3.4.75 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [375]

Optimal. Leaf size=237 \[ \frac {2 a^3 (803 B+710 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 B+710 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 B+710 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \]

[Out]

2/1155*a*(803*B+710*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/11*a*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(d*
x+c)/d+2/495*a^3*(803*B+710*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(209*B+194*C)*sec(d*x+c)^3*tan(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/3465*a^2*(803*B+710*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/99*a^2*(11*B+14*C
)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.51, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4157, 4103, 4101, 3885, 4086, 3877} \begin {gather*} \frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (803 B+710 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}-\frac {4 a^2 (803 B+710 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a (803 B+710 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(803*B + 710*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(209*B + 194*C)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(803*B + 710*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(3465*d) + (2*a^2*(11*B + 14*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(99*d) + (2*a*(803*B +
710*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*d) + (2*a*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Ta
n[c + d*x])/(11*d)

Rule 3877

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]

Rule 3885

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 4086

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 4101

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 4103

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] &&
NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 4157

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {2}{11} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (11 B+6 C)+\frac {1}{2} a (11 B+14 C) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {4}{99} \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (55 B+46 C)+\frac {1}{4} a^2 (209 B+194 C) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (803 B+710 C)\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 B+710 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {(2 a (803 B+710 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}{1155}\\ &=\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 B+710 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 B+710 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{495} \left (a^2 (803 B+710 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^3 (803 B+710 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 B+710 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 B+710 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 6.17, size = 117, normalized size = 0.49 \begin {gather*} \frac {2 a^3 \left (8 (803 B+710 C)+4 (803 B+710 C) \sec (c+d x)+3 (803 B+710 C) \sec ^2(c+d x)+5 (286 B+355 C) \sec ^3(c+d x)+35 (11 B+32 C) \sec ^4(c+d x)+315 C \sec ^5(c+d x)\right ) \tan (c+d x)}{3465 d \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(8*(803*B + 710*C) + 4*(803*B + 710*C)*Sec[c + d*x] + 3*(803*B + 710*C)*Sec[c + d*x]^2 + 5*(286*B + 355
*C)*Sec[c + d*x]^3 + 35*(11*B + 32*C)*Sec[c + d*x]^4 + 315*C*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 +
Sec[c + d*x])])

________________________________________________________________________________________

Maple [A]
time = 14.54, size = 163, normalized size = 0.69

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (6424 B \left (\cos ^{5}\left (d x +c \right )\right )+5680 C \left (\cos ^{5}\left (d x +c \right )\right )+3212 B \left (\cos ^{4}\left (d x +c \right )\right )+2840 C \left (\cos ^{4}\left (d x +c \right )\right )+2409 B \left (\cos ^{3}\left (d x +c \right )\right )+2130 C \left (\cos ^{3}\left (d x +c \right )\right )+1430 B \left (\cos ^{2}\left (d x +c \right )\right )+1775 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(6424*B*cos(d*x+c)^5+5680*C*cos(d*x+c)^5+3212*B*cos(d*x+c)^4+2840*C*cos(d*x+c)^4+240
9*B*cos(d*x+c)^3+2130*C*cos(d*x+c)^3+1430*B*cos(d*x+c)^2+1775*C*cos(d*x+c)^2+385*B*cos(d*x+c)+1120*C*cos(d*x+c
)+315*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)*a^2

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]
time = 2.21, size = 157, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (8 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, 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 )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(8*(803*B + 710*C)*a^2*cos(d*x + c)^5 + 4*(803*B + 710*C)*a^2*cos(d*x + c)^4 + 3*(803*B + 710*C)*a^2*co
s(d*x + c)^3 + 5*(286*B + 355*C)*a^2*cos(d*x + c)^2 + 35*(11*B + 32*C)*a^2*cos(d*x + c) + 315*C*a^2)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

