3.45 \(\int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=103 \[ \frac {a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {2 a b \cos ^5(c+d x)}{5 d}-\frac {b^2 \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 103, normalized size of antiderivative = 1.00, 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, 14} \[ \frac {a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {2 a b \cos ^5(c+d x)}{5 d}-\frac {b^2 \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 116, normalized size = 1.13 \[ \frac {150 a^2 \sin (c+d x)+25 a^2 \sin (3 (c+d x))+3 a^2 \sin (5 (c+d x))-60 a b \cos (c+d x)-30 a b \cos (3 (c+d x))-6 a b \cos (5 (c+d x))+30 b^2 \sin (c+d x)-5 b^2 \sin (3 (c+d x))-3 b^2 \sin (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(-60*a*b*Cos[c + d*x] - 30*a*b*Cos[3*(c + d*x)] - 6*a*b*Cos[5*(c + d*x)] + 150*a^2*Sin[c + d*x] + 30*b^2*Sin[c
+ d*x] + 25*a^2*Sin[3*(c + d*x)] - 5*b^2*Sin[3*(c + d*x)] + 3*a^2*Sin[5*(c + d*x)] - 3*b^2*Sin[5*(c + d*x)])/
(240*d)

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fricas [A]  time = 0.62, size = 74, normalized size = 0.72 \[ -\frac {6 \, a b \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/15*(6*a*b*cos(d*x + c)^5 - (3*(a^2 - b^2)*cos(d*x + c)^4 + (4*a^2 + b^2)*cos(d*x + c)^2 + 8*a^2 + 2*b^2)*si
n(d*x + c))/d

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giac [A]  time = 3.51, size = 114, normalized size = 1.11 \[ -\frac {a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac {a b \cos \left (d x + c\right )}{4 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/40*a*b*cos(5*d*x + 5*c)/d - 1/8*a*b*cos(3*d*x + 3*c)/d - 1/4*a*b*cos(d*x + c)/d + 1/80*(a^2 - b^2)*sin(5*d*
x + 5*c)/d + 1/48*(5*a^2 - b^2)*sin(3*d*x + 3*c)/d + 1/8*(5*a^2 + b^2)*sin(d*x + c)/d

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maple [A]  time = 1.76, size = 88, normalized size = 0.85 \[ \frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 77, normalized size = 0.75 \[ -\frac {6 \, a b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} + {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.61, size = 115, normalized size = 1.12 \[ \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^2-3\,a\,b\,{\cos \left (c+d\,x\right )}^5-\frac {3\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+\sin \left (c+d\,x\right )\,b^2\right )}{15\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a*cos(c + d*x) + b*sin(c + d*x))^2,x)

[Out]

(2*(4*a^2*sin(c + d*x) + b^2*sin(c + d*x) + 2*a^2*cos(c + d*x)^2*sin(c + d*x) + (3*a^2*cos(c + d*x)^4*sin(c +
d*x))/2 + (b^2*cos(c + d*x)^2*sin(c + d*x))/2 - (3*b^2*cos(c + d*x)^4*sin(c + d*x))/2 - 3*a*b*cos(c + d*x)^5))
/(15*d)

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sympy [A]  time = 1.75, size = 138, normalized size = 1.34 \[ \begin {cases} \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {2 a b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{2} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

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

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