3.1107 \(\int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=129 \[ -\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a b x}{8} \]

[Out]

(a*b*x)/8 - ((a^2 + 6*b^2)*Cos[c + d*x]^5)/(105*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*b*Cos[c + d*x]
^3*Sin[c + d*x])/(12*d) - (a*Cos[c + d*x]^5*(a + b*Sin[c + d*x]))/(21*d) - (Cos[c + d*x]^5*(a + b*Sin[c + d*x]
)^2)/(7*d)

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Rubi [A]  time = 0.180754, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a b x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(a*b*x)/8 - ((a^2 + 6*b^2)*Cos[c + d*x]^5)/(105*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*b*Cos[c + d*x]
^3*Sin[c + d*x])/(12*d) - (a*Cos[c + d*x]^5*(a + b*Sin[c + d*x]))/(21*d) - (Cos[c + d*x]^5*(a + b*Sin[c + d*x]
)^2)/(7*d)

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])

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 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 8

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

Rubi steps

\begin{align*} \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{7} \int \cos ^4(c+d x) (2 b+2 a \sin (c+d x)) (a+b \sin (c+d x)) \, dx\\ &=-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{42} \int \cos ^4(c+d x) \left (14 a b+2 \left (a^2+6 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{3} (a b) \int \cos ^4(c+d x) \, dx\\ &=-\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac{a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{4} (a b) \int \cos ^2(c+d x) \, dx\\ &=-\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac{a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{8} (a b) \int 1 \, dx\\ &=\frac{a b x}{8}-\frac{\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac{a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac{\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.42195, size = 132, normalized size = 1.02 \[ \frac{-105 \left (8 a^2+3 b^2\right ) \cos (c+d x)-105 \left (4 a^2+b^2\right ) \cos (3 (c+d x))-84 a^2 \cos (5 (c+d x))+210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))-70 a b \sin (6 (c+d x))+840 a b c+840 a b d x+21 b^2 \cos (5 (c+d x))+15 b^2 \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(840*a*b*c + 840*a*b*d*x - 105*(8*a^2 + 3*b^2)*Cos[c + d*x] - 105*(4*a^2 + b^2)*Cos[3*(c + d*x)] - 84*a^2*Cos[
5*(c + d*x)] + 21*b^2*Cos[5*(c + d*x)] + 15*b^2*Cos[7*(c + d*x)] + 210*a*b*Sin[2*(c + d*x)] - 210*a*b*Sin[4*(c
+ d*x)] - 70*a*b*Sin[6*(c + d*x)])/(6720*d)

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Maple [A]  time = 0.037, size = 105, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+2\,ab \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x)

[Out]

1/d*(-1/5*a^2*cos(d*x+c)^5+2*a*b*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1
/16*d*x+1/16*c)+b^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5))

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Maxima [A]  time = 0.999241, size = 109, normalized size = 0.84 \begin{align*} -\frac{672 \, a^{2} \cos \left (d x + c\right )^{5} - 35 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 96 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{2}}{3360 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3360*(672*a^2*cos(d*x + c)^5 - 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a*b - 96*(5*c
os(d*x + c)^7 - 7*cos(d*x + c)^5)*b^2)/d

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Fricas [A]  time = 1.79268, size = 224, normalized size = 1.74 \begin{align*} \frac{120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, a b d x - 35 \,{\left (8 \, a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + b^2)*cos(d*x + c)^5 + 105*a*b*d*x - 35*(8*a*b*cos(d*x + c)^5 - 2*a*
b*cos(d*x + c)^3 - 3*a*b*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 8.16441, size = 223, normalized size = 1.73 \begin{align*} \begin{cases} - \frac{a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac{3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac{a b \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a b \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{2 b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin{\left (c \right )} \cos ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((-a**2*cos(c + d*x)**5/(5*d) + a*b*x*sin(c + d*x)**6/8 + 3*a*b*x*sin(c + d*x)**4*cos(c + d*x)**2/8 +
3*a*b*x*sin(c + d*x)**2*cos(c + d*x)**4/8 + a*b*x*cos(c + d*x)**6/8 + a*b*sin(c + d*x)**5*cos(c + d*x)/(8*d)
+ a*b*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - a*b*sin(c + d*x)*cos(c + d*x)**5/(8*d) - b**2*sin(c + d*x)**2*co
s(c + d*x)**5/(5*d) - 2*b**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c)**4, True))

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Giac [A]  time = 1.22672, size = 190, normalized size = 1.47 \begin{align*} \frac{1}{8} \, a b x + \frac{b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac{{\left (4 \, a^{2} - b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*a*b*x + 1/448*b^2*cos(7*d*x + 7*c)/d - 1/96*a*b*sin(6*d*x + 6*c)/d - 1/32*a*b*sin(4*d*x + 4*c)/d + 1/32*a*
b*sin(2*d*x + 2*c)/d - 1/320*(4*a^2 - b^2)*cos(5*d*x + 5*c)/d - 1/64*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/64*(
8*a^2 + 3*b^2)*cos(d*x + c)/d