3.96 \(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=126 \[ \frac{5 a \left (a^2+b^2\right ) \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{16 d}+\frac{5}{16} a x \left (a^2+b^2\right )^2+\frac{\sin ^6(c+d x) (a \cot (c+d x)+b)^5}{6 d}+\frac{5 a \sin ^4(c+d x) (a \cot (c+d x)+b)^3 (a-b \cot (c+d x))}{24 d} \]

[Out]

(5*a*(a^2 + b^2)^2*x)/16 + (5*a*(a^2 + b^2)*(b + a*Cot[c + d*x])*(a - b*Cot[c + d*x])*Sin[c + d*x]^2)/(16*d) +
(5*a*(b + a*Cot[c + d*x])^3*(a - b*Cot[c + d*x])*Sin[c + d*x]^4)/(24*d) + ((b + a*Cot[c + d*x])^5*Sin[c + d*x
]^6)/(6*d)

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Rubi [A]  time = 0.0898664, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3088, 805, 723, 203} \[ \frac{5 a \left (a^2+b^2\right ) \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{16 d}+\frac{5}{16} a x \left (a^2+b^2\right )^2+\frac{\sin ^6(c+d x) (a \cot (c+d x)+b)^5}{6 d}+\frac{5 a \sin ^4(c+d x) (a \cot (c+d x)+b)^3 (a-b \cot (c+d x))}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(5*a*(a^2 + b^2)^2*x)/16 + (5*a*(a^2 + b^2)*(b + a*Cot[c + d*x])*(a - b*Cot[c + d*x])*Sin[c + d*x]^2)/(16*d) +
(5*a*(b + a*Cot[c + d*x])^3*(a - b*Cot[c + d*x])*Sin[c + d*x]^4)/(24*d) + ((b + a*Cot[c + d*x])^5*Sin[c + d*x
]^6)/(6*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))^5 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x (b+a x)^5}{\left (1+x^2\right )^4} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{(b+a x)^4}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{6 d}\\ &=\frac{5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac{(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac{\left (5 a \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{(b+a x)^2}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac{5 a \left (a^2+b^2\right ) (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{16 d}+\frac{5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac{(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac{\left (5 a \left (a^2+b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{16 d}\\ &=\frac{5}{16} a \left (a^2+b^2\right )^2 x+\frac{5 a \left (a^2+b^2\right ) (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{16 d}+\frac{5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac{(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.617089, size = 188, normalized size = 1.49 \[ \frac{60 a \left (a^2+b^2\right )^2 (c+d x)+15 a \left (2 a^2 b^2+3 a^4-b^4\right ) \sin (2 (c+d x))+3 a \left (-10 a^2 b^2+3 a^4-5 b^4\right ) \sin (4 (c+d x))+a \left (-10 a^2 b^2+a^4+5 b^4\right ) \sin (6 (c+d x))-15 b \left (6 a^2 b^2+5 a^4+b^4\right ) \cos (2 (c+d x))+6 b \left (b^4-5 a^4\right ) \cos (4 (c+d x))-b \left (-10 a^2 b^2+5 a^4+b^4\right ) \cos (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(60*a*(a^2 + b^2)^2*(c + d*x) - 15*b*(5*a^4 + 6*a^2*b^2 + b^4)*Cos[2*(c + d*x)] + 6*b*(-5*a^4 + b^4)*Cos[4*(c
+ d*x)] - b*(5*a^4 - 10*a^2*b^2 + b^4)*Cos[6*(c + d*x)] + 15*a*(3*a^4 + 2*a^2*b^2 - b^4)*Sin[2*(c + d*x)] + 3*
a*(3*a^4 - 10*a^2*b^2 - 5*b^4)*Sin[4*(c + d*x)] + a*(a^4 - 10*a^2*b^2 + 5*b^4)*Sin[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.18, size = 236, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6}}+5\,a{b}^{4} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1/8\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1/16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +10\,{a}^{2}{b}^{3} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-1/12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4} \right ) +10\,{a}^{3}{b}^{2} \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 ) -{\frac{5\,{a}^{4}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}}+{a}^{5} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \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))^5,x)

[Out]

