3.146 \(\int \frac {\sec ^5(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=232 \[ -\frac {32 (5 A+54 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {16 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(2 A+21 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

1/2*(2*A+21*C)*arctanh(sin(d*x+c))/a^4/d-32/105*(5*A+54*C)*tan(d*x+c)/a^4/d+1/2*(2*A+21*C)*sec(d*x+c)*tan(d*x+
c)/a^4/d-1/105*(10*A+129*C)*sec(d*x+c)^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-16/105*(5*A+54*C)*sec(d*x+c)^2*tan(
d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A+C)*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-2/5*C*sec(d*x+c)^4*tan(d*x+
c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.65, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4085, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac {32 (5 A+54 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {16 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(2 A+21 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

((2*A + 21*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (32*(5*A + 54*C)*Tan[c + d*x])/(105*a^4*d) + ((2*A + 21*C)*Se
c[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((10*A + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x]
)^2) - (16*(5*A + 54*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^5*
Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*C*Sec[c + d*x]^4*Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 4019

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

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 {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec ^5(c+d x) (-a (2 A-5 C)-a (2 A+9 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^4(c+d x) \left (56 a^2 C-a^2 (10 A+73 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x) \left (3 a^3 (10 A+129 C)-a^3 (50 A+477 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (5 A+54 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {\int \sec ^2(c+d x) \left (32 a^4 (5 A+54 C)-105 a^4 (2 A+21 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (5 A+54 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(2 A+21 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac {(32 (5 A+54 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac {(2 A+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (5 A+54 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(2 A+21 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac {(32 (5 A+54 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(2 A+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {32 (5 A+54 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (5 A+54 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 4.40, size = 746, normalized size = 3.22 \[ -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-17220 A \sin \left (c-\frac {d x}{2}\right )+17220 A \sin \left (c+\frac {d x}{2}\right )-14140 A \sin \left (2 c+\frac {d x}{2}\right )-9800 A \sin \left (c+\frac {3 d x}{2}\right )+15160 A \sin \left (2 c+\frac {3 d x}{2}\right )-9800 A \sin \left (3 c+\frac {3 d x}{2}\right )+10920 A \sin \left (c+\frac {5 d x}{2}\right )-4760 A \sin \left (2 c+\frac {5 d x}{2}\right )+10920 A \sin \left (3 c+\frac {5 d x}{2}\right )-4760 A \sin \left (4 c+\frac {5 d x}{2}\right )+5890 A \sin \left (2 c+\frac {7 d x}{2}\right )-1470 A \sin \left (3 c+\frac {7 d x}{2}\right )+5890 A \sin \left (4 c+\frac {7 d x}{2}\right )-1470 A \sin \left (5 c+\frac {7 d x}{2}\right )+2030 A \sin \left (3 c+\frac {9 d x}{2}\right )-210 A \sin \left (4 c+\frac {9 d x}{2}\right )+2030 A \sin \left (5 c+\frac {9 d x}{2}\right )-210 A \sin \left (6 c+\frac {9 d x}{2}\right )+320 A \sin \left (4 c+\frac {11 d x}{2}\right )+320 A \sin \left (6 c+\frac {11 d x}{2}\right )-14 (1010 A+5229 C) \sin \left (\frac {d x}{2}\right )+4 (3790 A+41667 C) \sin \left (\frac {3 d x}{2}\right )-183162 C \sin \left (c-\frac {d x}{2}\right )+100842 C \sin \left (c+\frac {d x}{2}\right )-155526 C \sin \left (2 c+\frac {d x}{2}\right )-37380 C \sin \left (c+\frac {3 d x}{2}\right )+101148 C \sin \left (2 c+\frac {3 d x}{2}\right )-102900 C \sin \left (3 c+\frac {3 d x}{2}\right )+119364 C \sin \left (c+\frac {5 d x}{2}\right )-8820 C \sin \left (2 c+\frac {5 d x}{2}\right )+78204 C \sin \left (3 c+\frac {5 d x}{2}\right )-49980 C \sin \left (4 c+\frac {5 d x}{2}\right )+64053 C \sin \left (2 c+\frac {7 d x}{2}\right )+3885 C \sin \left (3 c+\frac {7 d x}{2}\right )+44733 C \sin \left (4 c+\frac {7 d x}{2}\right )-15435 C \sin \left (5 c+\frac {7 d x}{2}\right )+21987 C \sin \left (3 c+\frac {9 d x}{2}\right )+3675 C \sin \left (4 c+\frac {9 d x}{2}\right )+16107 C \sin \left (5 c+\frac {9 d x}{2}\right )-2205 C \sin \left (6 c+\frac {9 d x}{2}\right )+3456 C \sin \left (4 c+\frac {11 d x}{2}\right )+840 C \sin \left (5 c+\frac {11 d x}{2}\right )+2616 C \sin \left (6 c+\frac {11 d x}{2}\right )\right )+53760 (2 A+21 C) \cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3360 a^4 d (\sec (c+d x)+1)^4 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-1/3360*(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(53760*(2*A + 21*C)*Cos[(c + d*x)/2]^7*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^
2*(-14*(1010*A + 5229*C)*Sin[(d*x)/2] + 4*(3790*A + 41667*C)*Sin[(3*d*x)/2] - 17220*A*Sin[c - (d*x)/2] - 18316
2*C*Sin[c - (d*x)/2] + 17220*A*Sin[c + (d*x)/2] + 100842*C*Sin[c + (d*x)/2] - 14140*A*Sin[2*c + (d*x)/2] - 155
526*C*Sin[2*c + (d*x)/2] - 9800*A*Sin[c + (3*d*x)/2] - 37380*C*Sin[c + (3*d*x)/2] + 15160*A*Sin[2*c + (3*d*x)/
2] + 101148*C*Sin[2*c + (3*d*x)/2] - 9800*A*Sin[3*c + (3*d*x)/2] - 102900*C*Sin[3*c + (3*d*x)/2] + 10920*A*Sin
[c + (5*d*x)/2] + 119364*C*Sin[c + (5*d*x)/2] - 4760*A*Sin[2*c + (5*d*x)/2] - 8820*C*Sin[2*c + (5*d*x)/2] + 10
920*A*Sin[3*c + (5*d*x)/2] + 78204*C*Sin[3*c + (5*d*x)/2] - 4760*A*Sin[4*c + (5*d*x)/2] - 49980*C*Sin[4*c + (5
*d*x)/2] + 5890*A*Sin[2*c + (7*d*x)/2] + 64053*C*Sin[2*c + (7*d*x)/2] - 1470*A*Sin[3*c + (7*d*x)/2] + 3885*C*S
in[3*c + (7*d*x)/2] + 5890*A*Sin[4*c + (7*d*x)/2] + 44733*C*Sin[4*c + (7*d*x)/2] - 1470*A*Sin[5*c + (7*d*x)/2]
- 15435*C*Sin[5*c + (7*d*x)/2] + 2030*A*Sin[3*c + (9*d*x)/2] + 21987*C*Sin[3*c + (9*d*x)/2] - 210*A*Sin[4*c +
(9*d*x)/2] + 3675*C*Sin[4*c + (9*d*x)/2] + 2030*A*Sin[5*c + (9*d*x)/2] + 16107*C*Sin[5*c + (9*d*x)/2] - 210*A
*Sin[6*c + (9*d*x)/2] - 2205*C*Sin[6*c + (9*d*x)/2] + 320*A*Sin[4*c + (11*d*x)/2] + 3456*C*Sin[4*c + (11*d*x)/
2] + 840*C*Sin[5*c + (11*d*x)/2] + 320*A*Sin[6*c + (11*d*x)/2] + 2616*C*Sin[6*c + (11*d*x)/2])))/(a^4*d*(A + 2
*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^4)

