\(\int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx\) [92]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (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, 2717} \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {d \sin (a+b x)}{8 b^2}+\frac {d \sin (3 a+3 b x)}{144 b^2}-\frac {d \sin (5 a+5 b x)}{400 b^2}-\frac {(c+d x) \cos (a+b x)}{8 b}-\frac {(c+d x) \cos (3 a+3 b x)}{48 b}+\frac {(c+d x) \cos (5 a+5 b x)}{80 b} \]

[In]

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

[Out]

-1/8*((c + d*x)*Cos[a + b*x])/b - ((c + d*x)*Cos[3*a + 3*b*x])/(48*b) + ((c + d*x)*Cos[5*a + 5*b*x])/(80*b) +
(d*Sin[a + b*x])/(8*b^2) + (d*Sin[3*a + 3*b*x])/(144*b^2) - (d*Sin[5*a + 5*b*x])/(400*b^2)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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) \sin (a+b x)+\frac {1}{16} (c+d x) \sin (3 a+3 b x)-\frac {1}{16} (c+d x) \sin (5 a+5 b x)\right ) \, dx \\ & = \frac {1}{16} \int (c+d x) \sin (3 a+3 b x) \, dx-\frac {1}{16} \int (c+d x) \sin (5 a+5 b x) \, dx+\frac {1}{8} \int (c+d x) \sin (a+b x) \, dx \\ & = -\frac {(c+d x) \cos (a+b x)}{8 b}-\frac {(c+d x) \cos (3 a+3 b x)}{48 b}+\frac {(c+d x) \cos (5 a+5 b x)}{80 b}-\frac {d \int \cos (5 a+5 b x) \, dx}{80 b}+\frac {d \int \cos (3 a+3 b x) \, dx}{48 b}+\frac {d \int \cos (a+b x) \, dx}{8 b} \\ & = -\frac {(c+d x) \cos (a+b x)}{8 b}-\frac {(c+d x) \cos (3 a+3 b x)}{48 b}+\frac {(c+d x) \cos (5 a+5 b x)}{80 b}+\frac {d \sin (a+b x)}{8 b^2}+\frac {d \sin (3 a+3 b x)}{144 b^2}-\frac {d \sin (5 a+5 b x)}{400 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {-450 b (c+d x) \cos (a+b x)-75 b (c+d x) \cos (3 (a+b x))+45 b c \cos (5 (a+b x))+45 b d x \cos (5 (a+b x))+450 d \sin (a+b x)+25 d \sin (3 (a+b x))-9 d \sin (5 (a+b x))}{3600 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {-75 b \left (d x +c \right ) \cos \left (3 x b +3 a \right )+45 b \left (d x +c \right ) \cos \left (5 x b +5 a \right )+25 \sin \left (3 x b +3 a \right ) d -9 d \sin \left (5 x b +5 a \right )-450 b \left (d x +c \right ) \cos \left (x b +a \right )-480 c b +450 d \sin \left (x b +a \right )}{3600 b^{2}}\)	\(92\)
risch	\(-\frac {\left (d x +c \right ) \cos \left (x b +a \right )}{8 b}-\frac {\left (d x +c \right ) \cos \left (3 x b +3 a \right )}{48 b}+\frac {\left (d x +c \right ) \cos \left (5 x b +5 a \right )}{80 b}+\frac {d \sin \left (x b +a \right )}{8 b^{2}}+\frac {d \sin \left (3 x b +3 a \right )}{144 b^{2}}-\frac {d \sin \left (5 x b +5 a \right )}{400 b^{2}}\)	\(98\)
derivativedivides	\(\frac {-\frac {d a \left (-\frac {\sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}{5}-\frac {2 \cos \left (x b +a \right )^{3}}{15}\right )}{b}+c \left (-\frac {\sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}{5}-\frac {2 \cos \left (x b +a \right )^{3}}{15}\right )+\frac {d \left (-\frac {\left (x b +a \right ) \left (2+\sin \left (x b +a \right )^{2}\right ) \cos \left (x b +a \right )}{3}+\frac {\sin \left (x b +a \right )^{3}}{45}+\frac {2 \sin \left (x b +a \right )}{15}+\frac {\left (x b +a \right ) \left (\frac {8}{3}+\sin \left (x b +a \right )^{4}+\frac {4 \sin \left (x b +a \right )^{2}}{3}\right ) \cos \left (x b +a \right )}{5}-\frac {\sin \left (x b +a \right )^{5}}{25}\right )}{b}}{b}\)	\(163\)
default	\(\frac {-\frac {d a \left (-\frac {\sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}{5}-\frac {2 \cos \left (x b +a \right )^{3}}{15}\right )}{b}+c \left (-\frac {\sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}{5}-\frac {2 \cos \left (x b +a \right )^{3}}{15}\right )+\frac {d \left (-\frac {\left (x b +a \right ) \left (2+\sin \left (x b +a \right )^{2}\right ) \cos \left (x b +a \right )}{3}+\frac {\sin \left (x b +a \right )^{3}}{45}+\frac {2 \sin \left (x b +a \right )}{15}+\frac {\left (x b +a \right ) \left (\frac {8}{3}+\sin \left (x b +a \right )^{4}+\frac {4 \sin \left (x b +a \right )^{2}}{3}\right ) \cos \left (x b +a \right )}{5}-\frac {\sin \left (x b +a \right )^{5}}{25}\right )}{b}}{b}\)	\(163\)
norman	\(\frac {-\frac {4 c}{15 b}+\frac {4 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{15 b^{2}}+\frac {56 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{45 b^{2}}+\frac {152 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{225 b^{2}}+\frac {56 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{45 b^{2}}+\frac {4 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{9}}{15 b^{2}}-\frac {2 d x}{15 b}-\frac {4 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{b}-\frac {4 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{3 b}+\frac {4 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{3 b}-\frac {2 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{3 b}+\frac {8 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{3 b}-\frac {8 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{3 b}+\frac {2 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}}{3 b}+\frac {2 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{10}}{15 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{5}}\)	\(255\)

