3.500 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=294 \[ \frac{2 a^3 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{1287 d}+\frac{2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt{a \sec (c+d x)+a}}-\frac{4 a^2 (10439 A+9230 B+8368 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{45045 d}+\frac{2 a (10439 A+9230 B+8368 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac{2 a (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \]

[Out]

(2*a^3*(10439*A + 9230*B + 8368*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(2717*A + 2522*B +
2224*C)*Sec[c + d*x]^3*Tan[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(10439*A + 9230*B + 8368*C)*S
qrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(45045*d) + (2*a^2*(143*A + 182*B + 136*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec
[c + d*x]]*Tan[c + d*x])/(1287*d) + (2*a*(10439*A + 9230*B + 8368*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/
(15015*d) + (2*a*(13*B + 5*C)*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(143*d) + (2*C*Sec[c + d
*x]^3*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(13*d)

________________________________________________________________________________________

Rubi [A]  time = 0.90489, antiderivative size = 294, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4088, 4018, 4016, 3800, 4001, 3792} \[ \frac{2 a^3 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{1287 d}+\frac{2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt{a \sec (c+d x)+a}}-\frac{4 a^2 (10439 A+9230 B+8368 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{45045 d}+\frac{2 a (10439 A+9230 B+8368 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac{2 a (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^3*(10439*A + 9230*B + 8368*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(2717*A + 2522*B +
2224*C)*Sec[c + d*x]^3*Tan[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(10439*A + 9230*B + 8368*C)*S
qrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(45045*d) + (2*a^2*(143*A + 182*B + 136*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec
[c + d*x]]*Tan[c + d*x])/(1287*d) + (2*a*(10439*A + 9230*B + 8368*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/
(15015*d) + (2*a*(13*B + 5*C)*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(143*d) + (2*C*Sec[c + d
*x]^3*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(13*d)

Rule 4088

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 4018

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

Rule 4016

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

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

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{2 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (13 A+6 C)+\frac{1}{2} a (13 B+5 C) \sec (c+d x)\right ) \, dx}{13 a}\\ &=\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{4 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (143 A+78 B+96 C)+\frac{1}{4} a^2 (143 A+182 B+136 C) \sec (c+d x)\right ) \, dx}{143 a}\\ &=\frac{2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{8 \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3}{8} a^3 (715 A+598 B+560 C)+\frac{1}{8} a^3 (2717 A+2522 B+2224 C) \sec (c+d x)\right ) \, dx}{1287 a}\\ &=\frac{2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{\left (a^2 (10439 A+9230 B+8368 C)\right ) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{3003}\\ &=\frac{2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac{2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{(2 a (10439 A+9230 B+8368 C)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{15015}\\ &=\frac{2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a^2 (10439 A+9230 B+8368 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac{2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac{2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac{\left (a^2 (10439 A+9230 B+8368 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx}{6435}\\ &=\frac{2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a^2 (10439 A+9230 B+8368 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac{2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac{2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac{2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}\\ \end{align*}

Mathematica [A]  time = 2.14288, size = 222, normalized size = 0.76 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \sqrt{a (\sec (c+d x)+1)} (70 (4576 A+5083 B+5552 C) \cos (c+d x)+14 (32747 A+31850 B+30334 C) \cos (2 (c+d x))+141570 A \cos (3 (c+d x))+156585 A \cos (4 (c+d x))+20878 A \cos (5 (c+d x))+20878 A \cos (6 (c+d x))+322751 A+138450 B \cos (3 (c+d x))+138450 B \cos (4 (c+d x))+18460 B \cos (5 (c+d x))+18460 B \cos (6 (c+d x))+325910 B+125520 C \cos (3 (c+d x))+125520 C \cos (4 (c+d x))+16736 C \cos (5 (c+d x))+16736 C \cos (6 (c+d x))+343612 C)}{180180 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(322751*A + 325910*B + 343612*C + 70*(4576*A + 5083*B + 5552*C)*Cos[c + d*x] + 14*(32747*A + 31850*B + 30
334*C)*Cos[2*(c + d*x)] + 141570*A*Cos[3*(c + d*x)] + 138450*B*Cos[3*(c + d*x)] + 125520*C*Cos[3*(c + d*x)] +
156585*A*Cos[4*(c + d*x)] + 138450*B*Cos[4*(c + d*x)] + 125520*C*Cos[4*(c + d*x)] + 20878*A*Cos[5*(c + d*x)] +
18460*B*Cos[5*(c + d*x)] + 16736*C*Cos[5*(c + d*x)] + 20878*A*Cos[6*(c + d*x)] + 18460*B*Cos[6*(c + d*x)] + 1
6736*C*Cos[6*(c + d*x)])*Sec[c + d*x]^6*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(180180*d)

