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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.377956, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4072, 4020, 3787, 2635, 8, 2637} \[ -\frac{2 (8 B-5 C) \sin (c+d x)}{3 a^2 d}+\frac{(7 B-4 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(8 B-5 C) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{x (7 B-4 C)}{2 a^2}-\frac{(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [B]  time = 0.754213, size = 315, normalized size = 2.2 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (36 d x (7 B-4 C) \cos \left (c+\frac{d x}{2}\right )+147 B \sin \left (c+\frac{d x}{2}\right )-239 B \sin \left (c+\frac{3 d x}{2}\right )-63 B \sin \left (2 c+\frac{3 d x}{2}\right )-15 B \sin \left (2 c+\frac{5 d x}{2}\right )-15 B \sin \left (3 c+\frac{5 d x}{2}\right )+3 B \sin \left (3 c+\frac{7 d x}{2}\right )+3 B \sin \left (4 c+\frac{7 d x}{2}\right )+84 B d x \cos \left (c+\frac{3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+36 d x (7 B-4 C) \cos \left (\frac{d x}{2}\right )-381 B \sin \left (\frac{d x}{2}\right )-120 C \sin \left (c+\frac{d x}{2}\right )+164 C \sin \left (c+\frac{3 d x}{2}\right )+36 C \sin \left (2 c+\frac{3 d x}{2}\right )+12 C \sin \left (2 c+\frac{5 d x}{2}\right )+12 C \sin \left (3 c+\frac{5 d x}{2}\right )-48 C d x \cos \left (c+\frac{3 d x}{2}\right )-48 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+264 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.091, size = 252, normalized size = 1.8 \begin{align*}{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-3\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+7\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.43435, size = 382, normalized size = 2.67 \begin{align*} -\frac{B{\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 )} - C{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(B*(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) - C*((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

________________________________________________________________________________________

Fricas [A]  time = 0.496026, size = 342, normalized size = 2.39 \begin{align*} \frac{3 \,{\left (7 \, B - 4 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (7 \, B - 4 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (7 \, B - 4 \, C\right )} d x +{\left (3 \, B \cos \left (d x + c\right )^{3} - 6 \,{\left (B - C\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, B - 28 \, C\right )} \cos \left (d x + c\right ) - 32 \, B + 20 \, C\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.13709, size = 221, normalized size = 1.55 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}{\left (7 \, B - 4 \, C\right )}}{a^{2}} - \frac{6 \,{\left (5 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C \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{B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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