \left (x b +a \right )^{4}}{4}+\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{16}-\frac {3 x b}{32}-\frac {3 a}{32}\right )}{b^{2}}+\frac {c^{2} \sin \left (x b +a \right )^{4}}{4}+\frac {2 c d \left (\frac {\left (x b +a \right ) \sin \left (x b +a \right )^{4}}{4}+\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{16}-\frac {3 x b}{32}-\frac {3 a}{32}\right )}{b}+\frac {d^{2} \left (\frac {\left (x b +a \right )^{2} \sin \left (x b +a \right )^{4}}{4}-\frac {\left (x b +a \right ) \left (-\frac {\left (\sin \left (x b +a \right )^{3}+\frac {3 \sin \left (x b +a \right )}{2}\right ) \cos \left (x b +a \right )}{4}+\frac {3 x b}{8}+\frac {3 a}{8}\right )}{2}+\frac {3 \left (x b +a \right )^{2}}{32}-\frac {\left (2 \cos \left (x b +a \right )^{2}-5\right )^{2}}{128}\right )}{b^{2}}}{b}\)	\(256\)
norman	\(\frac {-\frac {3 d^{2} x^{2}}{32 b}-\frac {3 d^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{8 b^{3}}-\frac {3 d^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{8 b^{3}}+\frac {\left (16 b^{2} c^{2}-5 d^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{4 b^{3}}+\frac {3 c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{8 b^{2}}+\frac {11 c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{8 b^{2}}-\frac {11 c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{8 b^{2}}-\frac {3 c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{8 b^{2}}-\frac {3 c d x}{16 b}+\frac {3 d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{8 b^{2}}+\frac {11 d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{8 b^{2}}-\frac {11 d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{8 b^{2}}-\frac {3 d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{8 b^{2}}-\frac {3 d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{8 b}+\frac {55 d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{16 b}-\frac {3 d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{8 b}-\frac {3 d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}}{32 b}-\frac {3 c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{4 b}+\frac {55 c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{8 b}-\frac {3 c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{4 b}-\frac {3 c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}}{16 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{4}}\)	\(417\)

[In]

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

[Out]

1/256*(16*(-2*(d*x+c)^2*b^2+d^2)*cos(2*b*x+2*a)+(8*(d*x+c)^2*b^2-d^2)*cos(4*b*x+4*a)+32*b*d*(d*x+c)*sin(2*b*x+
2*a)-4*b*d*(d*x+c)*sin(4*b*x+4*a)+24*b^2*c^2-15*d^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.19 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {5 \, b^{2} d^{2} x^{2} + 10 \, b^{2} c d x + {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{4} - {\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 5 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 2 \, {\left (2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 5 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{32 \, b^{3}} \]

[In]

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

[Out]

1/32*(5*b^2*d^2*x^2 + 10*b^2*c*d*x + (8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(b*x + a)^4 - (16*b^2
*d^2*x^2 + 32*b^2*c*d*x + 16*b^2*c^2 - 5*d^2)*cos(b*x + a)^2 - 2*(2*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 5*(b*d^
2*x + b*c*d)*cos(b*x + a))*sin(b*x + a))/b^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (129) = 258\).

Time = 0.43 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.39 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\begin {cases} \frac {c^{2} \sin ^{4}{\left (a + b x \right )}}{4 b} + \frac {5 c d x \sin ^{4}{\left (a + b x \right )}}{16 b} - \frac {3 c d x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b} - \frac {3 c d x \cos ^{4}{\left (a + b x \right )}}{16 b} + \frac {5 d^{2} x^{2} \sin ^{4}{\left (a + b x \right )}}{32 b} - \frac {3 d^{2} x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16 b} - \frac {3 d^{2} x^{2} \cos ^{4}{\left (a + b x \right )}}{32 b} + \frac {5 c d \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{16 b^{2}} + \frac {3 c d \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{16 b^{2}} + \frac {5 d^{2} x \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{16 b^{2}} + \frac {3 d^{2} x \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{16 b^{2}} - \frac {5 d^{2} \sin ^{4}{\left (a + b x \right )}}{64 b^{3}} + \frac {3 d^{2} \cos ^{4}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin ^{3}{\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((d*x+c)**2*cos(b*x+a)*sin(b*x+a)**3,x)

[Out]

Piecewise((c**2*sin(a + b*x)**4/(4*b) + 5*c*d*x*sin(a + b*x)**4/(16*b) - 3*c*d*x*sin(a + b*x)**2*cos(a + b*x)*
*2/(8*b) - 3*c*d*x*cos(a + b*x)**4/(16*b) + 5*d**2*x**2*sin(a + b*x)**4/(32*b) - 3*d**2*x**2*sin(a + b*x)**2*c
os(a + b*x)**2/(16*b) - 3*d**2*x**2*cos(a + b*x)**4/(32*b) + 5*c*d*sin(a + b*x)**3*cos(a + b*x)/(16*b**2) + 3*
c*d*sin(a + b*x)*cos(a + b*x)**3/(16*b**2) + 5*d**2*x*sin(a + b*x)**3*cos(a + b*x)/(16*b**2) + 3*d**2*x*sin(a
+ b*x)*cos(a + b*x)**3/(16*b**2) - 5*d**2*sin(a + b*x)**4/(64*b**3) + 3*d**2*cos(a + b*x)**4/(64*b**3), Ne(b,
0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)**3*cos(a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (120) = 240\).

