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

Optimal. Leaf size=78 \[ \frac{3}{8} a x \left (a^2+b^2\right )+\frac{\sin ^4(c+d x) (a \cot (c+d x)+b)^3}{4 d}+\frac{3 a \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{8 d} \]

[Out]

(3*a*(a^2 + b^2)*x)/8 + (3*a*(b + a*Cot[c + d*x])*(a - b*Cot[c + d*x])*Sin[c + d*x]^2)/(8*d) + ((b + a*Cot[c +
d*x])^3*Sin[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.0635502, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3088, 805, 723, 203} \[ \frac{3}{8} a x \left (a^2+b^2\right )+\frac{\sin ^4(c+d x) (a \cot (c+d x)+b)^3}{4 d}+\frac{3 a \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(a^2 + b^2)*x)/8 + (3*a*(b + a*Cot[c + d*x])*(a - b*Cot[c + d*x])*Sin[c + d*x]^2)/(8*d) + ((b + a*Cot[c +
d*x])^3*Sin[c + d*x]^4)/(4*d)

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 805

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x (b+a x)^3}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{(b+a \cot (c+d x))^3 \sin ^4(c+d x)}{4 d}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{(b+a x)^2}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{4 d}\\ &=\frac{3 a (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{8 d}+\frac{(b+a \cot (c+d x))^3 \sin ^4(c+d x)}{4 d}-\frac{\left (3 a \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac{3}{8} a \left (a^2+b^2\right ) x+\frac{3 a (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{8 d}+\frac{(b+a \cot (c+d x))^3 \sin ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.398133, size = 94, normalized size = 1.21 \[ \frac{12 a \left (a^2+b^2\right ) (c+d x)+a \left (a^2-3 b^2\right ) \sin (4 (c+d x))-4 \left (3 a^2 b+b^3\right ) \cos (2 (c+d x))+\left (b^3-3 a^2 b\right ) \cos (4 (c+d x))+8 a^3 \sin (2 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*(a^2 + b^2)*(c + d*x) - 4*(3*a^2*b + b^3)*Cos[2*(c + d*x)] + (-3*a^2*b + b^3)*Cos[4*(c + d*x)] + 8*a^3*S
in[2*(c + d*x)] + a*(a^2 - 3*b^2)*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.066, size = 114, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}+3\,a{b}^{2} \left ( -1/4\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1/8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/8\,dx+c/8 \right ) -{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}+{a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19002, size = 123, normalized size = 1.58 \begin{align*} -\frac{24 \, a^{2} b \cos \left (d x + c\right )^{4} - 8 \, b^{3} \sin \left (d x + c\right )^{4} -{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 3 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(24*a^2*b*cos(d*x + c)^4 - 8*b^3*sin(d*x + c)^4 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c)
)*a^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b^2)/d

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Fricas [A]  time = 0.495729, size = 228, normalized size = 2.92 \begin{align*} -\frac{4 \, b^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 3 \,{\left (a^{3} + a b^{2}\right )} d x -{\left (2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.58976, size = 299, normalized size = 3.83 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{3 a^{2} b \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac{3 a^{2} b \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{b^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{3} \cos{\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)*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.15493, size = 140, normalized size = 1.79 \begin{align*} \frac{a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{3}{8} \,{\left (a^{3} + a b^{2}\right )} x - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{8 \, d} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^3*sin(2*d*x + 2*c)/d + 3/8*(a^3 + a*b^2)*x - 1/32*(3*a^2*b - b^3)*cos(4*d*x + 4*c)/d - 1/8*(3*a^2*b + b^
3)*cos(2*d*x + 2*c)/d + 1/32*(a^3 - 3*a*b^2)*sin(4*d*x + 4*c)/d