3.12 \(\int \cos ^5(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=50 \[ -\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {(A+C) \sin (c+d x)}{d}+\frac {A \sin ^5(c+d x)}{5 d} \]

[Out]

(A+C)*sin(d*x+c)/d-1/3*(2*A+C)*sin(d*x+c)^3/d+1/5*A*sin(d*x+c)^5/d

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4044, 3013, 373} \[ -\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {(A+C) \sin (c+d x)}{d}+\frac {A \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

((A + C)*Sin[c + d*x])/d - ((2*A + C)*Sin[c + d*x]^3)/(3*d) + (A*Sin[c + d*x]^5)/(5*d)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3013

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 4044

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*} \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^3(c+d x) \left (C+A \cos ^2(c+d x)\right ) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \left (A+C-A x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (A \left (1+\frac {C}{A}\right )-(2 A+C) x^2+A x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {(A+C) \sin (c+d x)}{d}-\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {A \sin ^5(c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 71, normalized size = 1.42 \[ \frac {A \sin ^5(c+d x)}{5 d}-\frac {2 A \sin ^3(c+d x)}{3 d}+\frac {A \sin (c+d x)}{d}-\frac {C \sin ^3(c+d x)}{3 d}+\frac {C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

(A*Sin[c + d*x])/d + (C*Sin[c + d*x])/d - (2*A*Sin[c + d*x]^3)/(3*d) - (C*Sin[c + d*x]^3)/(3*d) + (A*Sin[c + d
*x]^5)/(5*d)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 45, normalized size = 0.90 \[ \frac {{\left (3 \, A \cos \left (d x + c\right )^{4} + {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, A + 10 \, C\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(3*A*cos(d*x + c)^4 + (4*A + 5*C)*cos(d*x + c)^2 + 8*A + 10*C)*sin(d*x + c)/d

________________________________________________________________________________________

giac [A]  time = 0.20, size = 57, normalized size = 1.14 \[ \frac {3 \, A \sin \left (d x + c\right )^{5} - 10 \, A \sin \left (d x + c\right )^{3} - 5 \, C \sin \left (d x + c\right )^{3} + 15 \, A \sin \left (d x + c\right ) + 15 \, C \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/15*(3*A*sin(d*x + c)^5 - 10*A*sin(d*x + c)^3 - 5*C*sin(d*x + c)^3 + 15*A*sin(d*x + c) + 15*C*sin(d*x + c))/d

________________________________________________________________________________________

maple [A]  time = 1.50, size = 54, normalized size = 1.08 \[ \frac {\frac {A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/5*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/3*C*(2+cos(d*x+c)^2)*sin(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 43, normalized size = 0.86 \[ \frac {3 \, A \sin \left (d x + c\right )^{5} - 5 \, {\left (2 \, A + C\right )} \sin \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/15*(3*A*sin(d*x + c)^5 - 5*(2*A + C)*sin(d*x + c)^3 + 15*(A + C)*sin(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 2.37, size = 43, normalized size = 0.86 \[ \frac {\frac {A\,{\sin \left (c+d\,x\right )}^5}{5}+\left (-\frac {2\,A}{3}-\frac {C}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (A+C\right )\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2),x)

[Out]

((A*sin(c + d*x)^5)/5 + sin(c + d*x)*(A + C) - sin(c + d*x)^3*((2*A)/3 + C/3))/d

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________