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

Optimal. Leaf size=22 \[ \frac{(a+b \sin (c+d x))^2}{2 b d} \]

[Out]

(a + b*Sin[c + d*x])^2/(2*b*d)

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Rubi [A]  time = 0.0164447, antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2668} \[ \frac{a \sin (c+d x)}{d}+\frac{b \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0110288, size = 39, normalized size = 1.77 \[ \frac{a \sin (c) \cos (d x)}{d}+\frac{a \cos (c) \sin (d x)}{d}-\frac{b \cos ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 25, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}b}{2}}+a\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.942197, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}{2 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*sin(d*x + c) + a)^2/(b*d)

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Fricas [A]  time = 2.04339, size = 62, normalized size = 2.82 \begin{align*} -\frac{b \cos \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.210159, size = 34, normalized size = 1.55 \begin{align*} \begin{cases} \frac{a \sin{\left (c + d x \right )}}{d} + \frac{b \sin ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right ) \cos{\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)*(a+b*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.06324, size = 34, normalized size = 1.55 \begin{align*} \frac{b \sin \left (d x + c\right )^{2} + 2 \, a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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