3.353 \(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=108 \[ -\frac {2 (11 B-C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {B x}{a^3}-\frac {(7 B-2 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

B*x/a^3-1/5*(B-C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(7*B-2*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-2/15*(11*B-
C)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]  time = 0.25, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {4072, 3922, 3919, 3794} \[ -\frac {2 (11 B-C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {B x}{a^3}-\frac {(7 B-2 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

(B*x)/a^3 - ((B - C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((7*B - 2*C)*Tan[c + d*x])/(15*a*d*(a + a*Se
c[c + d*x])^2) - (2*(11*B - C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3922

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

Rule 4072

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

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {B+C \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac {(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-5 a B+2 a (B-C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 B-2 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {15 a^2 B-a^2 (7 B-2 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {B x}{a^3}-\frac {(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 B-2 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(2 (11 B-C)) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {B x}{a^3}-\frac {(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 B-2 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {2 (11 B-C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 0.59, size = 241, normalized size = 2.23 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (270 B \sin \left (c+\frac {d x}{2}\right )-230 B \sin \left (c+\frac {3 d x}{2}\right )+90 B \sin \left (2 c+\frac {3 d x}{2}\right )-64 B \sin \left (2 c+\frac {5 d x}{2}\right )+150 B d x \cos \left (c+\frac {d x}{2}\right )+75 B d x \cos \left (c+\frac {3 d x}{2}\right )+75 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 B \sin \left (\frac {d x}{2}\right )+150 B d x \cos \left (\frac {d x}{2}\right )-60 C \sin \left (c+\frac {d x}{2}\right )+40 C \sin \left (c+\frac {3 d x}{2}\right )-30 C \sin \left (2 c+\frac {3 d x}{2}\right )+14 C \sin \left (2 c+\frac {5 d x}{2}\right )+80 C \sin \left (\frac {d x}{2}\right )\right )}{480 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*B*d*x*Cos[(d*x)/2] + 150*B*d*x*Cos[c + (d*x)/2] + 75*B*d*x*Cos[c + (3*d*x)/2
] + 75*B*d*x*Cos[2*c + (3*d*x)/2] + 15*B*d*x*Cos[2*c + (5*d*x)/2] + 15*B*d*x*Cos[3*c + (5*d*x)/2] - 370*B*Sin[
(d*x)/2] + 80*C*Sin[(d*x)/2] + 270*B*Sin[c + (d*x)/2] - 60*C*Sin[c + (d*x)/2] - 230*B*Sin[c + (3*d*x)/2] + 40*
C*Sin[c + (3*d*x)/2] + 90*B*Sin[2*c + (3*d*x)/2] - 30*C*Sin[2*c + (3*d*x)/2] - 64*B*Sin[2*c + (5*d*x)/2] + 14*
C*Sin[2*c + (5*d*x)/2]))/(480*a^3*d)

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fricas [A]  time = 0.49, size = 138, normalized size = 1.28 \[ \frac {15 \, B d x \cos \left (d x + c\right )^{3} + 45 \, B d x \cos \left (d x + c\right )^{2} + 45 \, B d x \cos \left (d x + c\right ) + 15 \, B d x - {\left ({\left (32 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, B - 2 \, C\right )} \cos \left (d x + c\right ) + 22 \, B - 2 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*B*d*x*cos(d*x + c)^3 + 45*B*d*x*cos(d*x + c)^2 + 45*B*d*x*cos(d*x + c) + 15*B*d*x - ((32*B - 7*C)*cos
(d*x + c)^2 + 3*(17*B - 2*C)*cos(d*x + c) + 22*B - 2*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x
+ c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 0.30, size = 121, normalized size = 1.12 \[ \frac {\frac {60 \, {\left (d x + c\right )} B}{a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*B/a^3 - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*B*a^12*tan(
1/2*d*x + 1/2*c)^3 + 10*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*B*a^12*tan(1/2*d*x + 1/2*c) - 15*C*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d

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maple [A]  time = 1.13, size = 137, normalized size = 1.27 \[ -\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{6 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*C+1/3/d/a^3*B*tan(1/2*d*x+1/2*c)^3-1/6/d/a^
3*tan(1/2*d*x+1/2*c)^3*C-7/4/d/a^3*B*tan(1/2*d*x+1/2*c)+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(tan(1/2*
d*x+1/2*c))*B

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maxima [A]  time = 0.60, size = 160, normalized size = 1.48 \[ -\frac {B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - C*(15*sin(d*x + c)/(cos(d*x + c) +
1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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mupad [B]  time = 3.06, size = 133, normalized size = 1.23 \[ \frac {B\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {7\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^3,x)

[Out]

(B*x)/a^3 + (cos(c/2 + (d*x)/2)^2*((B*sin(c/2 + (d*x)/2)^3)/3 - (C*sin(c/2 + (d*x)/2)^3)/6) - cos(c/2 + (d*x)/
2)^4*((7*B*sin(c/2 + (d*x)/2))/4 - (C*sin(c/2 + (d*x)/2))/4) - (B*sin(c/2 + (d*x)/2)^5)/20 + (C*sin(c/2 + (d*x
)/2)^5)/20)/(a^3*d*cos(c/2 + (d*x)/2)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(B*cos(c + d*x)*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integra
l(C*cos(c + d*x)*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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