3.13 \(\int (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=43 \[ \frac{(2 A+3 C) \tan (c+d x)}{3 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

((2*A + 3*C)*Tan[c + d*x])/(3*d) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.0375058, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3012, 3767, 8} \[ \frac{(2 A+3 C) \tan (c+d x)}{3 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

((2*A + 3*C)*Tan[c + d*x])/(3*d) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0872789, size = 36, normalized size = 0.84 \[ \frac{A \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

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

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Maple [A]  time = 0.067, size = 35, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -A \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +C\tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

1/d*(-A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*tan(d*x+c))

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Maxima [A]  time = 1.09077, size = 36, normalized size = 0.84 \begin{align*} \frac{A \tan \left (d x + c\right )^{3} + 3 \,{\left (A + C\right )} \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.545, size = 95, normalized size = 2.21 \begin{align*} \frac{{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

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

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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((A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.15448, size = 46, normalized size = 1.07 \begin{align*} \frac{A \tan \left (d x + c\right )^{3} + 3 \, A \tan \left (d x + c\right ) + 3 \, C \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

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