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

Optimal. Leaf size=52 \[ \frac{(a B+A b) \sin (c+d x)}{d}+\frac{1}{2} x (a A+2 b B)+\frac{a A \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0959616, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3996, 3787, 2637, 8} \[ \frac{(a B+A b) \sin (c+d x)}{d}+\frac{1}{2} x (a A+2 b B)+\frac{a A \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0822269, size = 51, normalized size = 0.98 \[ \frac{4 (a B+A b) \sin (c+d x)+a A \sin (2 (c+d x))+2 a A c+2 a A d x+4 b B d x}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 57, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( Aa \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Ab\sin \left ( dx+c \right ) +Ba\sin \left ( dx+c \right ) +Bb \left ( dx+c \right ) \right ) } \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))*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.959907, size = 74, normalized size = 1.42 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 4 \,{\left (d x + c\right )} B b + 4 \, B a \sin \left (d x + c\right ) + 4 \, 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))*(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 + 4*(d*x + c)*B*b + 4*B*a*sin(d*x + c) + 4*A*b*sin(d*x + c))/d

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Fricas [A]  time = 0.468127, size = 104, normalized size = 2. \begin{align*} \frac{{\left (A a + 2 \, B b\right )} d x +{\left (A a \cos \left (d x + c\right ) + 2 \, B a + 2 \, 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))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}\, dx \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))*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*cos(c + d*x)**2, x)

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Giac [B]  time = 1.16937, size = 163, normalized size = 3.13 \begin{align*} \frac{{\left (A a + 2 \, B b\right )}{\left (d x + c\right )} - \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, 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))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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