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

Optimal. Leaf size=156 \[ \frac {(4 A+3 C) \sin ^3(c+d x)}{3 a d}-\frac {(4 A+3 C) \sin (c+d x)}{a d}+\frac {(5 A+4 C) \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 (5 A+4 C) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 x (5 A+4 C)}{8 a} \]

[Out]

3/8*(5*A+4*C)*x/a-(4*A+3*C)*sin(d*x+c)/a/d+3/8*(5*A+4*C)*cos(d*x+c)*sin(d*x+c)/a/d+1/4*(5*A+4*C)*cos(d*x+c)^3*
sin(d*x+c)/a/d-(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*sec(d*x+c))+1/3*(4*A+3*C)*sin(d*x+c)^3/a/d

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Rubi [A]  time = 0.19, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4085, 3787, 2635, 8, 2633} \[ \frac {(4 A+3 C) \sin ^3(c+d x)}{3 a d}-\frac {(4 A+3 C) \sin (c+d x)}{a d}+\frac {(5 A+4 C) \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 (5 A+4 C) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 x (5 A+4 C)}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5*A + 4*C)*x)/(8*a) - ((4*A + 3*C)*Sin[c + d*x])/(a*d) + (3*(5*A + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*a*d)
+ ((5*A + 4*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) - ((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Sec[c +
d*x])) + ((4*A + 3*C)*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 4085

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

Rubi steps

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

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Mathematica [A]  time = 0.69, size = 283, normalized size = 1.81 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (72 d x (5 A+4 C) \cos \left (c+\frac {d x}{2}\right )-168 A \sin \left (c+\frac {d x}{2}\right )-120 A \sin \left (c+\frac {3 d x}{2}\right )-120 A \sin \left (2 c+\frac {3 d x}{2}\right )+40 A \sin \left (2 c+\frac {5 d x}{2}\right )+40 A \sin \left (3 c+\frac {5 d x}{2}\right )-5 A \sin \left (3 c+\frac {7 d x}{2}\right )-5 A \sin \left (4 c+\frac {7 d x}{2}\right )+3 A \sin \left (4 c+\frac {9 d x}{2}\right )+3 A \sin \left (5 c+\frac {9 d x}{2}\right )+72 d x (5 A+4 C) \cos \left (\frac {d x}{2}\right )-552 A \sin \left (\frac {d x}{2}\right )-96 C \sin \left (c+\frac {d x}{2}\right )-72 C \sin \left (c+\frac {3 d x}{2}\right )-72 C \sin \left (2 c+\frac {3 d x}{2}\right )+24 C \sin \left (2 c+\frac {5 d x}{2}\right )+24 C \sin \left (3 c+\frac {5 d x}{2}\right )-480 C \sin \left (\frac {d x}{2}\right )\right )}{192 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(72*(5*A + 4*C)*d*x*Cos[(d*x)/2] + 72*(5*A + 4*C)*d*x*Cos[c + (d*x)/2] - 552*A*Sin[
(d*x)/2] - 480*C*Sin[(d*x)/2] - 168*A*Sin[c + (d*x)/2] - 96*C*Sin[c + (d*x)/2] - 120*A*Sin[c + (3*d*x)/2] - 72
*C*Sin[c + (3*d*x)/2] - 120*A*Sin[2*c + (3*d*x)/2] - 72*C*Sin[2*c + (3*d*x)/2] + 40*A*Sin[2*c + (5*d*x)/2] + 2
4*C*Sin[2*c + (5*d*x)/2] + 40*A*Sin[3*c + (5*d*x)/2] + 24*C*Sin[3*c + (5*d*x)/2] - 5*A*Sin[3*c + (7*d*x)/2] -
5*A*Sin[4*c + (7*d*x)/2] + 3*A*Sin[4*c + (9*d*x)/2] + 3*A*Sin[5*c + (9*d*x)/2]))/(192*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.44, size = 113, normalized size = 0.72 \[ \frac {9 \, {\left (5 \, A + 4 \, C\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (5 \, A + 4 \, C\right )} d x + {\left (6 \, A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} + {\left (13 \, A + 12 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (19 \, A + 12 \, C\right )} \cos \left (d x + c\right ) - 64 \, A - 48 \, C\right )} \sin \left (d x + c\right )}{24 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(5*A + 4*C)*d*x*cos(d*x + c) + 9*(5*A + 4*C)*d*x + (6*A*cos(d*x + c)^4 - 2*A*cos(d*x + c)^3 + (13*A +
12*C)*cos(d*x + c)^2 - (19*A + 12*C)*cos(d*x + c) - 64*A - 48*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 0.73, size = 180, normalized size = 1.15 \[ \frac {\frac {9 \, {\left (d x + c\right )} {\left (5 \, A + 4 \, C\right )}}{a} - \frac {24 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (75 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 115 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 84 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 109 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(d*x + c)*(5*A + 4*C)/a - 24*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a - 2*(75*A*tan(1/2*d*x
+ 1/2*c)^7 + 36*C*tan(1/2*d*x + 1/2*c)^7 + 115*A*tan(1/2*d*x + 1/2*c)^5 + 84*C*tan(1/2*d*x + 1/2*c)^5 + 109*A
*tan(1/2*d*x + 1/2*c)^3 + 60*C*tan(1/2*d*x + 1/2*c)^3 + 21*A*tan(1/2*d*x + 1/2*c) + 12*C*tan(1/2*d*x + 1/2*c))
/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d

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maple [B]  time = 1.23, size = 352, normalized size = 2.26 \[ -\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {115 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{12 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {109 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{12 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {15 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-25/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^
7*A-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*C-115/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+
1/2*c)^5*A-7/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*C-109/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1
/2*d*x+1/2*c)^3*A-5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*C-7/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*A
*tan(1/2*d*x+1/2*c)-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*C*tan(1/2*d*x+1/2*c)+15/4/a/d*A*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 = 0.44, size = 351, normalized size = 2.25 \[ -\frac {A {\left (\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {12 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 12 \, C {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(A*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6
*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8) - 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 12*sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 12*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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mupad [B]  time = 2.75, size = 153, normalized size = 0.98 \[ \frac {15\,A\,x}{8\,a}+\frac {3\,C\,x}{2\,a}-\frac {7\,A\,\sin \left (c+d\,x\right )}{4\,a\,d}-\frac {C\,\sin \left (c+d\,x\right )}{a\,d}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{2\,a\,d}-\frac {A\,\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {A\,\sin \left (4\,c+4\,d\,x\right )}{32\,a\,d}-\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}+\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,a\,d}-\frac {C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*A*x)/(8*a) + (3*C*x)/(2*a) - (7*A*sin(c + d*x))/(4*a*d) - (C*sin(c + d*x))/(a*d) + (A*sin(2*c + 2*d*x))/(2
*a*d) - (A*sin(3*c + 3*d*x))/(12*a*d) + (A*sin(4*c + 4*d*x))/(32*a*d) - (A*tan(c/2 + (d*x)/2))/(a*d) + (C*sin(
2*c + 2*d*x))/(4*a*d) - (C*tan(c/2 + (d*x)/2))/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(A*cos(c + d*x)**4/(sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2/(sec(c + d*x)
+ 1), x))/a

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