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

Optimal. Leaf size=143 \[ -\frac {2 (8 A-5 B) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {x (7 A-4 B)}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

1/2*(7*A-4*B)*x/a^2-2/3*(8*A-5*B)*sin(d*x+c)/a^2/d+1/2*(7*A-4*B)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(8*A-5*B)*cos
(d*x+c)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]  time = 0.30, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4020, 3787, 2635, 8, 2637} \[ -\frac {2 (8 A-5 B) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {x (7 A-4 B)}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A - 4*B)*x)/(2*a^2) - (2*(8*A - 5*B)*Sin[c + d*x])/(3*a^2*d) + ((7*A - 4*B)*Cos[c + d*x]*Sin[c + d*x])/(2*
a^2*d) - ((8*A - 5*B)*Cos[c + d*x]*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]*Sin[c +
d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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 4020

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) (a (5 A-2 B)-3 a (A-B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(8 A-5 B) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \cos ^2(c+d x) \left (3 a^2 (7 A-4 B)-2 a^2 (8 A-5 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(8 A-5 B) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 (8 A-5 B)) \int \cos (c+d x) \, dx}{3 a^2}+\frac {(7 A-4 B) \int \cos ^2(c+d x) \, dx}{a^2}\\ &=-\frac {2 (8 A-5 B) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(7 A-4 B) \int 1 \, dx}{2 a^2}\\ &=\frac {(7 A-4 B) x}{2 a^2}-\frac {2 (8 A-5 B) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.78, size = 315, normalized size = 2.20 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (36 d x (7 A-4 B) \cos \left (c+\frac {d x}{2}\right )+36 d x (7 A-4 B) \cos \left (\frac {d x}{2}\right )+147 A \sin \left (c+\frac {d x}{2}\right )-239 A \sin \left (c+\frac {3 d x}{2}\right )-63 A \sin \left (2 c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {5 d x}{2}\right )-15 A \sin \left (3 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {7 d x}{2}\right )+3 A \sin \left (4 c+\frac {7 d x}{2}\right )+84 A d x \cos \left (c+\frac {3 d x}{2}\right )+84 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 A \sin \left (\frac {d x}{2}\right )-120 B \sin \left (c+\frac {d x}{2}\right )+164 B \sin \left (c+\frac {3 d x}{2}\right )+36 B \sin \left (2 c+\frac {3 d x}{2}\right )+12 B \sin \left (2 c+\frac {5 d x}{2}\right )+12 B \sin \left (3 c+\frac {5 d x}{2}\right )-48 B d x \cos \left (c+\frac {3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 B \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(7*A - 4*B)*d*x*Cos[(d*x)/2] + 36*(7*A - 4*B)*d*x*Cos[c + (d*x)/2] + 84*A*d*x*C
os[c + (3*d*x)/2] - 48*B*d*x*Cos[c + (3*d*x)/2] + 84*A*d*x*Cos[2*c + (3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2
] - 381*A*Sin[(d*x)/2] + 264*B*Sin[(d*x)/2] + 147*A*Sin[c + (d*x)/2] - 120*B*Sin[c + (d*x)/2] - 239*A*Sin[c +
(3*d*x)/2] + 164*B*Sin[c + (3*d*x)/2] - 63*A*Sin[2*c + (3*d*x)/2] + 36*B*Sin[2*c + (3*d*x)/2] - 15*A*Sin[2*c +
(5*d*x)/2] + 12*B*Sin[2*c + (5*d*x)/2] - 15*A*Sin[3*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] + 3*A*Sin[3*c
+ (7*d*x)/2] + 3*A*Sin[4*c + (7*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.43, size = 138, normalized size = 0.97 \[ \frac {3 \, {\left (7 \, A - 4 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (7 \, A - 4 \, B\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (7 \, A - 4 \, B\right )} d x + {\left (3 \, A \cos \left (d x + c\right )^{3} - 6 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 28 \, B\right )} \cos \left (d x + c\right ) - 32 \, A + 20 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(7*A - 4*B)*d*x*cos(d*x + c)^2 + 6*(7*A - 4*B)*d*x*cos(d*x + c) + 3*(7*A - 4*B)*d*x + (3*A*cos(d*x + c)
^3 - 6*(A - B)*cos(d*x + c)^2 - (43*A - 28*B)*cos(d*x + c) - 32*A + 20*B)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2
+ 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.26, size = 164, normalized size = 1.15 \[ \frac {\frac {3 \, {\left (d x + c\right )} {\left (7 \, A - 4 \, B\right )}}{a^{2}} - \frac {6 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(d*x + c)*(7*A - 4*B)/a^2 - 6*(5*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + 3*A*tan(1/2*d*
x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 -
B*a^4*tan(1/2*d*x + 1/2*c)^3 - 21*A*a^4*tan(1/2*d*x + 1/2*c) + 15*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 1.11, size = 252, normalized size = 1.76 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3-7/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*t
an(1/2*d*x+1/2*c)-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2
*B*tan(1/2*d*x+1/2*c)^3-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^
2)^2*B*tan(1/2*d*x+1/2*c)+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B

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maxima [B]  time = 0.44, size = 283, normalized size = 1.98 \[ -\frac {A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(A*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*((15*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 2.07, size = 154, normalized size = 1.08 \[ \frac {x\,\left (7\,A-4\,B\right )}{2\,a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^2}+\frac {4\,A-2\,B}{2\,a^2}\right )}{d}-\frac {\left (5\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(7*A - 4*B))/(2*a^2) - (tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^2) + (4*A - 2*B)/(2*a^2)))/d - (tan(c/2 + (d*x
)/2)^3*(5*A - 2*B) + tan(c/2 + (d*x)/2)*(3*A - 2*B))/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 + a^2*tan(c/2 + (d*x)/2)^4
+ a^2)) + (tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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