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

Optimal. Leaf size=174 \[ \frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^2 x}{16}-\frac{a b \cos ^6(c+d x)}{3 d}-\frac{b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b^2 x}{16} \]

[Out]

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

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Rubi [A]  time = 0.170358, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 2635, 8, 2565, 30, 2568} \[ \frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^2 x}{16}-\frac{a b \cos ^6(c+d x)}{3 d}-\frac{b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b^2 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 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]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2568

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

Rubi steps

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

Mathematica [A]  time = 0.24063, size = 147, normalized size = 0.84 \[ \frac{\left (5 a^2+b^2\right ) (c+d x)}{16 d}+\frac{\left (15 a^2+b^2\right ) \sin (2 (c+d x))}{64 d}+\frac{\left (3 a^2-b^2\right ) \sin (4 (c+d x))}{64 d}+\frac{\left (a^2-b^2\right ) \sin (6 (c+d x))}{192 d}-\frac{5 a b \cos (2 (c+d x))}{32 d}-\frac{a b \cos (4 (c+d x))}{16 d}-\frac{a b \cos (6 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^2 + b^2)*(c + d*x))/(16*d) - (5*a*b*Cos[2*(c + d*x)])/(32*d) - (a*b*Cos[4*(c + d*x)])/(16*d) - (a*b*Cos[
6*(c + d*x)])/(96*d) + ((15*a^2 + b^2)*Sin[2*(c + d*x)])/(64*d) + ((3*a^2 - b^2)*Sin[4*(c + d*x)])/(64*d) + ((
a^2 - b^2)*Sin[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.061, size = 118, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) -{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+{a}^{2} \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)^4*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

1/d*(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)-1/3*a*b*
cos(d*x+c)^6+a^2*(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 = 1.22801, size = 138, normalized size = 0.79 \begin{align*} -\frac{64 \, a b \cos \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^{2} -{\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 )} b^{2}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(64*a*b*cos(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^2 - (4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*b^2)/d

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Fricas [A]  time = 0.503055, size = 223, normalized size = 1.28 \begin{align*} -\frac{16 \, a b \cos \left (d x + c\right )^{6} - 3 \,{\left (5 \, a^{2} + b^{2}\right )} d x -{\left (8 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, a^{2} + b^{2}\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)^4*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/48*(16*a*b*cos(d*x + c)^6 - 3*(5*a^2 + b^2)*d*x - (8*(a^2 - b^2)*cos(d*x + c)^5 + 2*(5*a^2 + b^2)*cos(d*x +
c)^3 + 3*(5*a^2 + b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 4.89803, size = 384, normalized size = 2.21 \begin{align*} \begin{cases} \frac{5 a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{a b \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac{a b \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a b \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{b^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{b^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{2} \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*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.15153, size = 178, normalized size = 1.02 \begin{align*} \frac{1}{16} \,{\left (5 \, a^{2} + b^{2}\right )} x - \frac{a b \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac{5 \, a b \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac{{\left (a^{2} - b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (3 \, a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (15 \, a^{2} + b^{2}\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)^4*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(5*a^2 + b^2)*x - 1/96*a*b*cos(6*d*x + 6*c)/d - 1/16*a*b*cos(4*d*x + 4*c)/d - 5/32*a*b*cos(2*d*x + 2*c)/d
+ 1/192*(a^2 - b^2)*sin(6*d*x + 6*c)/d + 1/64*(3*a^2 - b^2)*sin(4*d*x + 4*c)/d + 1/64*(15*a^2 + b^2)*sin(2*d*
x + 2*c)/d