\(\int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) [243]

Optimal result

Integrand size = 28, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {\left (a^2-b^2\right ) (b+a \cos (c+d x))^4}{4 a^3 d}-\frac {2 b (b+a \cos (c+d x))^5}{5 a^3 d}+\frac {(b+a \cos (c+d x))^6}{6 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {4482, 2747, 711} \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {(a \cos (c+d x)+b)^6}{6 a^3 d}-\frac {2 b (a \cos (c+d x)+b)^5}{5 a^3 d}-\frac {\left (a^2-b^2\right ) (a \cos (c+d x)+b)^4}{4 a^3 d} \]

[In]

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

[Out]

-1/4*((a^2 - b^2)*(b + a*Cos[c + d*x])^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 711

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]

Rule 2747

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 4482

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

Rubi steps \begin{align*} \text {integral}& = \int (b+a \cos (c+d x))^3 \sin ^3(c+d x) \, dx \\ & = -\frac {\text {Subst}\left (\int (b+x)^3 \left (a^2-x^2\right ) \, dx,x,a \cos (c+d x)\right )}{a^3 d} \\ & = -\frac {\text {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] (verified)

Time = 0.97 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {-360 b \left (a^2+2 b^2\right ) \cos (c+d x)-45 \left (a^3+8 a b^2\right ) \cos (2 (c+d x))-60 a^2 b \cos (3 (c+d x))+80 b^3 \cos (3 (c+d x))+90 a b^2 \cos (4 (c+d x))+36 a^2 b \cos (5 (c+d x))+5 a^3 \cos (6 (c+d x))}{960 d} \]

[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)

Maple [A] (verified)

Time = 30.63 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {a^{3} \cos \left (d x +c \right )^{6}}{6}+\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \cos \left (d x +c \right )^{4}}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \cos \left (d x +c \right )^{3}}{3}-\frac {3 a \,b^{2} \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b^{3}}{d}\)	\(100\)
default	\(\frac {\frac {a^{3} \cos \left (d x +c \right )^{6}}{6}+\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \cos \left (d x +c \right )^{4}}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \cos \left (d x +c \right )^{3}}{3}-\frac {3 a \,b^{2} \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b^{3}}{d}\)	\(100\)
risch	\(-\frac {3 a^{2} b \cos \left (d x +c \right )}{8 d}-\frac {3 b^{3} \cos \left (d x +c \right )}{4 d}+\frac {a^{3} \cos \left (6 d x +6 c \right )}{192 d}+\frac {3 b \,a^{2} \cos \left (5 d x +5 c \right )}{80 d}+\frac {3 a \,b^{2} \cos \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{16 d}+\frac {b^{3} \cos \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{64 d}-\frac {3 a \cos \left (2 d x +2 c \right ) b^{2}}{8 d}\)	\(154\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {10 \, a^{3} \cos \left (d x + c\right )^{6} + 36 \, a^{2} b \cos \left (d x + c\right )^{5} - 90 \, a b^{2} \cos \left (d x + c\right )^{2} - 15 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 60 \, b^{3} \cos \left (d x + c\right ) - 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3}}{60 \, d} \]

[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

Sympy [F]

\[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {45 \, a b^{2} \sin \left (d x + c\right )^{4} - 5 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{3} + 12 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b + 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{60 \, d} \]

[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

Giac [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 22.71 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.94 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {32\,a^3}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}+\frac {4\,{\left (a-b\right )}^3}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {32\,a^2\,\left (5\,a-3\,b\right )}{5\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5}-\frac {8\,{\left (a-b\right )}^2\,\left (7\,a-b\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}+\frac {12\,a\,\left (3\,a^2-4\,a\,b+b^2\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

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

[Out]

(32*a^3)/(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^6) + (4*(a - b)^3)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^2) - (32*a^2*(5*a -
3*b))/(5*d*(tan(c/2 + (d*x)/2)^2 + 1)^5) - (8*(a - b)^2*(7*a - b))/(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^3) + (12*a*
(3*a^2 - 4*a*b + b^2))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)