3.1061 \(\int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=106 \[ -\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]

[Out]

1/4*a*b*x-1/30*(a^2+4*b^2)*cos(d*x+c)^3/d+1/4*a*b*cos(d*x+c)*sin(d*x+c)/d-1/10*a*cos(d*x+c)^3*(a+b*sin(d*x+c))
/d-1/5*cos(d*x+c)^3*(a+b*sin(d*x+c))^2/d

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Rubi [A]  time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*x)/4 - ((a^2 + 4*b^2)*Cos[c + d*x]^3)/(30*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(4*d) - (a*Cos[c + d*x]^3*
(a + b*Sin[c + d*x]))/(10*d) - (Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) (2 b+2 a \sin (c+d x)) (a+b \sin (c+d x)) \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{20} \int \cos ^2(c+d x) \left (10 a b+2 \left (a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{2} (a b) \int \cos ^2(c+d x) \, dx\\ &=-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{4} (a b) \int 1 \, dx\\ &=\frac {a b x}{4}-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 77, normalized size = 0.73 \[ \frac {-30 \left (2 a^2+b^2\right ) \cos (c+d x)-5 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+3 b (20 a (c+d x)-5 a \sin (4 (c+d x))+b \cos (5 (c+d x)))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*(2*a^2 + b^2)*Cos[c + d*x] - 5*(4*a^2 + b^2)*Cos[3*(c + d*x)] + 3*b*(20*a*(c + d*x) + b*Cos[5*(c + d*x)]
- 5*a*Sin[4*(c + d*x)]))/(240*d)

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fricas [A]  time = 0.86, size = 73, normalized size = 0.69 \[ \frac {12 \, b^{2} \cos \left (d x + c\right )^{5} + 15 \, a b d x - 20 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*b^2*cos(d*x + c)^5 + 15*a*b*d*x - 20*(a^2 + b^2)*cos(d*x + c)^3 - 15*(2*a*b*cos(d*x + c)^3 - a*b*cos(
d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.18, size = 82, normalized size = 0.77 \[ \frac {1}{4} \, a b x + \frac {b^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*b*x + 1/80*b^2*cos(5*d*x + 5*c)/d - 1/16*a*b*sin(4*d*x + 4*c)/d - 1/48*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d
- 1/8*(2*a^2 + b^2)*cos(d*x + c)/d

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maple [A]  time = 0.19, size = 94, normalized size = 0.89 \[ \frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 68, normalized size = 0.64 \[ -\frac {80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(80*a^2*cos(d*x + c)^3 - 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b - 16*(3*cos(d*x + c)^5 - 5*cos(d*x + c
)^3)*b^2)/d

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mupad [B]  time = 12.77, size = 180, normalized size = 1.70 \[ \frac {a\,b\,x}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{3}+\frac {4\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {8\,a^2}{3}-\frac {4\,b^2}{3}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {2\,a^2}{3}+\frac {4\,b^2}{15}-3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b*x)/4 - (tan(c/2 + (d*x)/2)^6*(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^2*((4*a^2)/3 + (4*b^2)/3) + tan(c/2 + (
d*x)/2)^4*((8*a^2)/3 - (4*b^2)/3) + 2*a^2*tan(c/2 + (d*x)/2)^8 + (2*a^2)/3 + (4*b^2)/15 - 3*a*b*tan(c/2 + (d*x
)/2)^3 + 3*a*b*tan(c/2 + (d*x)/2)^7 - (a*b*tan(c/2 + (d*x)/2)^9)/2 + (a*b*tan(c/2 + (d*x)/2))/2)/(d*(tan(c/2 +
(d*x)/2)^2 + 1)^5)

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sympy [A]  time = 1.96, size = 172, normalized size = 1.62 \[ \begin {cases} - \frac {a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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