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

Optimal. Leaf size=77 \[ -\frac{\left (a^2-b^2\right ) (a \cos (c+d x)+b)^4}{4 a^3 d}+\frac{(a \cos (c+d x)+b)^6}{6 a^3 d}-\frac{2 b (a \cos (c+d x)+b)^5}{5 a^3 d} \]

[Out]

-((a^2 - b^2)*(b + a*Cos[c + d*x])^4)/(4*a^3*d) - (2*b*(b + a*Cos[c + d*x])^5)/(5*a^3*d) + (b + a*Cos[c + d*x]
)^6/(6*a^3*d)

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Rubi [A]  time = 0.188703, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {4397, 2668, 697} \[ -\frac{\left (a^2-b^2\right ) (a \cos (c+d x)+b)^4}{4 a^3 d}+\frac{(a \cos (c+d x)+b)^6}{6 a^3 d}-\frac{2 b (a \cos (c+d x)+b)^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - b^2)*(b + a*Cos[c + d*x])^4)/(4*a^3*d) - (2*b*(b + a*Cos[c + d*x])^5)/(5*a^3*d) + (b + a*Cos[c + d*x]
)^6/(6*a^3*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

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]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx &=\int (b+a \cos (c+d x))^3 \sin ^3(c+d x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int (b+x)^3 \left (a^2-x^2\right ) \, dx,x,a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\left (a^2-b^2\right ) (b+x)^3+2 b (b+x)^4-(b+x)^5\right ) \, dx,x,a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac{\left (a^2-b^2\right ) (b+a \cos (c+d x))^4}{4 a^3 d}-\frac{2 b (b+a \cos (c+d x))^5}{5 a^3 d}+\frac{(b+a \cos (c+d x))^6}{6 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.248718, size = 114, normalized size = 1.48 \[ \frac{-360 b \left (a^2+2 b^2\right ) \cos (c+d x)-45 \left (a^3+8 a b^2\right ) \cos (2 (c+d x))-60 a^2 b \cos (3 (c+d x))+36 a^2 b \cos (5 (c+d x))+5 a^3 \cos (6 (c+d x))+90 a b^2 \cos (4 (c+d x))+80 b^3 \cos (3 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-360*b*(a^2 + 2*b^2)*Cos[c + d*x] - 45*(a^3 + 8*a*b^2)*Cos[2*(c + d*x)] - 60*a^2*b*Cos[3*(c + d*x)] + 80*b^3*
Cos[3*(c + d*x)] + 90*a*b^2*Cos[4*(c + d*x)] + 36*a^2*b*Cos[5*(c + d*x)] + 5*a^3*Cos[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.078, size = 109, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) +3\,{a}^{2}b \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}-{\frac{{b}^{3} \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+3*a^2*b*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d
*x+c)^3)+3/4*a*b^2*sin(d*x+c)^4-1/3*b^3*(2+sin(d*x+c)^2)*cos(d*x+c))

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Maxima [A]  time = 1.16065, size = 128, normalized size = 1.66 \begin{align*} \frac{45 \, a b^{2} \sin \left (d x + c\right )^{4} - 5 \,{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{3} + 12 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b + 20 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{60 \, d} \end{align*}

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))^3,x, algorithm="maxima")

[Out]

1/60*(45*a*b^2*sin(d*x + c)^4 - 5*(2*sin(d*x + c)^6 - 3*sin(d*x + c)^4)*a^3 + 12*(3*cos(d*x + c)^5 - 5*cos(d*x
+ c)^3)*a^2*b + 20*(cos(d*x + c)^3 - 3*cos(d*x + c))*b^3)/d

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Fricas [A]  time = 0.508109, size = 240, normalized size = 3.12 \begin{align*} \frac{10 \, a^{3} \cos \left (d x + c\right )^{6} + 36 \, a^{2} b \cos \left (d x + c\right )^{5} - 90 \, a b^{2} \cos \left (d x + c\right )^{2} - 15 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 60 \, b^{3} \cos \left (d x + c\right ) - 20 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3}}{60 \, d} \end{align*}

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))^3,x, algorithm="fricas")

[Out]

1/60*(10*a^3*cos(d*x + c)^6 + 36*a^2*b*cos(d*x + c)^5 - 90*a*b^2*cos(d*x + c)^2 - 15*(a^3 - 3*a*b^2)*cos(d*x +
c)^4 - 60*b^3*cos(d*x + c) - 20*(3*a^2*b - b^3)*cos(d*x + c)^3)/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^3,x, algorithm="giac")

[Out]

Timed out