\(\int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx\) [149]

Optimal result

Integrand size = 22, antiderivative size = 109 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=\frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b} \]

[Out]

1/8*d*cos(b*x+a)/b^2-1/144*d*cos(3*b*x+3*a)/b^2-1/400*d*cos(5*b*x+5*a)/b^2+1/8*(d*x+c)*sin(b*x+a)/b-1/48*(d*x+
c)*sin(3*b*x+3*a)/b-1/80*(d*x+c)*sin(5*b*x+5*a)/b

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4491, 3377, 2718} \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=\frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b} \]

[In]

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

[Out]

(d*Cos[a + b*x])/(8*b^2) - (d*Cos[3*a + 3*b*x])/(144*b^2) - (d*Cos[5*a + 5*b*x])/(400*b^2) + ((c + d*x)*Sin[a
+ b*x])/(8*b) - ((c + d*x)*Sin[3*a + 3*b*x])/(48*b) - ((c + d*x)*Sin[5*a + 5*b*x])/(80*b)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8} (c+d x) \cos (a+b x)-\frac {1}{16} (c+d x) \cos (3 a+3 b x)-\frac {1}{16} (c+d x) \cos (5 a+5 b x)\right ) \, dx \\ & = -\left (\frac {1}{16} \int (c+d x) \cos (3 a+3 b x) \, dx\right )-\frac {1}{16} \int (c+d x) \cos (5 a+5 b x) \, dx+\frac {1}{8} \int (c+d x) \cos (a+b x) \, dx \\ & = \frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b}+\frac {d \int \sin (5 a+5 b x) \, dx}{80 b}+\frac {d \int \sin (3 a+3 b x) \, dx}{48 b}-\frac {d \int \sin (a+b x) \, dx}{8 b} \\ & = \frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=-\frac {-450 d \cos (a+b x)+25 d \cos (3 (a+b x))+9 d \cos (5 (a+b x))-450 b c \sin (a+b x)-450 b d x \sin (a+b x)+75 b c \sin (3 (a+b x))+75 b d x \sin (3 (a+b x))+45 b c \sin (5 (a+b x))+45 b d x \sin (5 (a+b x))}{3600 b^2} \]

[In]

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

[Out]

-1/3600*(-450*d*Cos[a + b*x] + 25*d*Cos[3*(a + b*x)] + 9*d*Cos[5*(a + b*x)] - 450*b*c*Sin[a + b*x] - 450*b*d*x
*Sin[a + b*x] + 75*b*c*Sin[3*(a + b*x)] + 75*b*d*x*Sin[3*(a + b*x)] + 45*b*c*Sin[5*(a + b*x)] + 45*b*d*x*Sin[5
*(a + b*x)])/b^2

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\frac {-75 b \left (d x +c \right ) \sin \left (3 x b +3 a \right )-45 b \left (d x +c \right ) \sin \left (5 x b +5 a \right )-25 \cos \left (3 x b +3 a \right ) d -9 \cos \left (5 x b +5 a \right ) d +450 \left (d x +c \right ) b \sin \left (x b +a \right )+450 \cos \left (x b +a \right ) d +416 d}{3600 b^{2}}\)	\(91\)
risch	\(\frac {d \cos \left (x b +a \right )}{8 b^{2}}-\frac {d \cos \left (3 x b +3 a \right )}{144 b^{2}}-\frac {d \cos \left (5 x b +5 a \right )}{400 b^{2}}+\frac {\left (d x +c \right ) \sin \left (x b +a \right )}{8 b}-\frac {\left (d x +c \right ) \sin \left (3 x b +3 a \right )}{48 b}-\frac {\left (d x +c \right ) \sin \left (5 x b +5 a \right )}{80 b}\)	\(98\)
derivativedivides	\(\frac {-\frac {d a \left (-\frac {\cos \left (x b +a \right )^{4} \sin \left (x b +a \right )}{5}+\frac {\left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{15}\right )}{b}+c \left (-\frac {\cos \left (x b +a \right )^{4} \sin \left (x b +a \right )}{5}+\frac {\left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{15}\right )+\frac {d \left (\frac {\left (x b +a \right ) \left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{3}+\frac {\cos \left (x b +a \right )^{3}}{45}+\frac {2 \cos \left (x b +a \right )}{15}-\frac {\left (x b +a \right ) \left (\frac {8}{3}+\cos \left (x b +a \right )^{4}+\frac {4 \cos \left (x b +a \right )^{2}}{3}\right ) \sin \left (x b +a \right )}{5}-\frac {\cos \left (x b +a \right )^{5}}{25}\right )}{b}}{b}\)	\(175\)
default	\(\frac {-\frac {d a \left (-\frac {\cos \left (x b +a \right )^{4} \sin \left (x b +a \right )}{5}+\frac {\left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{15}\right )}{b}+c \left (-\frac {\cos \left (x b +a \right )^{4} \sin \left (x b +a \right )}{5}+\frac {\left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{15}\right )+\frac {d \left (\frac {\left (x b +a \right ) \left (2+\cos \left (x b +a \right )^{2}\right ) \sin \left (x b +a \right )}{3}+\frac {\cos \left (x b +a \right )^{3}}{45}+\frac {2 \cos \left (x b +a \right )}{15}-\frac {\left (x b +a \right ) \left (\frac {8}{3}+\cos \left (x b +a \right )^{4}+\frac {4 \cos \left (x b +a \right )^{2}}{3}\right ) \sin \left (x b +a \right )}{5}-\frac {\cos \left (x b +a \right )^{5}}{25}\right )}{b}}{b}\)	\(175\)
norman	\(\frac {\frac {52 d}{225 b^{2}}+\frac {8 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{3 b}-\frac {16 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{15 b}+\frac {8 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{3 b}+\frac {4 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{3 b^{2}}+\frac {44 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{45 b^{2}}+\frac {52 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{45 b^{2}}+\frac {8 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{3 b}-\frac {16 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{15 b}+\frac {8 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{3 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{5}}\)	\(180\)

