3.127 \(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=96 \[ -\frac {(2 A+C) \sin (c+d x)}{a d}+\frac {(3 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}+\frac {x (3 A+2 C)}{2 a} \]

[Out]

1/2*(3*A+2*C)*x/a-(2*A+C)*sin(d*x+c)/a/d+1/2*(3*A+2*C)*cos(d*x+c)*sin(d*x+c)/a/d-(A+C)*cos(d*x+c)*sin(d*x+c)/d
/(a+a*sec(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4085, 3787, 2635, 8, 2637} \[ -\frac {(2 A+C) \sin (c+d x)}{a d}+\frac {(3 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}+\frac {x (3 A+2 C)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 2*C)*x)/(2*a) - ((2*A + C)*Sin[c + d*x])/(a*d) + ((3*A + 2*C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((A
+ C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 8

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 4085

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

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 159, normalized size = 1.66 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (4 d x (3 A+2 C) \cos \left (c+\frac {d x}{2}\right )-4 A \sin \left (c+\frac {d x}{2}\right )-3 A \sin \left (c+\frac {3 d x}{2}\right )-3 A \sin \left (2 c+\frac {3 d x}{2}\right )+A \sin \left (2 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {5 d x}{2}\right )+4 d x (3 A+2 C) \cos \left (\frac {d x}{2}\right )-20 A \sin \left (\frac {d x}{2}\right )-16 C \sin \left (\frac {d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(4*(3*A + 2*C)*d*x*Cos[(d*x)/2] + 4*(3*A + 2*C)*d*x*Cos[c + (d*x)/2] - 20*A*Sin[(d*
x)/2] - 16*C*Sin[(d*x)/2] - 4*A*Sin[c + (d*x)/2] - 3*A*Sin[c + (3*d*x)/2] - 3*A*Sin[2*c + (3*d*x)/2] + A*Sin[2
*c + (5*d*x)/2] + A*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.45, size = 77, normalized size = 0.80 \[ \frac {{\left (3 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + {\left (3 \, A + 2 \, C\right )} d x + {\left (A \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) - 4 \, A - 2 \, C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*A + 2*C)*d*x*cos(d*x + c) + (3*A + 2*C)*d*x + (A*cos(d*x + c)^2 - A*cos(d*x + c) - 4*A - 2*C)*sin(d*x
+ c))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 0.29, size = 96, normalized size = 1.00 \[ \frac {\frac {{\left (d x + c\right )} {\left (3 \, A + 2 \, C\right )}}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.25, size = 144, normalized size = 1.50 \[ -\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {3 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 184, normalized size = 1.92 \[ -\frac {A {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(A*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x
+ c)/(a*(cos(d*x + c) + 1))) - C*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c)
+ 1))))/d

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mupad [B]  time = 2.61, size = 83, normalized size = 0.86 \[ \frac {3\,A\,x}{2\,a}+\frac {C\,x}{a}-\frac {A\,\sin \left (c+d\,x\right )}{a\,d}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,a\,d}-\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}-\frac {C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*x)/(2*a) + (C*x)/a - (A*sin(c + d*x))/(a*d) + (A*sin(2*c + 2*d*x))/(4*a*d) - (A*tan(c/2 + (d*x)/2))/(a*d)
- (C*tan(c/2 + (d*x)/2))/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(A*cos(c + d*x)**2/(sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(sec(c + d*x)
+ 1), x))/a

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