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

Optimal. Leaf size=194 \[ -\frac{(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac{(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{(25 A-88 B) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{(A-4 B) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}+\frac{(A-B) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(5 A-12 B) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

((A - 4*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((55*A - 244*B)*Tan[c + d*x])/(105*a^4*d) + ((25*A - 88*B)*Sec[c +
d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - ((A - 4*B)*Tan[c + d*x])/(a^4*d*(1 + Sec[c + d*x])) +
((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((5*A - 12*B)*Sec[c + d*x]^3*Tan[c + d*x
])/(35*a*d*(a + a*Sec[c + d*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.616241, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4019, 4008, 3787, 3770, 3767, 8} \[ -\frac{(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac{(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{(25 A-88 B) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{(A-4 B) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}+\frac{(A-B) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(5 A-12 B) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 4*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((55*A - 244*B)*Tan[c + d*x])/(105*a^4*d) + ((25*A - 88*B)*Sec[c +
d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - ((A - 4*B)*Tan[c + d*x])/(a^4*d*(1 + Sec[c + d*x])) +
((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((5*A - 12*B)*Sec[c + d*x]^3*Tan[c + d*x
])/(35*a*d*(a + a*Sec[c + d*x])^3)

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 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 3770

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

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{\int \frac{\sec ^4(c+d x) (4 a (A-B)-a (A-8 B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^3(c+d x) \left (3 a^2 (5 A-12 B)-2 a^2 (5 A-26 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a^3 (25 A-88 B)-a^3 (55 A-244 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac{(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{\int \sec (c+d x) \left (-105 a^4 (A-4 B)+a^4 (55 A-244 B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac{(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{(55 A-244 B) \int \sec ^2(c+d x) \, dx}{105 a^4}+\frac{(A-4 B) \int \sec (c+d x) \, dx}{a^4}\\ &=\frac{(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{(55 A-244 B) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac{(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac{(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.38583, size = 754, normalized size = 3.89 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^4(c+d x) \left (4795 A \sin \left (c-\frac{d x}{2}\right )-4795 A \sin \left (c+\frac{d x}{2}\right )+4165 A \sin \left (2 c+\frac{d x}{2}\right )+2275 A \sin \left (c+\frac{3 d x}{2}\right )-4445 A \sin \left (2 c+\frac{3 d x}{2}\right )+2275 A \sin \left (3 c+\frac{3 d x}{2}\right )-2785 A \sin \left (c+\frac{5 d x}{2}\right )+735 A \sin \left (2 c+\frac{5 d x}{2}\right )-2785 A \sin \left (3 c+\frac{5 d x}{2}\right )+735 A \sin \left (4 c+\frac{5 d x}{2}\right )-1015 A \sin \left (2 c+\frac{7 d x}{2}\right )+105 A \sin \left (3 c+\frac{7 d x}{2}\right )-1015 A \sin \left (4 c+\frac{7 d x}{2}\right )+105 A \sin \left (5 c+\frac{7 d x}{2}\right )-160 A \sin \left (3 c+\frac{9 d x}{2}\right )-160 A \sin \left (5 c+\frac{9 d x}{2}\right )+4165 A \sin \left (\frac{d x}{2}\right )-4445 A \sin \left (\frac{3 d x}{2}\right )-20524 B \sin \left (c-\frac{d x}{2}\right )+14644 B \sin \left (c+\frac{d x}{2}\right )-16660 B \sin \left (2 c+\frac{d x}{2}\right )-4690 B \sin \left (c+\frac{3 d x}{2}\right )+14378 B \sin \left (2 c+\frac{3 d x}{2}\right )-9100 B \sin \left (3 c+\frac{3 d x}{2}\right )+11668 B \sin \left (c+\frac{5 d x}{2}\right )-630 B \sin \left (2 c+\frac{5 d x}{2}\right )+9358 B \sin \left (3 c+\frac{5 d x}{2}\right )-2940 B \sin \left (4 c+\frac{5 d x}{2}\right )+4228 B \sin \left (2 c+\frac{7 d x}{2}\right )+315 B \sin \left (3 c+\frac{7 d x}{2}\right )+3493 B \sin \left (4 c+\frac{7 d x}{2}\right )-420 B \sin \left (5 c+\frac{7 d x}{2}\right )+664 B \sin \left (3 c+\frac{9 d x}{2}\right )+105 B \sin \left (4 c+\frac{9 d x}{2}\right )+559 B \sin \left (5 c+\frac{9 d x}{2}\right )-10780 B \sin \left (\frac{d x}{2}\right )+18788 B \sin \left (\frac{3 d x}{2}\right )\right ) (A+B \sec (c+d x))}{1680 d (a \sec (c+d x)+a)^4 (A \cos (c+d x)+B)}+\frac{16 (4 B-A) \cos ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) (A+B \sec (c+d x)) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^4 (A \cos (c+d x)+B)}-\frac{16 (4 B-A) \cos ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) (A+B \sec (c+d x)) \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^4 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*(-A + 4*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) - (16*(-A + 4*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]^4*(A + B*Sec[c + d*x])*(4165*A*Sin[(d*x)/2] - 10780*B*S
in[(d*x)/2] - 4445*A*Sin[(3*d*x)/2] + 18788*B*Sin[(3*d*x)/2] + 4795*A*Sin[c - (d*x)/2] - 20524*B*Sin[c - (d*x)
/2] - 4795*A*Sin[c + (d*x)/2] + 14644*B*Sin[c + (d*x)/2] + 4165*A*Sin[2*c + (d*x)/2] - 16660*B*Sin[2*c + (d*x)
/2] + 2275*A*Sin[c + (3*d*x)/2] - 4690*B*Sin[c + (3*d*x)/2] - 4445*A*Sin[2*c + (3*d*x)/2] + 14378*B*Sin[2*c +
(3*d*x)/2] + 2275*A*Sin[3*c + (3*d*x)/2] - 9100*B*Sin[3*c + (3*d*x)/2] - 2785*A*Sin[c + (5*d*x)/2] + 11668*B*S
in[c + (5*d*x)/2] + 735*A*Sin[2*c + (5*d*x)/2] - 630*B*Sin[2*c + (5*d*x)/2] - 2785*A*Sin[3*c + (5*d*x)/2] + 93
58*B*Sin[3*c + (5*d*x)/2] + 735*A*Sin[4*c + (5*d*x)/2] - 2940*B*Sin[4*c + (5*d*x)/2] - 1015*A*Sin[2*c + (7*d*x
)/2] + 4228*B*Sin[2*c + (7*d*x)/2] + 105*A*Sin[3*c + (7*d*x)/2] + 315*B*Sin[3*c + (7*d*x)/2] - 1015*A*Sin[4*c
+ (7*d*x)/2] + 3493*B*Sin[4*c + (7*d*x)/2] + 105*A*Sin[5*c + (7*d*x)/2] - 420*B*Sin[5*c + (7*d*x)/2] - 160*A*S
in[3*c + (9*d*x)/2] + 664*B*Sin[3*c + (9*d*x)/2] + 105*B*Sin[4*c + (9*d*x)/2] - 160*A*Sin[5*c + (9*d*x)/2] + 5
59*B*Sin[5*c + (9*d*x)/2]))/(1680*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.062, size = 285, normalized size = 1.5 \begin{align*} -{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{7\,B}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{11\,A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{23\,B}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{15\,A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{49\,B}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{A}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{4}}}-{\frac{B}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{4}}}-{\frac{B}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \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))/(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-1/8/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-11/24/d/a^4*A*tan(1/2*d*x+1/2*c)^3+23/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*A*t
an(1/2*d*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)+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-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-1/d/a^4/(tan(1/2*d*x+1/2*c)-1)*B