[In]

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

[Out]

1/3600*(-75*b*(d*x+c)*sin(3*b*x+3*a)-45*b*(d*x+c)*sin(5*b*x+5*a)-25*cos(3*b*x+3*a)*d-9*cos(5*b*x+5*a)*d+450*(d
*x+c)*b*sin(b*x+a)+450*cos(b*x+a)*d+416*d)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=-\frac {9 \, d \cos \left (b x + a\right )^{5} - 5 \, d \cos \left (b x + a\right )^{3} - 30 \, d \cos \left (b x + a\right ) + 15 \, {\left (3 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} - 2 \, b d x - {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} - 2 \, b c\right )} \sin \left (b x + a\right )}{225 \, b^{2}} \]

[In]

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

[Out]

-1/225*(9*d*cos(b*x + a)^5 - 5*d*cos(b*x + a)^3 - 30*d*cos(b*x + a) + 15*(3*(b*d*x + b*c)*cos(b*x + a)^4 - 2*b
*d*x - (b*d*x + b*c)*cos(b*x + a)^2 - 2*b*c)*sin(b*x + a))/b^2

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.50 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=\begin {cases} \frac {2 c \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {c \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {2 d x \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {d x \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {2 d \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{15 b^{2}} + \frac {13 d \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac {26 d \cos ^{5}{\left (a + b x \right )}}{225 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\left (a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*c*sin(a + b*x)**5/(15*b) + c*sin(a + b*x)**3*cos(a + b*x)**2/(3*b) + 2*d*x*sin(a + b*x)**5/(15*b)
+ d*x*sin(a + b*x)**3*cos(a + b*x)**2/(3*b) + 2*d*sin(a + b*x)**4*cos(a + b*x)/(15*b**2) + 13*d*sin(a + b*x)*
*2*cos(a + b*x)**3/(45*b**2) + 26*d*cos(a + b*x)**5/(225*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)**2*cos(a)*
*3, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=-\frac {240 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} c - \frac {240 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} a d}{b} + \frac {{\left (45 \, {\left (b x + a\right )} \sin \left (5 \, b x + 5 \, a\right ) + 75 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + 9 \, \cos \left (5 \, b x + 5 \, a\right ) + 25 \, \cos \left (3 \, b x + 3 \, a\right ) - 450 \, \cos \left (b x + a\right )\right )} d}{b}}{3600 \, b} \]

[In]

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

[Out]

-1/3600*(240*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)*c - 240*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)*a*d/b + (45*(
b*x + a)*sin(5*b*x + 5*a) + 75*(b*x + a)*sin(3*b*x + 3*a) - 450*(b*x + a)*sin(b*x + a) + 9*cos(5*b*x + 5*a) +
25*cos(3*b*x + 3*a) - 450*cos(b*x + a))*d/b)/b

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=-\frac {d \cos \left (5 \, b x + 5 \, a\right )}{400 \, b^{2}} - \frac {d \cos \left (3 \, b x + 3 \, a\right )}{144 \, b^{2}} + \frac {d \cos \left (b x + a\right )}{8 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (5 \, b x + 5 \, a\right )}{80 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (3 \, b x + 3 \, a\right )}{48 \, b^{2}} + \frac {{\left (b d x + b c\right )} \sin \left (b x + a\right )}{8 \, b^{2}} \]

[In]

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

[Out]

-1/400*d*cos(5*b*x + 5*a)/b^2 - 1/144*d*cos(3*b*x + 3*a)/b^2 + 1/8*d*cos(b*x + a)/b^2 - 1/80*(b*d*x + b*c)*sin
(5*b*x + 5*a)/b^2 - 1/48*(b*d*x + b*c)*sin(3*b*x + 3*a)/b^2 + 1/8*(b*d*x + b*c)*sin(b*x + a)/b^2

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09 \[ \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx=\frac {26\,d\,{\cos \left (a+b\,x\right )}^5+65\,d\,{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^2+30\,d\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^4+30\,b\,c\,{\sin \left (a+b\,x\right )}^5+30\,b\,d\,x\,{\sin \left (a+b\,x\right )}^5+75\,b\,c\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3+75\,b\,d\,x\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{225\,b^2} \]

[In]

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

[Out]

(26*d*cos(a + b*x)^5 + 65*d*cos(a + b*x)^3*sin(a + b*x)^2 + 30*d*cos(a + b*x)*sin(a + b*x)^4 + 30*b*c*sin(a +
b*x)^5 + 30*b*d*x*sin(a + b*x)^5 + 75*b*c*cos(a + b*x)^2*sin(a + b*x)^3 + 75*b*d*x*cos(a + b*x)^2*sin(a + b*x)
^3)/(225*b^2)