________________________________________________________________________________________

Giac [A]
time = 1.37, size = 306, normalized size = 1.29 \begin {gather*} \frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 231 \, \sqrt {2} {\left (69 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 65 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (9 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3465*((((4*(2*sqrt(2)*(143*B*a^8*sgn(cos(d*x + c)) + 125*C*a^8*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 1
1*sqrt(2)*(143*B*a^8*sgn(cos(d*x + c)) + 125*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 99*sqrt(2)*(14
3*B*a^8*sgn(cos(d*x + c)) + 125*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 231*sqrt(2)*(69*B*a^8*sgn(c
os(d*x + c)) + 65*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 1155*sqrt(2)*(9*B*a^8*sgn(cos(d*x + c)) +
7*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 3465*sqrt(2)*(B*a^8*sgn(cos(d*x + c)) + C*a^8*sgn(cos(d*
x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

________________________________________________________________________________________

Mupad [B]
time = 15.40, size = 855, normalized size = 3.61 \begin {gather*} \frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {B\,a^2\,8{}\mathrm {i}}{3\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (803\,B+710\,C\right )\,8{}\mathrm {i}}{3465\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {a^2\,\left (5\,B+2\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,B+16\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (11\,B+50\,C\right )\,32{}\mathrm {i}}{693\,d}\right )+\frac {B\,a^2\,24{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (9\,B+10\,C\right )\,8{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,a^2\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,40{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,8{}\mathrm {i}}{11\,d}\right )+\frac {B\,a^2\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,40{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,8{}\mathrm {i}}{11\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {C\,a^2\,64{}\mathrm {i}}{99\,d}-\frac {a^2\,\left (B-8\,C\right )\,8{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,8{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,B+9\,C\right )\,16{}\mathrm {i}}{9\,d}\right )+\frac {B\,a^2\,8{}\mathrm {i}}{9\,d}+\frac {C\,a^2\,64{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (B+2\,C\right )\,40{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (B+C\right )\,80{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (5\,B+2\,C\right )\,8{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (44\,B-31\,C\right )\,16{}\mathrm {i}}{1155\,d}\right )-\frac {B\,a^2\,8{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (4\,B+5\,C\right )\,16{}\mathrm {i}}{5\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (803\,B+710\,C\right )\,16{}\mathrm {i}}{3465\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*8i)/(3*d) - (a^2*exp(c*1i + d*x*1i)*(80
3*B + 710*C)*8i)/(3465*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - ((a + a/(exp(- c*1i - d*x*1i
)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*24i)/(7*d) - exp(c*1i + d*x*1i)*((a^2*(5*B + 16*C)*8i)/(7*d) - (a^2
*(5*B + 2*C)*8i)/(7*d) + (a^2*(11*B + 50*C)*32i)/(693*d)) + (a^2*(9*B + 10*C)*8i)/(7*d)))/((exp(c*1i + d*x*1i)
+ 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i +
d*x*1i)*((B*a^2*8i)/(11*d) + (a^2*(3*B + 4*C)*40i)/(11*d) - (a^2*(5*B + 2*C)*8i)/(11*d) - (a^2*(11*B + 10*C)*
8i)/(11*d)) + (B*a^2*8i)/(11*d) + (a^2*(3*B + 4*C)*40i)/(11*d) - (a^2*(5*B + 2*C)*8i)/(11*d) - (a^2*(11*B + 10
*C)*8i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^5) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp
(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*a^2*64i)/(99*d) - (a^2*(B - 8*C)*8i)/(9*d) - (a^2*(5*B + 2*C
)*8i)/(9*d) + (a^2*(5*B + 9*C)*16i)/(9*d)) + (B*a^2*8i)/(9*d) + (C*a^2*64i)/(9*d) + (a^2*(B + 2*C)*40i)/(9*d)
- (a^2*(B + C)*80i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(- c*1i - d*x*
1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(5*B + 2*C)*8i)/(5*d) + (a^2*(44*B - 31*C)*16i)
/(1155*d)) - (B*a^2*8i)/(5*d) + (a^2*(4*B + 5*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) +
1)^2) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(803*B + 710*C)*
16i)/(3465*d*(exp(c*1i + d*x*1i) + 1))

________________________________________________________________________________________