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

Optimal. Leaf size=254 \[ -\frac{8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac{(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{4 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac{(2 A-8 B+21 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

((2*A - 8*B + 21*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(20*A - 83*B + 216*C)*Tan[c + d*x])/(105*a^4*d) + ((
2*A - 8*B + 21*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((10*A - 52*B + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(
105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(20*A - 83*B + 216*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c
+ d*x])) - ((A - B + C)*Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((B - 2*C)*Sec[c + d*x]^4*
Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.689982, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4084, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac{8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac{(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{4 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac{(2 A-8 B+21 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac{(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A - 8*B + 21*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(20*A - 83*B + 216*C)*Tan[c + d*x])/(105*a^4*d) + ((
2*A - 8*B + 21*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((10*A - 52*B + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(
105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(20*A - 83*B + 216*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c
+ d*x])) - ((A - B + C)*Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((B - 2*C)*Sec[c + d*x]^4*
Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 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 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 8

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

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]

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B+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 B-5 C)+a (2 A-2 B+9 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-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 (28 a^2 (B-2 C)+a^2 (10 A-24 B+73 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-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-52 B+129 C)+a^3 (50 A-176 B+477 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (20 A-83 B+216 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 (-8 a^4 (20 A-83 B+216 C)+105 a^4 (2 A-8 B+21 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac{(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (20 A-83 B+216 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-8 B+21 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac{(8 (20 A-83 B+216 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac{(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (20 A-83 B+216 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-8 B+21 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac{(8 (20 A-83 B+216 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac{(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac{(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac{(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.47611, size = 1322, normalized size = 5.2 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A +
B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^
4) + (16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(
A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*
x])^4) - (4*Cos[c/2 + (d*x)/2]^2*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Sin[c/2] -
B*Sin[c/2] + C*Sin[c/2]))/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (8
*Cos[c/2 + (d*x)/2]^4*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(10*A*Sin[c/2] - 17*B*Si
n[c/2] + 24*C*Sin[c/2]))/(35*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (16
*Cos[c/2 + (d*x)/2]^6*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(55*A*Sin[c/2] - 139*B*S
in[c/2] + 258*C*Sin[c/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) -
(4*Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] - B*Sin[
(d*x)/2] + C*Sin[(d*x)/2]))/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (
8*Cos[c/2 + (d*x)/2]^3*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(10*A*Sin[(d*x)/2] - 17
*B*Sin[(d*x)/2] + 24*C*Sin[(d*x)/2]))/(35*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d
*x])^4) - (16*Cos[c/2 + (d*x)/2]^5*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(55*A*Sin[(
d*x)/2] - 139*B*Sin[(d*x)/2] + 258*C*Sin[(d*x)/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(
a + a*Sec[c + d*x])^4) - (32*Cos[c/2 + (d*x)/2]^7*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]
^2)*(160*A*Sin[(d*x)/2] - 559*B*Sin[(d*x)/2] + 1308*C*Sin[(d*x)/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Co
s[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) + (16*C*Cos[c/2 + (d*x)/2]^8*Sec[c]*Sec[c + d*x]^4*(A + B*Sec[c + d*x]
+ C*Sec[c + d*x]^2)*Sin[d*x])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) +
(16*Cos[c/2 + (d*x)/2]^8*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(C*Sin[c] + 2*B*Sin[d*x
] - 8*C*Sin[d*x]))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4)

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Maple [B]  time = 0.086, size = 493, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+B*sec(d*x+c)+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*tan(1/2*d*x+1/2*c)^7*B-1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7-1/8/d/a
^4*tan(1/2*d*x+1/2*c)^5*A+7/40/d/a^4*tan(1/2*d*x+1/2*c)^5*B-9/40/d/a^4*C*tan(1/2*d*x+1/2*c)^5-11/24/d/a^4*A*ta
n(1/2*d*x+1/2*c)^3+23/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3-13/8/d/a^4*C*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*A*tan(1/2*d
*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)-111/8/d/a^4*C*tan(1/2*d*x+1/2*c)+9/2/d/a^4*C/(tan(1/2*d*x+1/2*c)+1)-
1/d/a^4/(tan(1/2*d*x+1/2*c)+1)*B+1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*A-4/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*B+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
/(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A+4/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*B-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

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Maxima [B]  time = 1.00343, size = 751, normalized size = 2.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+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) - B*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))
+ (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^
4 + 3360*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*sin(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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Fricas [A]  time = 0.547221, size = 1062, normalized size = 4.18 \begin{align*} \frac{105 \,{\left ({\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 8 \, B + 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 - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (20 \, A - 83 \, B + 216 \, C\right )} \cos \left (d x + c\right )^{5} +{\left (1070 \, A - 4472 \, B + 11619 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (310 \, A - 1318 \, B + 3411 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (130 \, A - 592 \, B + 1509 \, C\right )} \cos \left (d x + c\right )^{2} - 210 \,{\left (B - 2 \, C\right )} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(105*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^5 + 6*(2*A - 8*B + 21*C)*cos
(d*x + c)^4 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) -
105*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^5 + 6*(2*A - 8*B + 21*C)*cos(d*x +
c)^4 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(1
6*(20*A - 83*B + 216*C)*cos(d*x + c)^5 + (1070*A - 4472*B + 11619*C)*cos(d*x + c)^4 + 4*(310*A - 1318*B + 3411
*C)*cos(d*x + c)^3 + 4*(130*A - 592*B + 1509*C)*cos(d*x + c)^2 - 210*(B - 2*C)*cos(d*x + c) - 105*C)*sin(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{B \sec ^{6}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(A+B*sec(d*x+c)+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(B*sec(c + d*x)**6/(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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Giac [A]  time = 1.32497, size = 458, normalized size = 1.8 \begin{align*} \frac{\frac{420 \,{\left (2 \, A - 8 \, B + 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 - 8 \, B + 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 (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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 \, B 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} - 147 \, B 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} - 805 \, B 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 ) - 5145 \, B 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(420*(2*A - 8*B + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(2*A - 8*B + 21*C)*log(abs(tan(1/2*
d*x + 1/2*c) - 1))/a^4 - 840*(2*B*tan(1/2*d*x + 1/2*c)^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*
c) + 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*B
*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 - 147*B*a^
24*tan(1/2*d*x + 1/2*c)^5 + 189*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 - 805*B*a^24
*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*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