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

Optimal. Leaf size=80 \[ \frac{1}{2} x \left (a^2 A+4 a b B+2 A b^2\right )+\frac{a^2 A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a (a B+2 A b) \sin (c+d x)}{d}+\frac{b^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

((a^2*A + 2*A*b^2 + 4*a*b*B)*x)/2 + (b^2*B*ArcTanh[Sin[c + d*x]])/d + (a*(2*A*b + a*B)*Sin[c + d*x])/d + (a^2*
A*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.173926, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4024, 4047, 8, 4045, 3770} \[ \frac{1}{2} x \left (a^2 A+4 a b B+2 A b^2\right )+\frac{a^2 A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a (a B+2 A b) \sin (c+d x)}{d}+\frac{b^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*A + 2*A*b^2 + 4*a*b*B)*x)/2 + (b^2*B*ArcTanh[Sin[c + d*x]])/d + (a*(2*A*b + a*B)*Sin[c + d*x])/d + (a^2*
A*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 4024

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

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 8

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -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))^2 (A+B \sec (c+d x)) \, dx &=\frac{a^2 A \cos (c+d x) \sin (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) \left (-2 a (2 A b+a B)+\left (A \left (-a^2-2 b^2\right )-4 a b B\right ) \sec (c+d x)-2 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 A \cos (c+d x) \sin (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) \left (-2 a (2 A b+a B)-2 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac{1}{2} \left (-a^2 A-2 A b^2-4 a b B\right ) \int 1 \, dx\\ &=\frac{1}{2} \left (a^2 A+2 A b^2+4 a b B\right ) x+\frac{a (2 A b+a B) \sin (c+d x)}{d}+\frac{a^2 A \cos (c+d x) \sin (c+d x)}{2 d}+\left (b^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (a^2 A+2 A b^2+4 a b B\right ) x+\frac{b^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (2 A b+a B) \sin (c+d x)}{d}+\frac{a^2 A \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.21232, size = 120, normalized size = 1.5 \[ \frac{2 (c+d x) \left (a^2 A+4 a b B+2 A b^2\right )+a^2 A \sin (2 (c+d x))+4 a (a B+2 A b) \sin (c+d x)-4 b^2 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 b^2 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2*A + 2*A*b^2 + 4*a*b*B)*(c + d*x) - 4*b^2*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*b^2*B*Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*a*(2*A*b + a*B)*Sin[c + d*x] + a^2*A*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.054, size = 120, normalized size = 1.5 \begin{align*}{\frac{{a}^{2}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}Ax}{2}}+{\frac{{a}^{2}Ac}{2\,d}}+{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{Aab\sin \left ( dx+c \right ) }{d}}+2\,Babx+2\,{\frac{Babc}{d}}+A{b}^{2}x+{\frac{A{b}^{2}c}{d}}+{\frac{B{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{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))^2*(A+B*sec(d*x+c)),x)

[Out]

1/2*a^2*A*cos(d*x+c)*sin(d*x+c)/d+1/2*a^2*A*x+1/2/d*A*a^2*c+1/d*B*a^2*sin(d*x+c)+2/d*A*a*b*sin(d*x+c)+2*B*a*b*
x+2/d*B*a*b*c+A*b^2*x+1/d*A*b^2*c+1/d*B*b^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.959678, size = 134, normalized size = 1.68 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 8 \,{\left (d x + c\right )} B a b + 4 \,{\left (d x + c\right )} A b^{2} + 2 \, B b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \sin \left (d x + c\right ) + 8 \, A a b \sin \left (d x + c\right )}{4 \, 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))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 8*(d*x + c)*B*a*b + 4*(d*x + c)*A*b^2 + 2*B*b^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^2*sin(d*x + c) + 8*A*a*b*sin(d*x + c))/d

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Fricas [A]  time = 0.510376, size = 213, normalized size = 2.66 \begin{align*} \frac{B b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} d x +{\left (A a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2} + 4 \, A a b\right )} \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))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(B*b^2*log(sin(d*x + c) + 1) - B*b^2*log(-sin(d*x + c) + 1) + (A*a^2 + 4*B*a*b + 2*A*b^2)*d*x + (A*a^2*cos
(d*x + c) + 2*B*a^2 + 4*A*a*b)*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))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.19371, size = 240, normalized size = 3. \begin{align*} \frac{2 \, B b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*B*b^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*B*b^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (A*a^2 + 4*B*a*
b + 2*A*b^2)*(d*x + c) - 2*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 4*A*a*b*tan(1/2*d*
x + 1/2*c)^3 - A*a^2*tan(1/2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c) - 4*A*a*b*tan(1/2*d*x + 1/2*c))/(tan(
1/2*d*x + 1/2*c)^2 + 1)^2)/d