3.43 \(\int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=137 \[ -\frac{a^2 \sin ^7(c+d x)}{7 d}+\frac{3 a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x)}{d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^7(c+d x)}{7 d}+\frac{b^2 \sin ^7(c+d x)}{7 d}-\frac{2 b^2 \sin ^5(c+d x)}{5 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.13825, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 2633, 2565, 30, 2564, 270} \[ -\frac{a^2 \sin ^7(c+d x)}{7 d}+\frac{3 a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x)}{d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^7(c+d x)}{7 d}+\frac{b^2 \sin ^7(c+d x)}{7 d}-\frac{2 b^2 \sin ^5(c+d x)}{5 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(-2*a*b*Cos[c + d*x]^7)/(7*d) + (a^2*Sin[c + d*x])/d - (a^2*Sin[c + d*x]^3)/d + (b^2*Sin[c + d*x]^3)/(3*d) + (
3*a^2*Sin[c + d*x]^5)/(5*d) - (2*b^2*Sin[c + d*x]^5)/(5*d) - (a^2*Sin[c + d*x]^7)/(7*d) + (b^2*Sin[c + d*x]^7)
/(7*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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 2564

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.377821, size = 154, normalized size = 1.12 \[ -\frac{-3675 a^2 \sin (c+d x)-735 a^2 \sin (3 (c+d x))-147 a^2 \sin (5 (c+d x))-15 a^2 \sin (7 (c+d x))+1050 a b \cos (c+d x)+630 a b \cos (3 (c+d x))+210 a b \cos (5 (c+d x))+30 a b \cos (7 (c+d x))-525 b^2 \sin (c+d x)+35 b^2 \sin (3 (c+d x))+63 b^2 \sin (5 (c+d x))+15 b^2 \sin (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

-(1050*a*b*Cos[c + d*x] + 630*a*b*Cos[3*(c + d*x)] + 210*a*b*Cos[5*(c + d*x)] + 30*a*b*Cos[7*(c + d*x)] - 3675
*a^2*Sin[c + d*x] - 525*b^2*Sin[c + d*x] - 735*a^2*Sin[3*(c + d*x)] + 35*b^2*Sin[3*(c + d*x)] - 147*a^2*Sin[5*
(c + d*x)] + 63*b^2*Sin[5*(c + d*x)] - 15*a^2*Sin[7*(c + d*x)] + 15*b^2*Sin[7*(c + d*x)])/(6720*d)

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Maple [A]  time = 0.064, size = 108, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) -{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

1/d*(b^2*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-2/7*a*b*cos(d*x+c)
^7+1/7*a^2*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.14375, size = 132, normalized size = 0.96 \begin{align*} -\frac{30 \, a b \cos \left (d x + c\right )^{7} + 3 \,{\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{2} -{\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(30*a*b*cos(d*x + c)^7 + 3*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))
*a^2 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*b^2)/d

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Fricas [A]  time = 0.499819, size = 221, normalized size = 1.61 \begin{align*} -\frac{30 \, a b \cos \left (d x + c\right )^{7} -{\left (15 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(30*a*b*cos(d*x + c)^7 - (15*(a^2 - b^2)*cos(d*x + c)^6 + 3*(6*a^2 + b^2)*cos(d*x + c)^4 + 4*(6*a^2 + b
^2)*cos(d*x + c)^2 + 48*a^2 + 8*b^2)*sin(d*x + c))/d

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Sympy [A]  time = 7.90145, size = 187, normalized size = 1.36 \begin{align*} \begin{cases} \frac{16 a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac{2 a b \cos ^{7}{\left (c + d x \right )}}{7 d} + \frac{8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{2} \cos ^{5}{\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)**5*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Piecewise((16*a**2*sin(c + d*x)**7/(35*d) + 8*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*a**2*sin(c + d*x)
**3*cos(c + d*x)**4/d + a**2*sin(c + d*x)*cos(c + d*x)**6/d - 2*a*b*cos(c + d*x)**7/(7*d) + 8*b**2*sin(c + d*x
)**7/(105*d) + 4*b**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + b**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d), Ne(
d, 0)), (x*(a*cos(c) + b*sin(c))**2*cos(c)**5, True))

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Giac [A]  time = 1.16666, size = 209, normalized size = 1.53 \begin{align*} -\frac{a b \cos \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac{a b \cos \left (5 \, d x + 5 \, c\right )}{32 \, d} - \frac{3 \, a b \cos \left (3 \, d x + 3 \, c\right )}{32 \, d} - \frac{5 \, a b \cos \left (d x + c\right )}{32 \, d} + \frac{{\left (a^{2} - b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{{\left (7 \, a^{2} - 3 \, b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (21 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \,{\left (7 \, a^{2} + b^{2}\right )} \sin \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)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/224*a*b*cos(7*d*x + 7*c)/d - 1/32*a*b*cos(5*d*x + 5*c)/d - 3/32*a*b*cos(3*d*x + 3*c)/d - 5/32*a*b*cos(d*x +
c)/d + 1/448*(a^2 - b^2)*sin(7*d*x + 7*c)/d + 1/320*(7*a^2 - 3*b^2)*sin(5*d*x + 5*c)/d + 1/192*(21*a^2 - b^2)
*sin(3*d*x + 3*c)/d + 5/64*(7*a^2 + b^2)*sin(d*x + c)/d