∫ cos3(c + dx)(a sin(c + dx) + b tan(c + dx)) dx

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

Optimal. Leaf size=33 \[ -\frac {a \cos ^4(c+d x)}{4 d}-\frac {b \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A] time = 0.06, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4377, 12, 2565, 30} \[ -\frac {a \cos ^4(c+d x)}{4 d}-\frac {b \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[

Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]

&& !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +

b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[

Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] && !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&

(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x)) \, dx &=a \int \cos ^3(c+d x) \sin (c+d x) \, dx+\int b \cos ^2(c+d x) \sin (c+d x) \, dx\\ &=b \int \cos ^2(c+d x) \sin (c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a \cos ^4(c+d x)}{4 d}-\frac {b \operatorname {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {b \cos ^3(c+d x)}{3 d}-\frac {a \cos ^4(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A] time = 0.01, size = 33, normalized size = 1.00 \[ -\frac {a \cos ^4(c+d x)}{4 d}-\frac {b \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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fricas [A] time = 0.41, size = 28, normalized size = 0.85 \[ -\frac {3 \, a \cos \left (d x + c\right )^{4} + 4 \, b \cos \left (d x + c\right )^{3}}{12 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/