3.108 \(\int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=238 \[ \frac {8 (83 A-216 B) \tan (c+d x)}{105 a^4 d}-\frac {(8 A-21 B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {4 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(8 A-21 B) \tan (c+d x) \sec (c+d x)}{2 a^4 d}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

-1/2*(8*A-21*B)*arctanh(sin(d*x+c))/a^4/d+8/105*(83*A-216*B)*tan(d*x+c)/a^4/d-1/2*(8*A-21*B)*sec(d*x+c)*tan(d*
x+c)/a^4/d+1/105*(52*A-129*B)*sec(d*x+c)^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+4/105*(83*A-216*B)*sec(d*x+c)^2*t
an(d*x+c)/a^4/d/(1+sec(d*x+c))+1/7*(A-B)*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/5*(A-2*B)*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 = 0.66, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4019, 3787, 3767, 8, 3768, 3770} \[ \frac {8 (83 A-216 B) \tan (c+d x)}{105 a^4 d}-\frac {(8 A-21 B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {4 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(8 A-21 B) \tan (c+d x) \sec (c+d x)}{2 a^4 d}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^6*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=\frac {(A-B) \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) (5 a (A-B)-a (2 A-9 B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \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 (A-2 B)-a^2 (24 A-73 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \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 (52 A-129 B)-a^3 (176 A-477 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {4 (83 A-216 B) \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 (83 A-216 B)-105 a^4 (8 A-21 B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {4 (83 A-216 B) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(8 (83 A-216 B)) \int \sec ^2(c+d x) \, dx}{105 a^4}-\frac {(8 A-21 B) \int \sec ^3(c+d x) \, dx}{a^4}\\ &=-\frac {(8 A-21 B) \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {4 (83 A-216 B) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {(8 A-21 B) \int \sec (c+d x) \, dx}{2 a^4}-\frac {(8 (83 A-216 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=-\frac {(8 A-21 B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {8 (83 A-216 B) \tan (c+d x)}{105 a^4 d}-\frac {(8 A-21 B) \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {4 (83 A-216 B) \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 = 6.52, size = 880, normalized size = 3.70 \[ -\frac {8 (21 B-8 A) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^3(c+d x) (A+B \sec (c+d x)) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^4}+\frac {8 (21 B-8 A) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^3(c+d x) (A+B \sec (c+d x)) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^4}+\frac {\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^5(c+d x) (A+B \sec (c+d x)) \left (-38668 A \sin \left (\frac {d x}{2}\right )+73206 B \sin \left (\frac {d x}{2}\right )+64384 A \sin \left (\frac {3 d x}{2}\right )-166668 B \sin \left (\frac {3 d x}{2}\right )-70896 A \sin \left (c-\frac {d x}{2}\right )+183162 B \sin \left (c-\frac {d x}{2}\right )+50316 A \sin \left (c+\frac {d x}{2}\right )-100842 B \sin \left (c+\frac {d x}{2}\right )-59248 A \sin \left (2 c+\frac {d x}{2}\right )+155526 B \sin \left (2 c+\frac {d x}{2}\right )-22820 A \sin \left (c+\frac {3 d x}{2}\right )+37380 B \sin \left (c+\frac {3 d x}{2}\right )+48004 A \sin \left (2 c+\frac {3 d x}{2}\right )-101148 B \sin \left (2 c+\frac {3 d x}{2}\right )-39200 A \sin \left (3 c+\frac {3 d x}{2}\right )+102900 B \sin \left (3 c+\frac {3 d x}{2}\right )+46032 A \sin \left (c+\frac {5 d x}{2}\right )-119364 B \sin \left (c+\frac {5 d x}{2}\right )-8750 A \sin \left (2 c+\frac {5 d x}{2}\right )+8820 B \sin \left (2 c+\frac {5 d x}{2}\right )+35742 A \sin \left (3 c+\frac {5 d x}{2}\right )-78204 B \sin \left (3 c+\frac {5 d x}{2}\right )-19040 A \sin \left (4 c+\frac {5 d x}{2}\right )+49980 B \sin \left (4 c+\frac {5 d x}{2}\right )+24664 A \sin \left (2 c+\frac {7 d x}{2}\right )-64053 B \sin \left (2 c+\frac {7 d x}{2}\right )-1050 A \sin \left (3 c+\frac {7 d x}{2}\right )-3885 B \sin \left (3 c+\frac {7 d x}{2}\right )+19834 A \sin \left (4 c+\frac {7 d x}{2}\right )-44733 B \sin \left (4 c+\frac {7 d x}{2}\right )-5880 A \sin \left (5 c+\frac {7 d x}{2}\right )+15435 B \sin \left (5 c+\frac {7 d x}{2}\right )+8456 A \sin \left (3 c+\frac {9 d x}{2}\right )-21987 B \sin \left (3 c+\frac {9 d x}{2}\right )+630 A \sin \left (4 c+\frac {9 d x}{2}\right )-3675 B \sin \left (4 c+\frac {9 d x}{2}\right )+6986 A \sin \left (5 c+\frac {9 d x}{2}\right )-16107 B \sin \left (5 c+\frac {9 d x}{2}\right )-840 A \sin \left (6 c+\frac {9 d x}{2}\right )+2205 B \sin \left (6 c+\frac {9 d x}{2}\right )+1328 A \sin \left (4 c+\frac {11 d x}{2}\right )-3456 B \sin \left (4 c+\frac {11 d x}{2}\right )+210 A \sin \left (5 c+\frac {11 d x}{2}\right )-840 B \sin \left (5 c+\frac {11 d x}{2}\right )+1118 A \sin \left (6 c+\frac {11 d x}{2}\right )-2616 B \sin \left (6 c+\frac {11 d x}{2}\right )\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{6720 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^6*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

(-8*(-8*A + 21*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[
c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4) + (8*(-8*A + 21*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2
+ (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c +
d*x])^4) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^5*(A + B*Sec[c + d*x])*(-38668*A*Sin[(d*x)/2] + 73
206*B*Sin[(d*x)/2] + 64384*A*Sin[(3*d*x)/2] - 166668*B*Sin[(3*d*x)/2] - 70896*A*Sin[c - (d*x)/2] + 183162*B*Si
n[c - (d*x)/2] + 50316*A*Sin[c + (d*x)/2] - 100842*B*Sin[c + (d*x)/2] - 59248*A*Sin[2*c + (d*x)/2] + 155526*B*
Sin[2*c + (d*x)/2] - 22820*A*Sin[c + (3*d*x)/2] + 37380*B*Sin[c + (3*d*x)/2] + 48004*A*Sin[2*c + (3*d*x)/2] -
101148*B*Sin[2*c + (3*d*x)/2] - 39200*A*Sin[3*c + (3*d*x)/2] + 102900*B*Sin[3*c + (3*d*x)/2] + 46032*A*Sin[c +
(5*d*x)/2] - 119364*B*Sin[c + (5*d*x)/2] - 8750*A*Sin[2*c + (5*d*x)/2] + 8820*B*Sin[2*c + (5*d*x)/2] + 35742*
A*Sin[3*c + (5*d*x)/2] - 78204*B*Sin[3*c + (5*d*x)/2] - 19040*A*Sin[4*c + (5*d*x)/2] + 49980*B*Sin[4*c + (5*d*
x)/2] + 24664*A*Sin[2*c + (7*d*x)/2] - 64053*B*Sin[2*c + (7*d*x)/2] - 1050*A*Sin[3*c + (7*d*x)/2] - 3885*B*Sin
[3*c + (7*d*x)/2] + 19834*A*Sin[4*c + (7*d*x)/2] - 44733*B*Sin[4*c + (7*d*x)/2] - 5880*A*Sin[5*c + (7*d*x)/2]
+ 15435*B*Sin[5*c + (7*d*x)/2] + 8456*A*Sin[3*c + (9*d*x)/2] - 21987*B*Sin[3*c + (9*d*x)/2] + 630*A*Sin[4*c +
(9*d*x)/2] - 3675*B*Sin[4*c + (9*d*x)/2] + 6986*A*Sin[5*c + (9*d*x)/2] - 16107*B*Sin[5*c + (9*d*x)/2] - 840*A*
Sin[6*c + (9*d*x)/2] + 2205*B*Sin[6*c + (9*d*x)/2] + 1328*A*Sin[4*c + (11*d*x)/2] - 3456*B*Sin[4*c + (11*d*x)/
2] + 210*A*Sin[5*c + (11*d*x)/2] - 840*B*Sin[5*c + (11*d*x)/2] + 1118*A*Sin[6*c + (11*d*x)/2] - 2616*B*Sin[6*c
+ (11*d*x)/2]))/(6720*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4)

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fricas [A]  time = 0.49, size = 358, normalized size = 1.50 \[ -\frac {105 \, {\left ({\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (83 \, A - 216 \, B\right )} \cos \left (d x + c\right )^{5} + {\left (4472 \, A - 11619 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (1318 \, A - 3411 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (592 \, A - 1509 \, B\right )} \cos \left (d x + c\right )^{2} + 210 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) + 105 \, B\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/420*(105*((8*A - 21*B)*cos(d*x + c)^6 + 4*(8*A - 21*B)*cos(d*x + c)^5 + 6*(8*A - 21*B)*cos(d*x + c)^4 + 4*(
8*A - 21*B)*cos(d*x + c)^3 + (8*A - 21*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 105*((8*A - 21*B)*cos(d*x +
