\(\int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) [245]

Optimal result

Integrand size = 26, antiderivative size = 112 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {a^2 b \cos ^3(c+d x)}{d}+\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4482, 2916, 12, 908} \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a^2 b \cos ^3(c+d x)}{d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[In]

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

[Out]

-((b*(3*a^2 - b^2)*Cos[c + d*x])/d) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^2)/(2*d) + (a^2*b*Cos[c + d*x]^3)/d + (a^3
*Cos[c + d*x]^4)/(4*d) - (3*a*b^2*Log[Cos[c + d*x]])/d + (b^3*Sec[c + d*x])/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 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(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 (c+d x) \tan ^2(c+d x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {a^2 (b+x)^3 \left (a^2-x^2\right )}{x^2} \, dx,x,a \cos (c+d x)\right )}{a^3 d} \\ & = -\frac {\text {Subst}\left (\int \frac {(b+x)^3 \left (a^2-x^2\right )}{x^2} \, dx,x,a \cos (c+d x)\right )}{a d} \\ & = -\frac {\text {Subst}\left (\int \left (3 a^2 b \left (1-\frac {b^2}{3 a^2}\right )+\frac {a^2 b^3}{x^2}+\frac {3 a^2 b^2}{x}+\left (a^2-3 b^2\right ) x-3 b x^2-x^3\right ) \, dx,x,a \cos (c+d x)\right )}{a d} \\ & = -\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {a^2 b \cos ^3(c+d x)}{d}+\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {8 b \left (-9 a^2+4 b^2\right ) \cos (c+d x)-4 \left (a^3-6 a b^2\right ) \cos (2 (c+d x))+8 a^2 b \cos (3 (c+d x))+a^3 \cos (4 (c+d x))-96 a b^2 \log (\cos (c+d x))+32 b^3 \sec (c+d x)}{32 d} \]

[In]

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

[Out]

(8*b*(-9*a^2 + 4*b^2)*Cos[c + d*x] - 4*(a^3 - 6*a*b^2)*Cos[2*(c + d*x)] + 8*a^2*b*Cos[3*(c + d*x)] + a^3*Cos[4
*(c + d*x)] - 96*a*b^2*Log[Cos[c + d*x]] + 32*b^3*Sec[c + d*x])/(32*d)

Maple [A] (verified)

Time = 9.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {a^{3} \sin \left (d x +c \right )^{4}}{4}-a^{2} b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(106\)
default	\(\frac {\frac {a^{3} \sin \left (d x +c \right )^{4}}{4}-a^{2} b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(106\)
risch	\(3 i x a \,b^{2}-\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}-\frac {9 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {9 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {6 i a \,b^{2} c}{d}+\frac {2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{3} \cos \left (4 d x +4 c \right )}{32 d}+\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{4 d}\)	\(247\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {8 \, a^{3} \cos \left (d x + c\right )^{5} + 32 \, a^{2} b \cos \left (d x + c\right )^{4} - 96 \, a b^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 16 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 32 \, b^{3} - 32 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{3} - 24 \, a b^{2}\right )} \cos \left (d x + c\right )}{32 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/32*(8*a^3*cos(d*x + c)^5 + 32*a^2*b*cos(d*x + c)^4 - 96*a*b^2*cos(d*x + c)*log(-cos(d*x + c)) - 16*(a^3 - 3*
a*b^2)*cos(d*x + c)^3 + 32*b^3 - 32*(3*a^2*b - b^3)*cos(d*x + c)^2 + (5*a^3 - 24*a*b^2)*cos(d*x + c))/(d*cos(d
*x + c))

Sympy [F]

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

[In]

integrate(cos(d*x+c)*(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), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{4} + 4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b - 6 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a b^{2} + 4 \, b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{4 \, d} \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Modgcd: no suitable evaluation pointindex.cc index_m operator + Error: Bad Argument ValueDone

Mupad [B] (verification not implemented)

Time = 25.90 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.01 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^3+4\,a^2\,b-6\,a\,b^2+12\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^2\,b+6\,a\,b^2-12\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-4\,a^3+12\,a^2\,b+6\,a\,b^2+4\,b^3\right )-4\,a^2\,b+4\,b^3+6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {6\,a\,b^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^4*(4*a^2*b - 6*a*b^2 + 4*a^3 + 12*b^3) - tan(c/2 + (d*x)/2)^2*(6*a*b^2 + 12*a^2*b - 12*b^3
) + tan(c/2 + (d*x)/2)^6*(6*a*b^2 + 12*a^2*b - 4*a^3 + 4*b^3) - 4*a^2*b + 4*b^3 + 6*a*b^2*tan(c/2 + (d*x)/2)^8
)/(d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(
c/2 + (d*x)/2)^10 + 1)) + (6*a*b^2*atanh(tan(c/2 + (d*x)/2)^2))/d