3.18 \(\int \cos (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=22 \[ \frac {(a \sin (c+d x)+a)^3}{3 a d} \]

[Out]

1/3*(a+a*sin(d*x+c))^3/a/d

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2667, 32} \[ \frac {(a \sin (c+d x)+a)^3}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + a*Sin[c + d*x])^3/(3*a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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))^2 \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^2 \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {(a+a \sin (c+d x))^3}{3 a d}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 47, normalized size = 2.14 \[ \frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^2(c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sin[c + d*x])/d + (a^2*Sin[c + d*x]^2)/d + (a^2*Sin[c + d*x]^3)/(3*d)

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fricas [B]  time = 0.69, size = 44, normalized size = 2.00 \[ -\frac {3 \, a^{2} \cos \left (d x + c\right )^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.74, size = 20, normalized size = 0.91 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{3 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*sin(d*x + c) + a)^3/(a*d)

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maple [A]  time = 0.07, size = 21, normalized size = 0.95 \[ \frac {\left (a +a \sin \left (d x +c \right )\right )^{3}}{3 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a+a*sin(d*x+c))^3/d/a

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maxima [A]  time = 0.34, size = 20, normalized size = 0.91 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{3 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*sin(d*x + c) + a)^3/(a*d)

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mupad [B]  time = 4.55, size = 32, normalized size = 1.45 \[ \frac {a^2\,\sin \left (c+d\,x\right )\,\left ({\sin \left (c+d\,x\right )}^2+3\,\sin \left (c+d\,x\right )+3\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*sin(c + d*x)*(3*sin(c + d*x) + sin(c + d*x)^2 + 3))/(3*d)

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sympy [A]  time = 0.82, size = 53, normalized size = 2.41 \[ \begin {cases} \frac {a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin ^{2}{\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)**2/d + a**2*sin(c + d*x)/d, Ne(d, 0)), (x*(a*sin(c)
+ a)**2*cos(c), True))

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