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Maple [A]  time = 0.37, size = 240, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 83512\,A \left ( \cos \left ( dx+c \right ) \right ) ^{6}+73840\,B \left ( \cos \left ( dx+c \right ) \right ) ^{6}+66944\,C \left ( \cos \left ( dx+c \right ) \right ) ^{6}+41756\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+36920\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+33472\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+31317\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+27690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+25104\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+18590\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+23075\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+20920\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+5005\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+14560\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18305\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4095\,B\cos \left ( dx+c \right ) +11970\,C\cos \left ( dx+c \right ) +3465\,C \right ) }{45045\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/45045/d*a^2*(-1+cos(d*x+c))*(83512*A*cos(d*x+c)^6+73840*B*cos(d*x+c)^6+66944*C*cos(d*x+c)^6+41756*A*cos(d*x
+c)^5+36920*B*cos(d*x+c)^5+33472*C*cos(d*x+c)^5+31317*A*cos(d*x+c)^4+27690*B*cos(d*x+c)^4+25104*C*cos(d*x+c)^4
+18590*A*cos(d*x+c)^3+23075*B*cos(d*x+c)^3+20920*C*cos(d*x+c)^3+5005*A*cos(d*x+c)^2+14560*B*cos(d*x+c)^2+18305
*C*cos(d*x+c)^2+4095*B*cos(d*x+c)+11970*C*cos(d*x+c)+3465*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^6/
sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.570552, size = 541, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (8 \,{\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \,{\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \,{\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \,{\left (3718 \, A + 4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \,{\left (143 \, A + 416 \, B + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \,{\left (13 \, B + 38 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/45045*(8*(10439*A + 9230*B + 8368*C)*a^2*cos(d*x + c)^6 + 4*(10439*A + 9230*B + 8368*C)*a^2*cos(d*x + c)^5 +
3*(10439*A + 9230*B + 8368*C)*a^2*cos(d*x + c)^4 + 5*(3718*A + 4615*B + 4184*C)*a^2*cos(d*x + c)^3 + 35*(143*
A + 416*B + 523*C)*a^2*cos(d*x + c)^2 + 315*(13*B + 38*C)*a^2*cos(d*x + c) + 3465*C*a^2)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 5.7069, size = 637, normalized size = 2.17 \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)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/45045*(45045*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 45045*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 45045*sqrt(2)*C*a^9*s
gn(cos(d*x + c)) - (180180*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 150150*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 120120*s
qrt(2)*C*a^9*sgn(cos(d*x + c)) - (342342*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 300300*sqrt(2)*B*a^9*sgn(cos(d*x +
c)) + 294294*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - (391248*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 356070*sqrt(2)*B*a^9*
sgn(cos(d*x + c)) + 310596*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - (265837*sqrt(2)*A*a^9*sgn(cos(d*x + c)) + 232375*
sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 212069*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - 4*(24167*sqrt(2)*A*a^9*sgn(cos(d*x
+ c)) + 21125*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 19279*sqrt(2)*C*a^9*sgn(cos(d*x + c)) - 2*(1859*sqrt(2)*A*a^9*
sgn(cos(d*x + c)) + 1625*sqrt(2)*B*a^9*sgn(cos(d*x + c)) + 1483*sqrt(2)*C*a^9*sgn(cos(d*x + c)))*tan(1/2*d*x +
1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan
(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^6*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a
)*d)