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fricas [A]  time = 0.45, size = 354, normalized size = 1.53 \[ \frac {105 \, {\left ({\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (64 \, {\left (5 \, A + 54 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (1070 \, A + 11619 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (310 \, A + 3411 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (130 \, A + 1509 \, C\right )} \cos \left (d x + c\right )^{2} + 420 \, C \cos \left (d x + c\right ) - 105 \, C\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(105*((2*A + 21*C)*cos(d*x + c)^6 + 4*(2*A + 21*C)*cos(d*x + c)^5 + 6*(2*A + 21*C)*cos(d*x + c)^4 + 4*(2
*A + 21*C)*cos(d*x + c)^3 + (2*A + 21*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 105*((2*A + 21*C)*cos(d*x + c
)^6 + 4*(2*A + 21*C)*cos(d*x + c)^5 + 6*(2*A + 21*C)*cos(d*x + c)^4 + 4*(2*A + 21*C)*cos(d*x + c)^3 + (2*A + 2
1*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(64*(5*A + 54*C)*cos(d*x + c)^5 + (1070*A + 11619*C)*cos(d*x +
c)^4 + 4*(310*A + 3411*C)*cos(d*x + c)^3 + 4*(130*A + 1509*C)*cos(d*x + c)^2 + 420*C*cos(d*x + c) - 105*C)*si
n(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 +
a^4*d*cos(d*x + c)^2)

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giac [A]  time = 1.51, size = 241, normalized size = 1.04 \[ \frac {\frac {420 \, {\left (2 \, A + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (2 \, A + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(420*(2*A + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(2*A + 21*C)*log(abs(tan(1/2*d*x + 1/2*c)
- 1))/a^4 + 840*(9*C*tan(1/2*d*x + 1/2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4)
- (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 1
89*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 15
75*A*a^24*tan(1/2*d*x + 1/2*c) + 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.67, size = 329, normalized size = 1.42 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {9 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {13 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{d \,a^{4}}-\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 d \,a^{4}}+\frac {C}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {9 C}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{d \,a^{4}}+\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 d \,a^{4}}-\frac {C}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {9 C}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7-1/8/d/a^4*A*tan(1/2*d*x+1/2*c)^5-9/40/d/a
^4*C*tan(1/2*d*x+1/2*c)^5-11/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A-13/8/d/a^4*C*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*A*ta
n(1/2*d*x+1/2*c)-111/8/d/a^4*C*tan(1/2*d*x+1/2*c)-1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A-21/2/d/a^4*ln(tan(1/2*d*x
+1/2*c)-1)*C+1/2/d/a^4*C/(tan(1/2*d*x+1/2*c)-1)^2+9/2/d/a^4*C/(tan(1/2*d*x+1/2*c)-1)+1/d/a^4*ln(tan(1/2*d*x+1/
2*c)+1)*A+21/2/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^4*C/(tan(1/2*d*x+1/2*c)+1)^2+9/2/d/a^4*C/(tan(1/2*d*x+
1/2*c)+1)

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maxima [A]  time = 0.48, size = 372, normalized size = 1.60 \[ -\frac {3 \, C {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x +
c) + 1) - 1)/a^4) + 5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*si
n(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x
+ c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/d

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mupad [B]  time = 2.64, size = 256, normalized size = 1.10 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {21\,C}{2}\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A+C\right )}{40\,a^4}+\frac {2\,A+6\,C}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^4}-\frac {A-15\,C}{24\,a^4}+\frac {2\,A+6\,C}{8\,a^4}\right )}{d}-\frac {7\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{4\,a^4}-\frac {3\,\left (A-15\,C\right )}{8\,a^4}+\frac {3\,\left (2\,A+6\,C\right )}{4\,a^4}-\frac {4\,A-20\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^5*(a + a/cos(c + d*x))^4),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A + (21*C)/2))/(a^4*d) - (tan(c/2 + (d*x)/2)^5*((3*(A + C))/(40*a^4) + (2*A + 6*
C)/(40*a^4)))/d - (tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^4) - (A - 15*C)/(24*a^4) + (2*A + 6*C)/(8*a^4)))/d - (7*
C*tan(c/2 + (d*x)/2) - 9*C*tan(c/2 + (d*x)/2)^3)/(d*(a^4*tan(c/2 + (d*x)/2)^4 - 2*a^4*tan(c/2 + (d*x)/2)^2 + a
^4)) - (tan(c/2 + (d*x)/2)*((5*(A + C))/(4*a^4) - (3*(A - 15*C))/(8*a^4) + (3*(2*A + 6*C))/(4*a^4) - (4*A - 20
*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{7}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(C*sec(c + d*x)**7/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1),
x))/a**4

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