3.34 \(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=43 \[ \frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}+\frac {b \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3090, 2635, 8, 2564, 30} \[ \frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}+\frac {b \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]  time = 0.05, size = 46, normalized size = 1.07 \[ \frac {a (c+d x)}{2 d}+\frac {a \sin (2 (c+d x))}{4 d}-\frac {b \cos ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 35, normalized size = 0.81 \[ \frac {a d x - b \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.99, size = 35, normalized size = 0.81 \[ \frac {1}{2} \, a x - \frac {b \cos \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

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)),x, algorithm="giac")

[Out]

1/2*a*x - 1/4*b*cos(2*d*x + 2*c)/d + 1/4*a*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.71, size = 41, normalized size = 0.95 \[ \frac {a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (\cos ^{2}\left (d x +c \right )\right ) b}{2}}{d} \]

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)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 37, normalized size = 0.86 \[ -\frac {2 \, b \cos \left (d x + c\right )^{2} - {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a}{4 \, d} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.43, size = 35, normalized size = 0.81 \[ \frac {a\,x}{2}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

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)),x)

[Out]

(a*x)/2 - (b*cos(2*c + 2*d*x))/(4*d) + (a*sin(2*c + 2*d*x))/(4*d)

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sympy [A]  time = 0.21, size = 73, normalized size = 1.70 \[ \begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {b \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

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)),x)

[Out]

Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 + a*sin(c + d*x)*cos(c + d*x)/(2*d) - b*cos(c + d*x)*
*2/(2*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))*cos(c), True))

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