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

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

[Out]

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

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Rubi [A]  time = 0.02, 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 = {2667} \[ \frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.77 \[ -\frac {a \cos ^2(c+d x)}{2 d}+\frac {a \sin (c) \cos (d x)}{d}+\frac {a \cos (c) \sin (d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*Cos[c + d*x]^2)/d + (a*Cos[d*x]*Sin[c])/d + (a*Cos[c]*Sin[d*x])/d

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fricas [A]  time = 0.69, size = 25, normalized size = 1.14 \[ -\frac {a \cos \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.50, size = 25, normalized size = 1.14 \[ \frac {a \sin \left (d x + c\right )^{2} + 2 \, a \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 1.14 \[ \frac {\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 20, normalized size = 0.91 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*sin(d*x + c) + a)^2/(a*d)

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mupad [B]  time = 0.04, size = 20, normalized size = 0.91 \[ \frac {a\,\sin \left (c+d\,x\right )\,\left (\sin \left (c+d\,x\right )+2\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*sin(c + d*x)*(sin(c + d*x) + 2))/(2*d)

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sympy [A]  time = 0.22, size = 34, normalized size = 1.55 \[ \begin {cases} \frac {a \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {a \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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