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

Optimal. Leaf size=136 \[ \frac{2 (36 B-11 C) \sin (c+d x)}{15 a^3 d}-\frac{(3 B-C) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (3 B-C)}{a^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(B-C) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

-(((3*B - C)*x)/a^3) + (2*(36*B - 11*C)*Sin[c + d*x])/(15*a^3*d) - ((B - C)*Sin[c + d*x])/(5*d*(a + a*Sec[c +
d*x])^3) - ((9*B - 4*C)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((3*B - C)*Sin[c + d*x])/(d*(a^3 + a^3
*Sec[c + d*x]))

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Rubi [A]  time = 0.44351, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4072, 4020, 3787, 2637, 8} \[ \frac{2 (36 B-11 C) \sin (c+d x)}{15 a^3 d}-\frac{(3 B-C) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (3 B-C)}{a^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(B-C) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*B - C)*x)/a^3) + (2*(36*B - 11*C)*Sin[c + d*x])/(15*a^3*d) - ((B - C)*Sin[c + d*x])/(5*d*(a + a*Sec[c +
d*x])^3) - ((9*B - 4*C)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((3*B - C)*Sin[c + d*x])/(d*(a^3 + a^3
*Sec[c + d*x]))

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]

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]

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 \frac{\cos ^2(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{\cos (c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{(B-C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\cos (c+d x) (a (6 B-C)-3 a (B-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(B-C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos (c+d x) \left (a^2 (27 B-7 C)-2 a^2 (9 B-4 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(B-C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 B-C) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \cos (c+d x) \left (2 a^3 (36 B-11 C)-15 a^3 (3 B-C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(B-C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 B-C) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(2 (36 B-11 C)) \int \cos (c+d x) \, dx}{15 a^3}-\frac{(3 B-C) \int 1 \, dx}{a^3}\\ &=-\frac{(3 B-C) x}{a^3}+\frac{2 (36 B-11 C) \sin (c+d x)}{15 a^3 d}-\frac{(B-C) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 B-4 C) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 B-C) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.00983, size = 365, normalized size = 2.68 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-300 d x (3 B-C) \cos \left (c+\frac{d x}{2}\right )-1125 B \sin \left (c+\frac{d x}{2}\right )+1215 B \sin \left (c+\frac{3 d x}{2}\right )-225 B \sin \left (2 c+\frac{3 d x}{2}\right )+363 B \sin \left (2 c+\frac{5 d x}{2}\right )+75 B \sin \left (3 c+\frac{5 d x}{2}\right )+15 B \sin \left (3 c+\frac{7 d x}{2}\right )+15 B \sin \left (4 c+\frac{7 d x}{2}\right )-450 B d x \cos \left (c+\frac{3 d x}{2}\right )-450 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 B d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 B d x \cos \left (3 c+\frac{5 d x}{2}\right )-300 d x (3 B-C) \cos \left (\frac{d x}{2}\right )+1755 B \sin \left (\frac{d x}{2}\right )+540 C \sin \left (c+\frac{d x}{2}\right )-460 C \sin \left (c+\frac{3 d x}{2}\right )+180 C \sin \left (2 c+\frac{3 d x}{2}\right )-128 C \sin \left (2 c+\frac{5 d x}{2}\right )+150 C d x \cos \left (c+\frac{3 d x}{2}\right )+150 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-740 C \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-300*(3*B - C)*d*x*Cos[(d*x)/2] - 300*(3*B - C)*d*x*Cos[c + (d*x)/2] - 450*B*d*x*C
os[c + (3*d*x)/2] + 150*C*d*x*Cos[c + (3*d*x)/2] - 450*B*d*x*Cos[2*c + (3*d*x)/2] + 150*C*d*x*Cos[2*c + (3*d*x
)/2] - 90*B*d*x*Cos[2*c + (5*d*x)/2] + 30*C*d*x*Cos[2*c + (5*d*x)/2] - 90*B*d*x*Cos[3*c + (5*d*x)/2] + 30*C*d*
x*Cos[3*c + (5*d*x)/2] + 1755*B*Sin[(d*x)/2] - 740*C*Sin[(d*x)/2] - 1125*B*Sin[c + (d*x)/2] + 540*C*Sin[c + (d
*x)/2] + 1215*B*Sin[c + (3*d*x)/2] - 460*C*Sin[c + (3*d*x)/2] - 225*B*Sin[2*c + (3*d*x)/2] + 180*C*Sin[2*c + (
3*d*x)/2] + 363*B*Sin[2*c + (5*d*x)/2] - 128*C*Sin[2*c + (5*d*x)/2] + 75*B*Sin[3*c + (5*d*x)/2] + 15*B*Sin[3*c
+ (7*d*x)/2] + 15*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.109, size = 189, normalized size = 1.4 \begin{align*}{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(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*tan(1/2*d*x+1/2*c)^5*B-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*B+1/3/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*B*tan(1/2*d*x+1
/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^3*B*arctan(tan(1/2*d*x+1/2*c))+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.44386, size = 312, normalized size = 2.29 \begin{align*} \frac{3 \, B{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \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 )}} + \frac{\frac{85 \, \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{\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 )} - C{\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 )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*B*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - C*((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/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3))/d

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Fricas [A]  time = 0.493415, size = 431, normalized size = 3.17 \begin{align*} -\frac{15 \,{\left (3 \, B - C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (3 \, B - C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (3 \, B - C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (3 \, B - C\right )} d x -{\left (15 \, B \cos \left (d x + c\right )^{3} +{\left (117 \, B - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (57 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 72 \, B - 22 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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*(3*B - C)*d*x*cos(d*x + c)^3 + 45*(3*B - C)*d*x*cos(d*x + c)^2 + 45*(3*B - C)*d*x*cos(d*x + c) + 15*
(3*B - C)*d*x - (15*B*cos(d*x + c)^3 + (117*B - 32*C)*cos(d*x + c)^2 + 3*(57*B - 17*C)*cos(d*x + c) + 72*B - 2
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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1791, size = 212, normalized size = 1.56 \begin{align*} -\frac{\frac{60 \,{\left (d x + c\right )}{\left (3 \, B - C\right )}}{a^{3}} - \frac{120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} 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} - 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)*(3*B - C)/a^3 - 120*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*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 - 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/
2*d*x + 1/2*c)^3 + 255*B*a^12*tan(1/2*d*x + 1/2*c) - 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d