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

Optimal. Leaf size=109 \[ \frac {d \sin (a+b x)}{8 b^2}+\frac {d \sin (3 a+3 b x)}{144 b^2}-\frac {d \sin (5 a+5 b x)}{400 b^2}-\frac {(c+d x) \cos (a+b x)}{8 b}-\frac {(c+d x) \cos (3 a+3 b x)}{48 b}+\frac {(c+d x) \cos (5 a+5 b x)}{80 b} \]

[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

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Rubi [A]  time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4406, 3296, 2637} \[ \frac {d \sin (a+b x)}{8 b^2}+\frac {d \sin (3 a+3 b x)}{144 b^2}-\frac {d \sin (5 a+5 b x)}{400 b^2}-\frac {(c+d x) \cos (a+b x)}{8 b}-\frac {(c+d x) \cos (3 a+3 b x)}{48 b}+\frac {(c+d x) \cos (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)*Cos[a + b*x])/(8*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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 4406

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*} \int (c+d x) \cos ^2(a+b x) \sin ^3(a+b x) \, dx &=\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*}

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Mathematica [A]  time = 0.29, size = 94, normalized size = 0.86 \[ \frac {-450 b (c+d x) \cos (a+b x)-75 b (c+d x) \cos (3 (a+b x))+45 b c \cos (5 (a+b x))+450 d \sin (a+b x)+25 d \sin (3 (a+b x))-9 d \sin (5 (a+b x))+45 b d x \cos (5 (a+b x))}{3600 b^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.93, size = 76, normalized size = 0.70 \[ \frac {45 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{5} - 75 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (9 \, d \cos \left (b x + a\right )^{4} - 13 \, d \cos \left (b x + a\right )^{2} - 26 \, d\right )} \sin \left (b x + a\right )}{225 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 1.17, size = 106, normalized size = 0.97 \[ \frac {{\left (b d x + b c\right )} \cos \left (5 \, b x + 5 \, a\right )}{80 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (3 \, b x + 3 \, a\right )}{48 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (b x + a\right )}{8 \, b^{2}} - \frac {d \sin \left (5 \, b x + 5 \, a\right )}{400 \, b^{2}} + \frac {d \sin \left (3 \, b x + 3 \, a\right )}{144 \, b^{2}} + \frac {d \sin \left (b x + a\right )}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 163, normalized size = 1.50 \[ \frac {\frac {d \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \sin \left (b x +a \right )}{15}+\frac {\left (b x +a \right ) \left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{5}-\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{25}\right )}{b}-\frac {d a \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{15}\right )}{b}+c \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{15}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b*d*(-1/3*(b*x+a)*(2+sin(b*x+a)^2)*cos(b*x+a)+1/45*sin(b*x+a)^3+2/15*sin(b*x+a)+1/5*(b*x+a)*(8/3+sin(b*
x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a)-1/25*sin(b*x+a)^5)-1/b*d*a*(-1/5*sin(b*x+a)^2*cos(b*x+a)^3-2/15*cos(b*x+a)
^3)+c*(-1/5*sin(b*x+a)^2*cos(b*x+a)^3-2/15*cos(b*x+a)^3))

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maxima [A]  time = 0.34, size = 139, normalized size = 1.28 \[ \frac {240 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} c - \frac {240 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} a d}{b} + \frac {{\left (45 \, {\left (b x + a\right )} \cos \left (5 \, b x + 5 \, a\right ) - 75 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - 9 \, \sin \left (5 \, b x + 5 \, a\right ) + 25 \, \sin \left (3 \, b x + 3 \, a\right ) + 450 \, \sin \left (b x + a\right )\right )} d}{b}}{3600 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.24, size = 99, normalized size = 0.91 \[ \frac {26\,d\,\sin \left (a+b\,x\right )-75\,b\,c\,{\cos \left (a+b\,x\right )}^3+45\,b\,c\,{\cos \left (a+b\,x\right )}^5+13\,d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )-9\,d\,{\cos \left (a+b\,x\right )}^4\,\sin \left (a+b\,x\right )-75\,b\,d\,x\,{\cos \left (a+b\,x\right )}^3+45\,b\,d\,x\,{\cos \left (a+b\,x\right )}^5}{225\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 3.05, size = 163, normalized size = 1.50 \[ \begin {cases} - \frac {c \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 c \cos ^{5}{\left (a + b x \right )}}{15 b} - \frac {d x \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 d x \cos ^{5}{\left (a + b x \right )}}{15 b} + \frac {26 d \sin ^{5}{\left (a + b x \right )}}{225 b^{2}} + \frac {13 d \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {2 d \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{15 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{3}{\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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