3.283 \(\int \frac {\cos (c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=11 \[ \frac {B \sin (c+d x)}{d} \]

[Out]

B*sin(d*x+c)/d

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {21, 2637} \[ \frac {B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sin[c + d*x])/d

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx &=B \int \cos (c+d x) \, dx\\ &=\frac {B \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 23, normalized size = 2.09 \[ B \left (\frac {\sin (c) \cos (d x)}{d}+\frac {\cos (c) \sin (d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

B*((Cos[d*x]*Sin[c])/d + (Cos[c]*Sin[d*x])/d)

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fricas [A]  time = 0.93, size = 11, normalized size = 1.00 \[ \frac {B \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

B*sin(d*x + c)/d

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giac [A]  time = 0.35, size = 11, normalized size = 1.00 \[ \frac {B \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

B*sin(d*x + c)/d

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maple [A]  time = 0.06, size = 12, normalized size = 1.09 \[ \frac {B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x)

[Out]

B*sin(d*x+c)/d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 0.47, size = 11, normalized size = 1.00 \[ \frac {B\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(B*a + B*b*cos(c + d*x)))/(a + b*cos(c + d*x)),x)

[Out]

(B*sin(c + d*x))/d

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sympy [A]  time = 0.60, size = 31, normalized size = 2.82 \[ \begin {cases} \frac {B \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\\frac {x \left (B a + B b \cos {\relax (c )}\right ) \cos {\relax (c )}}{a + b \cos {\relax (c )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x)

[Out]

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

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