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

Optimal. Leaf size=91 \[ \frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a^2 x}{4} \]

[Out]

1/4*a^2*x-2/3*a^2*cos(d*x+c)^3/d+1/5*a^2*cos(d*x+c)^5/d+1/4*a^2*cos(d*x+c)*sin(d*x+c)/d-1/2*a^2*cos(d*x+c)^3*s
in(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 105, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac {a^2 \cos ^3(c+d x)}{6 d}-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{10 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a^2 x}{4}-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/4 - (a^2*Cos[c + d*x]^3)/(6*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(4*d) - (Cos[c + d*x]^3*(a + a*Sin[c
+ d*x])^2)/(5*d) - (Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x]))/(10*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 2678

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

Rule 2860

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[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {2}{5} \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{2} a \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {a^2 \cos ^3(c+d x)}{6 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{2} a^2 \int \cos ^2(c+d x) \, dx\\ &=-\frac {a^2 \cos ^3(c+d x)}{6 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{4} a^2 \int 1 \, dx\\ &=\frac {a^2 x}{4}-\frac {a^2 \cos ^3(c+d x)}{6 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 57, normalized size = 0.63 \[ \frac {a^2 (-90 \cos (c+d x)-25 \cos (3 (c+d x))+3 (-5 \sin (4 (c+d x))+\cos (5 (c+d x))+20 c+20 d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-90*Cos[c + d*x] - 25*Cos[3*(c + d*x)] + 3*(20*c + 20*d*x + Cos[5*(c + d*x)] - 5*Sin[4*(c + d*x)])))/(24
0*d)

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fricas [A]  time = 0.51, size = 72, normalized size = 0.79 \[ \frac {12 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} d x - 15 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^2*cos(d*x + c)^5 - 40*a^2*cos(d*x + c)^3 + 15*a^2*d*x - 15*(2*a^2*cos(d*x + c)^3 - a^2*cos(d*x + c)
)*sin(d*x + c))/d

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giac [A]  time = 0.17, size = 72, normalized size = 0.79 \[ \frac {1}{4} \, a^{2} x + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, a^{2} \cos \left (d x + c\right )}{8 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 95, normalized size = 1.04 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+2 a^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 69, normalized size = 0.76 \[ -\frac {80 \, a^{2} \cos \left (d x + c\right )^{3} - 16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.13, size = 262, normalized size = 2.88 \[ \frac {a^2\,x}{4}-\frac {\frac {a^2\,\left (c+d\,x\right )}{4}-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-\frac {a^2\,\left (15\,c+15\,d\,x-56\right )}{60}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{4}-\frac {a^2\,\left (75\,c+75\,d\,x-120\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{4}-\frac {a^2\,\left (75\,c+75\,d\,x-160\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (150\,c+150\,d\,x-80\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (150\,c+150\,d\,x-480\right )}{60}\right )+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x)/4 - ((a^2*(c + d*x))/4 - 3*a^2*tan(c/2 + (d*x)/2)^3 + 3*a^2*tan(c/2 + (d*x)/2)^7 - (a^2*tan(c/2 + (d*x
)/2)^9)/2 - (a^2*(15*c + 15*d*x - 56))/60 + tan(c/2 + (d*x)/2)^8*((5*a^2*(c + d*x))/4 - (a^2*(75*c + 75*d*x -
120))/60) + tan(c/2 + (d*x)/2)^2*((5*a^2*(c + d*x))/4 - (a^2*(75*c + 75*d*x - 160))/60) + tan(c/2 + (d*x)/2)^4
*((5*a^2*(c + d*x))/2 - (a^2*(150*c + 150*d*x - 80))/60) + tan(c/2 + (d*x)/2)^6*((5*a^2*(c + d*x))/2 - (a^2*(1
50*c + 150*d*x - 480))/60) + (a^2*tan(c/2 + (d*x)/2))/2)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)

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sympy [A]  time = 2.48, size = 172, normalized size = 1.89 \[ \begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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