3.339 \(\int \frac{\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=122 \[ -\frac{(4 B-3 C) \sin ^3(c+d x)}{3 a d}+\frac{(4 B-3 C) \sin (c+d x)}{a d}-\frac{3 (B-C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{3 x (B-C)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.241275, antiderivative size = 122, 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, 2633, 2635, 8} \[ -\frac{(4 B-3 C) \sin ^3(c+d x)}{3 a d}+\frac{(4 B-3 C) \sin (c+d x)}{a d}-\frac{3 (B-C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{3 x (B-C)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

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

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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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]

Rubi steps

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

Mathematica [B]  time = 0.603744, size = 249, normalized size = 2.04 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-36 d x (B-C) \cos \left (c+\frac{d x}{2}\right )+21 B \sin \left (c+\frac{d x}{2}\right )+18 B \sin \left (c+\frac{3 d x}{2}\right )+18 B \sin \left (2 c+\frac{3 d x}{2}\right )-2 B \sin \left (2 c+\frac{5 d x}{2}\right )-2 B \sin \left (3 c+\frac{5 d x}{2}\right )+B \sin \left (3 c+\frac{7 d x}{2}\right )+B \sin \left (4 c+\frac{7 d x}{2}\right )-36 d x (B-C) \cos \left (\frac{d x}{2}\right )+69 B \sin \left (\frac{d x}{2}\right )-12 C \sin \left (c+\frac{d x}{2}\right )-9 C \sin \left (c+\frac{3 d x}{2}\right )-9 C \sin \left (2 c+\frac{3 d x}{2}\right )+3 C \sin \left (2 c+\frac{5 d x}{2}\right )+3 C \sin \left (3 c+\frac{5 d x}{2}\right )-60 C \sin \left (\frac{d x}{2}\right )\right )}{24 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(B - C)*d*x*Cos[(d*x)/2] - 36*(B - C)*d*x*Cos[c + (d*x)/2] + 69*B*Sin[(d*x)/2]
- 60*C*Sin[(d*x)/2] + 21*B*Sin[c + (d*x)/2] - 12*C*Sin[c + (d*x)/2] + 18*B*Sin[c + (3*d*x)/2] - 9*C*Sin[c + (
3*d*x)/2] + 18*B*Sin[2*c + (3*d*x)/2] - 9*C*Sin[2*c + (3*d*x)/2] - 2*B*Sin[2*c + (5*d*x)/2] + 3*C*Sin[2*c + (5
*d*x)/2] - 2*B*Sin[3*c + (5*d*x)/2] + 3*C*Sin[3*c + (5*d*x)/2] + B*Sin[3*c + (7*d*x)/2] + B*Sin[4*c + (7*d*x)/
2]))/(24*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 0.096, size = 281, normalized size = 2.3 \begin{align*}{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-3\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}B}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-4\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{16\,B}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+3\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-3\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)

[Out]

1/a/d*B*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)^3*tan(1/2*d*x+1/2*c)^5*C+
5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B-4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*
C+16/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*B-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/
2*c)+3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)-3/a/d*B*arctan(tan(1/2*d*x+1/2*c))+3/a/d*arctan(tan
(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 1.43261, size = 419, normalized size = 3.43 \begin{align*} \frac{B{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{9 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{3 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, C{\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 )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/3*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x +
c) + 1))) - 3*C*((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))))/d

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Fricas [A]  time = 0.493189, size = 243, normalized size = 1.99 \begin{align*} -\frac{9 \,{\left (B - C\right )} d x \cos \left (d x + c\right ) + 9 \,{\left (B - C\right )} d x -{\left (2 \, B \cos \left (d x + c\right )^{3} -{\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B - 12 \, C\right )} \sin \left (d x + c\right )}{6 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(9*(B - C)*d*x*cos(d*x + c) + 9*(B - C)*d*x - (2*B*cos(d*x + c)^3 - (B - 3*C)*cos(d*x + c)^2 + (7*B - 3*C
)*cos(d*x + c) + 16*B - 12*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*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)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.15318, size = 204, normalized size = 1.67 \begin{align*} -\frac{\frac{9 \,{\left (d x + c\right )}{\left (B - C\right )}}{a} - \frac{6 \,{\left (B \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 (15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 16 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, 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 )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(9*(d*x + c)*(B - C)/a - 6*(B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a - 2*(15*B*tan(1/2*d*x + 1/
2*c)^5 - 9*C*tan(1/2*d*x + 1/2*c)^5 + 16*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 9*B*tan(1/2*
d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d