3.377 \(\int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=44 \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2668, 641} \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sin[c + d*x]),x]

[Out]

-(b*Cos[c + d*x]^4)/(4*d) + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -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 ^3(c+d x) (a+b \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a+x) \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \operatorname {Subst}\left (\int \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 44, normalized size = 1.00 \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sin[c + d*x]),x]

[Out]

-1/4*(b*Cos[c + d*x]^4)/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

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fricas [A]  time = 0.47, size = 39, normalized size = 0.89 \[ -\frac {3 \, b \cos \left (d x + c\right )^{4} - 4 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(3*b*cos(d*x + c)^4 - 4*(a*cos(d*x + c)^2 + 2*a)*sin(d*x + c))/d

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giac [A]  time = 0.44, size = 48, normalized size = 1.09 \[ -\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, b \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 6*b*sin(d*x + c)^2 - 12*a*sin(d*x + c))/d

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maple [A]  time = 0.14, size = 36, normalized size = 0.82 \[ \frac {-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) b}{4}+\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 48, normalized size = 1.09 \[ -\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, b \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 6*b*sin(d*x + c)^2 - 12*a*sin(d*x + c))/d

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mupad [B]  time = 0.06, size = 46, normalized size = 1.05 \[ \frac {-\frac {b\,{\sin \left (c+d\,x\right )}^4}{4}-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + b*sin(c + d*x)),x)

[Out]

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

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sympy [A]  time = 1.23, size = 60, normalized size = 1.36 \[ \begin {cases} \frac {2 a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right ) \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*sin(d*x+c)),x)

[Out]

Piecewise((2*a*sin(c + d*x)**3/(3*d) + a*sin(c + d*x)*cos(c + d*x)**2/d - b*cos(c + d*x)**4/(4*d), Ne(d, 0)),
(x*(a + b*sin(c))*cos(c)**3, True))

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