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]

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

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 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])

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 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 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])

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*}

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Mathematica [A]  time = 0.39, size = 94, normalized size = 1.21 \[ \frac {8 a^3 \sin (2 (c+d x))-4 \left (3 a^2 b+b^3\right ) \cos (2 (c+d x))+\left (b^3-3 a^2 b\right ) \cos (4 (c+d x))+12 a \left (a^2+b^2\right ) (c+d x)+a \left (a^2-3 b^2\right ) \sin (4 (c+d x))}{32 d} \]

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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fricas [A]  time = 0.69, size = 100, normalized size = 1.28 \[ -\frac {4 \, b^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (a^{3} + a b^{2}\right )} d x - {\left (2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

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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giac [A]  time = 0.24, size = 104, normalized size = 1.33 \[ \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3}{8} \, {\left (a^{3} + a b^{2}\right )} x - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{8 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

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

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maple [A]  time = 1.38, size = 114, normalized size = 1.46 \[ \frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+3 b^{2} a \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

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*b^2*a*(-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)-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 = 0.33, size = 91, normalized size = 1.17 \[ -\frac {24 \, a^{2} b \cos \left (d x + c\right )^{4} - 8 \, b^{3} \sin \left (d x + c\right )^{4} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2}}{32 \, d} \]

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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mupad [B]  time = 1.63, size = 281, normalized size = 3.60 \[ \frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{4}-\frac {5\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,a\,b^2}{4}-\frac {5\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {21\,a\,b^2}{4}-\frac {3\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {21\,a\,b^2}{4}-\frac {3\,a^3}{4}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (a^2+b^2\right )}{4\,d}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+b^2\right )}{4\,\left (\frac {3\,a^3}{4}+\frac {3\,a\,b^2}{4}\right )}\right )\,\left (a^2+b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^3*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)*((3*a*b^2)/4 - (5*a^3)/4) + tan(c/2 + (d*x)/2)^7*((3*a*b^2)/4
- (5*a^3)/4) + tan(c/2 + (d*x)/2)^3*((21*a*b^2)/4 - (3*a^3)/4) - tan(c/2 + (d*x)/2)^5*((21*a*b^2)/4 - (3*a^3)
/4) + 6*a^2*b*tan(c/2 + (d*x)/2)^2 + 6*a^2*b*tan(c/2 + (d*x)/2)^6)/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d
*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (3*a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(a^
2 + b^2))/(4*d) + (3*a*atan((3*a*tan(c/2 + (d*x)/2)*(a^2 + b^2))/(4*((3*a*b^2)/4 + (3*a^3)/4)))*(a^2 + b^2))/(
4*d)

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sympy [A]  time = 1.12, size = 272, normalized size = 3.49 \[ \begin {cases} \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {3 a^{2} b \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {b^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{3} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

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*cos(c + d
*x)**4/(4*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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