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

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Rubi [A]  time = 0.0676302, antiderivative size = 50, normalized size of antiderivative = 1., 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 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]

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 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]

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 = 0.021858, 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)

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Maple [A]  time = 0.051, size = 54, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{A\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+{\frac{C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

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))

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Maxima [A]  time = 0.921677, size = 58, normalized size = 1.16 \begin{align*} \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} \end{align*}

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

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Fricas [A]  time = 0.476398, size = 113, normalized size = 2.26 \begin{align*} \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} \end{align*}

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

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

[Out]

Timed out

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Giac [A]  time = 1.21075, size = 77, normalized size = 1.54 \begin{align*} \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} \end{align*}

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