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

Optimal. Leaf size=30 \[ \frac{(A+C) \sin (c+d x)}{d}-\frac{A \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0473099, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4044, 3013} \[ \frac{(A+C) \sin (c+d x)}{d}-\frac{A \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) \left (C+A \cos ^2(c+d x)\right ) \, dx\\ &=-\frac{\operatorname{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]  time = 0.0180238, size = 50, normalized size = 1.67 \[ -\frac{A \sin ^3(c+d x)}{3 d}+\frac{A \sin (c+d x)}{d}+\frac{C \sin (c) \cos (d x)}{d}+\frac{C \cos (c) \sin (d x)}{d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.05, size = 33, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+C\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.921315, size = 36, normalized size = 1.2 \begin{align*} -\frac{A \sin \left (d x + c\right )^{3} - 3 \,{\left (A + C\right )} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.471226, size = 69, normalized size = 2.3 \begin{align*} \frac{{\left (A \cos \left (d x + c\right )^{2} + 2 \, A + 3 \, C\right )} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

Timed out

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Giac [A]  time = 1.12817, size = 46, normalized size = 1.53 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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