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]

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

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Rubi [A]  time = 0.10, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 8

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

Rule 2637

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

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 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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.46, size = 42, normalized size = 0.81 \[ \frac {{\left (A a + 2 \, B b\right )} d x + {\left (A a \cos \left (d x + c\right ) + 2 \, B a + 2 \, A b\right )} \sin \left (d x + c\right )}{2 \, d} \]

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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giac [B]  time = 0.22, size = 121, normalized size = 2.33 \[ \frac {{\left (A a + 2 \, B b\right )} {\left (d x + c\right )} - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.78, size = 57, normalized size = 1.10 \[ \frac {a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A b \sin \left (d x +c \right )+a B \sin \left (d x +c \right )+B \left (d x +c \right ) b}{d} \]

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)+a*B*sin(d*x+c)+B*(d*x+c)*b)

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maxima [A]  time = 0.78, size = 55, normalized size = 1.06 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 4 \, {\left (d x + c\right )} B b + 4 \, B a \sin \left (d x + c\right ) + 4 \, A b \sin \left (d x + c\right )}{4 \, d} \]

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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mupad [B]  time = 2.03, size = 50, normalized size = 0.96 \[ \frac {A\,a\,x}{2}+B\,b\,x+\frac {A\,b\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + B/cos(c + d*x))*(a + b/cos(c + d*x)),x)

[Out]

(A*a*x)/2 + B*b*x + (A*b*sin(c + d*x))/d + (B*a*sin(c + d*x))/d + (A*a*sin(2*c + 2*d*x))/(4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}\, dx \]

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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