\(\int \cos ^3(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [10]

Optimal result

Integrand size = 21, antiderivative size = 30 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(A+C) \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d} \]

[Out]

(A+C)*sin(d*x+c)/d-1/3*A*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4129, 3092} \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(A+C) \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]

[Out]

((A + C)*Sin[c + d*x])/d - (A*Sin[c + d*x]^3)/(3*d)

Rule 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {C \cos (d x) \sin (c)}{d}+\frac {C \cos (c) \sin (d x)}{d}+\frac {A \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d} \]

[In]

Integrate[Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(C*Cos[d*x]*Sin[c])/d + (C*Cos[c]*Sin[d*x])/d + (A*Sin[c + d*x])/d - (A*Sin[c + d*x]^3)/(3*d)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {A \sin \left (3 d x +3 c \right )+9 \left (A +\frac {4 C}{3}\right ) \sin \left (d x +c \right )}{12 d}\)	\(31\)
derivativedivides	\(\frac {\frac {A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \sin \left (d x +c \right )}{d}\)	\(33\)
default	\(\frac {\frac {A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \sin \left (d x +c \right )}{d}\)	\(33\)
risch	\(\frac {3 A \sin \left (d x +c \right )}{4 d}+\frac {C \sin \left (d x +c \right )}{d}+\frac {A \sin \left (3 d x +3 c \right )}{12 d}\)	\(40\)
norman	\(\frac {\frac {2 \left (A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 \left (A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(111\)

[In]

int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/12*(A*sin(3*d*x+3*c)+9*(A+4/3*C)*sin(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (A \cos \left (d x + c\right )^{2} + 2 \, A + 3 \, C\right )} \sin \left (d x + c\right )}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/3*(A*cos(d*x + c)^2 + 2*A + 3*C)*sin(d*x + c)/d

Sympy [F]

\[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**3*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {A \sin \left (d x + c\right )^{3} - 3 \, {\left (A + C\right )} \sin \left (d x + c\right )}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/3*(A*sin(d*x + c)^3 - 3*(A + C)*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {A \sin \left (d x + c\right )^{3} - 3 \, A \sin \left (d x + c\right ) - 3 \, C \sin \left (d x + c\right )}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/3*(A*sin(d*x + c)^3 - 3*A*sin(d*x + c) - 3*C*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {A\,{\sin \left (c+d\,x\right )}^3}{3}-\sin \left (c+d\,x\right )\,\left (A+C\right )}{d} \]

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2),x)

[Out]

-((A*sin(c + d*x)^3)/3 - sin(c + d*x)*(A + C))/d