3.770 \(\int \cos ^2(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=35 \[ x (a C+b B)+\frac{a B \sin (c+d x)}{d}+\frac{b C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(b*B + a*C)*x + (b*C*ArcTanh[Sin[c + d*x]])/d + (a*B*Sin[c + d*x])/d

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Rubi [A]  time = 0.101763, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {4072, 3996, 3770} \[ x (a C+b B)+\frac{a B \sin (c+d x)}{d}+\frac{b C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(b*B + a*C)*x + (b*C*ArcTanh[Sin[c + d*x]])/d + (a*B*Sin[c + d*x])/d

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \sin (c+d x)}{d}-\int (-b B-a C-b C \sec (c+d x)) \, dx\\ &=(b B+a C) x+\frac{a B \sin (c+d x)}{d}+(b C) \int \sec (c+d x) \, dx\\ &=(b B+a C) x+\frac{b C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.0274246, size = 46, normalized size = 1.31 \[ \frac{a B \sin (c) \cos (d x)}{d}+\frac{a B \cos (c) \sin (d x)}{d}+a C x+b B x+\frac{b C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

b*B*x + a*C*x + (b*C*ArcTanh[Sin[c + d*x]])/d + (a*B*Cos[d*x]*Sin[c])/d + (a*B*Cos[c]*Sin[d*x])/d

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Maple [A]  time = 0.055, size = 56, normalized size = 1.6 \begin{align*} Bbx+aCx+{\frac{B\sin \left ( dx+c \right ) a}{d}}+{\frac{Bbc}{d}}+{\frac{Cb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Cac}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

B*b*x+a*C*x+a*B*sin(d*x+c)/d+1/d*B*b*c+1/d*C*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*a*c

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Maxima [A]  time = 0.969817, size = 78, normalized size = 2.23 \begin{align*} \frac{2 \,{\left (d x + c\right )} C a + 2 \,{\left (d x + c\right )} B b + C b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2*(2*(d*x + c)*C*a + 2*(d*x + c)*B*b + C*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*B*a*sin(d*x +
c))/d

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Fricas [A]  time = 0.510087, size = 142, normalized size = 4.06 \begin{align*} \frac{2 \,{\left (C a + B b\right )} d x + C b \log \left (\sin \left (d x + c\right ) + 1\right ) - C b \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*(2*(C*a + B*b)*d*x + C*b*log(sin(d*x + c) + 1) - C*b*log(-sin(d*x + c) + 1) + 2*B*a*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)**2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.21137, size = 107, normalized size = 3.06 \begin{align*} \frac{C b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - C b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (C a + B b\right )}{\left (d x + c\right )} + \frac{2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

(C*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - C*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (C*a + B*b)*(d*x + c) + 2*B
*a*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1))/d