1/d*(1/6*b^5*sin(d*x+c)^6+5*a*b^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*sin(d*x+c)*cos(d*x+c)^3+1/16*cos(d*x+c)*
sin(d*x+c)+1/16*d*x+1/16*c)+10*a^2*b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+10*a^3*b^2*(-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)-5/6*a^4*b*cos(d*x+c)^6+a^5
*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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Maxima [A]  time = 1.09425, size = 252, normalized size = 2. \begin{align*} -\frac{160 \, a^{4} b \cos \left (d x + c\right )^{6} - 32 \, b^{5} \sin \left (d x + c\right )^{6} +{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{5} - 10 \,{\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^{3} b^{2} + 160 \,{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{2} b^{3} + 5 \,{\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^{4}}{192 \, 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))^5,x, algorithm="maxima")

[Out]

-1/192*(160*a^4*b*cos(d*x + c)^6 - 32*b^5*sin(d*x + c)^6 + (4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x
+ 4*c) - 48*sin(2*d*x + 2*c))*a^5 - 10*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3*b^2 +
160*(2*sin(d*x + c)^6 - 3*sin(d*x + c)^4)*a^2*b^3 + 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*
c))*a*b^4)/d

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Fricas [A]  time = 0.536616, size = 417, normalized size = 3.31 \begin{align*} -\frac{24 \, b^{5} \cos \left (d x + c\right )^{2} + 8 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{6} + 24 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 15 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x -{\left (8 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 10 \,{\left (a^{5} + 2 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, 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))^5,x, algorithm="fricas")

[Out]

-1/48*(24*b^5*cos(d*x + c)^2 + 8*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^6 + 24*(5*a^2*b^3 - b^5)*cos(d*x +
c)^4 - 15*(a^5 + 2*a^3*b^2 + a*b^4)*d*x - (8*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^5 + 10*(a^5 + 2*a^3*b^2
- 7*a*b^4)*cos(d*x + c)^3 + 15*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 9.86799, size = 663, normalized size = 5.26 \begin{align*} \text{result too large to display} \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))**5,x)

[Out]

Piecewise((5*a**5*x*sin(c + d*x)**6/16 + 15*a**5*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**5*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**5*x*cos(c + d*x)**6/16 + 5*a**5*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**5*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**5*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 5*a**4*b*sin(c + d*x)**6/(6
*d) + 5*a**4*b*sin(c + d*x)**4*cos(c + d*x)**2/(2*d) + 5*a**4*b*sin(c + d*x)**2*cos(c + d*x)**4/(2*d) + 5*a**3
*b**2*x*sin(c + d*x)**6/8 + 15*a**3*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 15*a**3*b**2*x*sin(c + d*x)**2*
cos(c + d*x)**4/8 + 5*a**3*b**2*x*cos(c + d*x)**6/8 + 5*a**3*b**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + 5*a**3*
b**2*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - 5*a**3*b**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) + 5*a**2*b**3*sin(
c + d*x)**6/(6*d) + 5*a**2*b**3*sin(c + d*x)**4*cos(c + d*x)**2/(2*d) + 5*a*b**4*x*sin(c + d*x)**6/16 + 15*a*b
**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*a*b**4*x*cos(c +
d*x)**6/16 + 5*a*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 5*a*b**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) -
5*a*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) + b**5*sin(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))
**5*cos(c), True))

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Giac [A]  time = 1.32448, size = 285, normalized size = 2.26 \begin{align*} \frac{5}{16} \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} x - \frac{{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{{\left (5 \, a^{4} b - b^{5}\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{5 \,{\left (5 \, a^{4} b + 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (3 \, a^{5} - 10 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{5 \,{\left (3 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, 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))^5,x, algorithm="giac")

[Out]

5/16*(a^5 + 2*a^3*b^2 + a*b^4)*x - 1/192*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(6*d*x + 6*c)/d - 1/32*(5*a^4*b - b^5
)*cos(4*d*x + 4*c)/d - 5/64*(5*a^4*b + 6*a^2*b^3 + b^5)*cos(2*d*x + 2*c)/d + 1/192*(a^5 - 10*a^3*b^2 + 5*a*b^4
)*sin(6*d*x + 6*c)/d + 1/64*(3*a^5 - 10*a^3*b^2 - 5*a*b^4)*sin(4*d*x + 4*c)/d + 5/64*(3*a^5 + 2*a^3*b^2 - a*b^
4)*sin(2*d*x + 2*c)/d