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

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

[Out]

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

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Rubi [A]  time = 0.03, 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 = {2668, 32} \[ \frac {(a+b \sin (c+d x))^3}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 0.01, size = 46, normalized size = 2.09 \[ \frac {a^2 \sin (c+d x)}{d}+\frac {a b \sin ^2(c+d x)}{d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 48, normalized size = 2.18 \[ -\frac {3 \, a b \cos \left (d x + c\right )^{2} + {\left (b^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.66, size = 20, normalized size = 0.91 \[ \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*sin(d*x + c) + a)^3/(b*d)

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maple [A]  time = 0.08, size = 21, normalized size = 0.95 \[ \frac {\left (a +b \sin \left (d x +c \right )\right )^{3}}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 20, normalized size = 0.91 \[ \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*sin(d*x + c) + a)^3/(b*d)

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mupad [B]  time = 0.06, size = 39, normalized size = 1.77 \[ \frac {a^2\,\sin \left (c+d\,x\right )+a\,b\,{\sin \left (c+d\,x\right )}^2+\frac {b^2\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.77, size = 53, normalized size = 2.41 \[ \begin {cases} \frac {a^{2} \sin {\left (c + d x \right )}}{d} + \frac {a b \sin ^{2}{\left (c + d x \right )}}{d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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