________________________________________________________________________________________

Maxima [A]  time = 1.01956, size = 440, normalized size = 2.27 \begin{align*} \frac{B{\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 )} - 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} \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))/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(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*l
og(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

________________________________________________________________________________________

Fricas [A]  time = 0.504279, size = 837, normalized size = 4.31 \begin{align*} \frac{105 \,{\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (A - 4 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (A - 4 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (8 \,{\left (20 \, A - 83 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (535 \, A - 2236 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (155 \, A - 659 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (65 \, A - 296 \, B\right )} \cos \left (d x + c\right ) - 105 \, B\right )} \sin \left (d x + c\right )}{210 \,{\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\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))/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/210*(105*((A - 4*B)*cos(d*x + c)^5 + 4*(A - 4*B)*cos(d*x + c)^4 + 6*(A - 4*B)*cos(d*x + c)^3 + 4*(A - 4*B)*c
os(d*x + c)^2 + (A - 4*B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 105*((A - 4*B)*cos(d*x + c)^5 + 4*(A - 4*B)*co
s(d*x + c)^4 + 6*(A - 4*B)*cos(d*x + c)^3 + 4*(A - 4*B)*cos(d*x + c)^2 + (A - 4*B)*cos(d*x + c))*log(-sin(d*x
+ c) + 1) - 2*(8*(20*A - 83*B)*cos(d*x + c)^4 + (535*A - 2236*B)*cos(d*x + c)^3 + 4*(155*A - 659*B)*cos(d*x +
c)^2 + 4*(65*A - 296*B)*cos(d*x + c) - 105*B)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6
*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

________________________________________________________________________________________

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}{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))/(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))/a**4

________________________________________________________________________________________

Giac [A]  time = 1.33136, size = 297, normalized size = 1.53 \begin{align*} \frac{\frac{840 \,{\left (A - 4 \, 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 (A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{1680 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} 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} + 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} + 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} + 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 )}{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))/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*(A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
/a^4 - 1680*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*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 + 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 + 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 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*a
^24*tan(1/2*d*x + 1/2*c))/a^28)/d