[In]

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

[Out]

1/3600*(-75*b*(d*x+c)*cos(3*b*x+3*a)+45*b*(d*x+c)*cos(5*b*x+5*a)+25*sin(3*b*x+3*a)*d-9*d*sin(5*b*x+5*a)-450*b*
(d*x+c)*cos(b*x+a)-480*c*b+450*d*sin(b*x+a))/b^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {45 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{5} - 75 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (9 \, d \cos \left (b x + a\right )^{4} - 13 \, d \cos \left (b x + a\right )^{2} - 26 \, d\right )} \sin \left (b x + a\right )}{225 \, b^{2}} \]

[In]

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

[Out]

1/225*(45*(b*d*x + b*c)*cos(b*x + a)^5 - 75*(b*d*x + b*c)*cos(b*x + a)^3 - (9*d*cos(b*x + a)^4 - 13*d*cos(b*x
+ a)^2 - 26*d)*sin(b*x + a))/b^2

Sympy [A] (verification not implemented)

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {240 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} c - \frac {240 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} a d}{b} + \frac {{\left (45 \, {\left (b x + a\right )} \cos \left (5 \, b x + 5 \, a\right ) - 75 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - 9 \, \sin \left (5 \, b x + 5 \, a\right ) + 25 \, \sin \left (3 \, b x + 3 \, a\right ) + 450 \, \sin \left (b x + a\right )\right )} d}{b}}{3600 \, b} \]

[In]

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

[Out]

1/3600*(240*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)*c - 240*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)*a*d/b + (45*(b
*x + a)*cos(5*b*x + 5*a) - 75*(b*x + a)*cos(3*b*x + 3*a) - 450*(b*x + a)*cos(b*x + a) - 9*sin(5*b*x + 5*a) + 2
5*sin(3*b*x + 3*a) + 450*sin(b*x + a))*d/b)/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {{\left (b d x + b c\right )} \cos \left (5 \, b x + 5 \, a\right )}{80 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (3 \, b x + 3 \, a\right )}{48 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (b x + a\right )}{8 \, b^{2}} - \frac {d \sin \left (5 \, b x + 5 \, a\right )}{400 \, b^{2}} + \frac {d \sin \left (3 \, b x + 3 \, a\right )}{144 \, b^{2}} + \frac {d \sin \left (b x + a\right )}{8 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 23.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {26\,d\,\sin \left (a+b\,x\right )-75\,b\,c\,{\cos \left (a+b\,x\right )}^3+45\,b\,c\,{\cos \left (a+b\,x\right )}^5+13\,d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )-9\,d\,{\cos \left (a+b\,x\right )}^4\,\sin \left (a+b\,x\right )-75\,b\,d\,x\,{\cos \left (a+b\,x\right )}^3+45\,b\,d\,x\,{\cos \left (a+b\,x\right )}^5}{225\,b^2} \]

[In]

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

[Out]

(26*d*sin(a + b*x) - 75*b*c*cos(a + b*x)^3 + 45*b*c*cos(a + b*x)^5 + 13*d*cos(a + b*x)^2*sin(a + b*x) - 9*d*co
s(a + b*x)^4*sin(a + b*x) - 75*b*d*x*cos(a + b*x)^3 + 45*b*d*x*cos(a + b*x)^5)/(225*b^2)