c)^6 + 4*(8*A - 21*B)*cos(d*x + c)^5 + 6*(8*A - 21*B)*cos(d*x + c)^4 + 4*(8*A - 21*B)*cos(d*x + c)^3 + (8*A -
21*B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(83*A - 216*B)*cos(d*x + c)^5 + (4472*A - 11619*B)*cos(d*
x + c)^4 + 4*(1318*A - 3411*B)*cos(d*x + c)^3 + 4*(592*A - 1509*B)*cos(d*x + c)^2 + 210*(A - 2*B)*cos(d*x + c)
+ 105*B)*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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giac [A]  time = 0.87, size = 267, normalized size = 1.12 \[ -\frac {\frac {420 \, {\left (8 \, A - 21 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (8 \, A - 21 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 189 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1365 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11655 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(420*(8*A - 21*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(8*A - 21*B)*log(abs(tan(1/2*d*x + 1/2*c
) - 1))/a^4 + 840*(2*A*tan(1/2*d*x + 1/2*c)^3 - 9*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 7*B*ta
n(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 + 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 189*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2
*d*x + 1/2*c)^3 - 1365*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*tan(1/2*d*x + 1/2*c) - 11655*B*a^24*tan(1/2
*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.63, size = 374, normalized size = 1.57 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {9 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {13 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {111 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{d \,a^{4}}-\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 d \,a^{4}}+\frac {9 B}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {9 B}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{d \,a^{4}}+\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 d \,a^{4}}-\frac {B}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(A+B*sec(d*x+c))/(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*B*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-9/40/d/a
^4*B*tan(1/2*d*x+1/2*c)^5+23/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A-13/8/d/a^4*B*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*A*ta
n(1/2*d*x+1/2*c)-111/8/d/a^4*B*tan(1/2*d*x+1/2*c)+4/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)*B+9/2/d/a^4/(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^4/(tan(1/2*d*x+1/2*c)-1)*A+1/2/d/a^4*B/(tan(1/2*d*x+1/2*
c)-1)^2+9/2/d/a^4/(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^4/(tan(1/2*d*x+1/2*c)+1)*A-4/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)*B-1/2/d/a^4*B/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [A]  time = 0.36, size = 419, normalized size = 1.76 \[ -\frac {3 \, B {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(3*B*(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) - A*(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))/d

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mupad [B]  time = 2.05, size = 272, normalized size = 1.14 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^4}+\frac {4\,A-6\,B}{8\,a^4}+\frac {5\,A-15\,B}{24\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A-B\right )}{4\,a^4}-\frac {5\,B}{2\,a^4}+\frac {3\,\left (4\,A-6\,B\right )}{4\,a^4}+\frac {3\,\left (5\,A-15\,B\right )}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-9\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-7\,B\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A-B\right )}{40\,a^4}+\frac {4\,A-6\,B}{40\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (8\,A-21\,B\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^6*(a + a/cos(c + d*x))^4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*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(B*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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