3.89 \(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 122, 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, 2633, 2635, 8} \[ -\frac {(4 A-3 B) \sin ^3(c+d x)}{3 a d}+\frac {(4 A-3 B) \sin (c+d x)}{a d}-\frac {3 (A-B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac {3 x (A-B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx &=-\frac {(A-B) \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 A-3 B)-3 a (A-B) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(4 A-3 B) \int \cos ^3(c+d x) \, dx}{a}-\frac {(3 (A-B)) \int \cos ^2(c+d x) \, dx}{a}\\ &=-\frac {3 (A-B) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 (A-B)) \int 1 \, dx}{2 a}-\frac {(4 A-3 B) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d}\\ &=-\frac {3 (A-B) x}{2 a}+\frac {(4 A-3 B) \sin (c+d x)}{a d}-\frac {3 (A-B) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(4 A-3 B) \sin ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B]  time = 0.75, size = 249, normalized size = 2.04 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-36 d x (A-B) \cos \left (c+\frac {d x}{2}\right )-36 d x (A-B) \cos \left (\frac {d x}{2}\right )+21 A \sin \left (c+\frac {d x}{2}\right )+18 A \sin \left (c+\frac {3 d x}{2}\right )+18 A \sin \left (2 c+\frac {3 d x}{2}\right )-2 A \sin \left (2 c+\frac {5 d x}{2}\right )-2 A \sin \left (3 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {7 d x}{2}\right )+69 A \sin \left (\frac {d x}{2}\right )-12 B \sin \left (c+\frac {d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )-60 B \sin \left (\frac {d x}{2}\right )\right )}{24 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 97, normalized size = 0.80 \[ -\frac {9 \, {\left (A - B\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (A - B\right )} d x - {\left (2 \, A \cos \left (d x + c\right )^{3} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 16 \, A - 12 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.82, size = 151, normalized size = 1.24 \[ -\frac {\frac {9 \, {\left (d x + c\right )} {\left (A - B\right )}}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.42, size = 281, normalized size = 2.30 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 310, normalized size = 2.54 \[ \frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*((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*B*((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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mupad [B]  time = 3.02, size = 138, normalized size = 1.13 \[ \frac {\left (5\,A-3\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {16\,A}{3}-4\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {3\,x\,\left (A-B\right )}{2\,a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*(5*A - 3*B) + tan(c/2 + (d*x)/2)^3*((16*A)/3 - 4*B) + tan(c/2 + (d*x)/2)*(3*A - B))/(d*(
a + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6)) - (3*x*(A - B))/(2*a) + (ta
n(c/2 + (d*x)/2)*(A - B))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*cos(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)/(sec(c + d*x) + 1
), x))/a

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