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

Optimal. Leaf size=33 \[ \frac{b \sin ^2(c+d x)}{2 d}-\frac{a \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0538762, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4377, 12, 2564, 30, 2565} \[ \frac{b \sin ^2(c+d x)}{2 d}-\frac{a \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 12

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

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])

Rubi steps

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

Mathematica [A]  time = 0.11502, size = 38, normalized size = 1.15 \[ -\frac{3 a \cos (c+d x)+a \cos (3 (c+d x))+3 b \cos (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*a*Cos[c + d*x] + 3*b*Cos[2*(c + d*x)] + a*Cos[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.03, size = 29, normalized size = 0.9 \begin{align*} -{\frac{1}{d} \left ({\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}a}{3}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06575, size = 38, normalized size = 1.15 \begin{align*} -\frac{2 \, a \cos \left (d x + c\right )^{3} - 3 \, b \sin \left (d x + c\right )^{2}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.486196, size = 68, normalized size = 2.06 \begin{align*} -\frac{2 \, a \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )^{2}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21127, size = 134, normalized size = 4.06 \begin{align*} -\frac{a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{a \cos \left (d x + c\right )}{4 \, d} - \frac{b \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - b \tan \left (d x\right )^{2} - 4 \, b \tan \left (d x\right ) \tan \left (c\right ) - b \tan \left (c\right )^{2} + b}{4 \,{\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + d \tan \left (d x\right )^{2} + d \tan \left (c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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