Time = 0.22 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.96 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {64 \, c^{2} \sin \left (b x + a\right )^{4} - \frac {128 \, a c d \sin \left (b x + a\right )^{4}}{b} + \frac {64 \, a^{2} d^{2} \sin \left (b x + a\right )^{4}}{b^{2}} + \frac {4 \, {\left (4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) - 16 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} c d}{b} - \frac {4 \, {\left (4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) - 16 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left (8 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (4 \, b x + 4 \, a\right ) - 16 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{256 \, b} \]

[In]

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

[Out]

1/256*(64*c^2*sin(b*x + a)^4 - 128*a*c*d*sin(b*x + a)^4/b + 64*a^2*d^2*sin(b*x + a)^4/b^2 + 4*(4*(b*x + a)*cos
(4*b*x + 4*a) - 16*(b*x + a)*cos(2*b*x + 2*a) - sin(4*b*x + 4*a) + 8*sin(2*b*x + 2*a))*c*d/b - 4*(4*(b*x + a)*
cos(4*b*x + 4*a) - 16*(b*x + a)*cos(2*b*x + 2*a) - sin(4*b*x + 4*a) + 8*sin(2*b*x + 2*a))*a*d^2/b^2 + ((8*(b*x
+ a)^2 - 1)*cos(4*b*x + 4*a) - 16*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 4*(b*x + a)*sin(4*b*x + 4*a) + 32*(b
*x + a)*sin(2*b*x + 2*a))*d^2/b^2)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} - \frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{16 \, b^{3}} - \frac {{\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )}{64 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]

[In]

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

[Out]

1/256*(8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(4*b*x + 4*a)/b^3 - 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*
x + 2*b^2*c^2 - d^2)*cos(2*b*x + 2*a)/b^3 - 1/64*(b*d^2*x + b*c*d)*sin(4*b*x + 4*a)/b^3 + 1/8*(b*d^2*x + b*c*d
)*sin(2*b*x + 2*a)/b^3

Mupad [B] (verification not implemented)

Time = 24.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.51 \[ \int (c+d x)^2 \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {8\,d^2\,\cos \left (2\,a+2\,b\,x\right )-\frac {d^2\,\cos \left (4\,a+4\,b\,x\right )}{2}-16\,b^2\,c^2\,\cos \left (2\,a+2\,b\,x\right )+4\,b^2\,c^2\,\cos \left (4\,a+4\,b\,x\right )+16\,b\,c\,d\,\sin \left (2\,a+2\,b\,x\right )-2\,b\,c\,d\,\sin \left (4\,a+4\,b\,x\right )-16\,b^2\,d^2\,x^2\,\cos \left (2\,a+2\,b\,x\right )+4\,b^2\,d^2\,x^2\,\cos \left (4\,a+4\,b\,x\right )+16\,b\,d^2\,x\,\sin \left (2\,a+2\,b\,x\right )-2\,b\,d^2\,x\,\sin \left (4\,a+4\,b\,x\right )-32\,b^2\,c\,d\,x\,\cos \left (2\,a+2\,b\,x\right )+8\,b^2\,c\,d\,x\,\cos \left (4\,a+4\,b\,x\right )}{128\,b^3} \]

[In]

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

[Out]

(8*d^2*cos(2*a + 2*b*x) - (d^2*cos(4*a + 4*b*x))/2 - 16*b^2*c^2*cos(2*a + 2*b*x) + 4*b^2*c^2*cos(4*a + 4*b*x)
+ 16*b*c*d*sin(2*a + 2*b*x) - 2*b*c*d*sin(4*a + 4*b*x) - 16*b^2*d^2*x^2*cos(2*a + 2*b*x) + 4*b^2*d^2*x^2*cos(4
*a + 4*b*x) + 16*b*d^2*x*sin(2*a + 2*b*x) - 2*b*d^2*x*sin(4*a + 4*b*x) - 32*b^2*c*d*x*cos(2*a + 2*b*x) + 8*b^2
*c*d*x*cos(4*a + 4*b*x))/(128*b^3)