3.356 \(\int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=230 \[ \frac {2 a (11 B+10 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a (11 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {32 (11 B+10 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {64 (11 B+10 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a C \tan (c+d x) \sec ^5(c+d x)}{11 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

32/1155*(11*B+10*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+32/495*a*(11*B+10*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(
1/2)+16/693*a*(11*B+10*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a*(11*B+10*C)*sec(d*x+c)^4*tan
(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/11*a*C*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-64/3465*(11*B+10*C)
*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.49, antiderivative size = 230, 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 = {4072, 4016, 3803, 3800, 4001, 3792} \[ \frac {2 a (11 B+10 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a (11 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {32 (11 B+10 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {64 (11 B+10 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a C \tan (c+d x) \sec ^5(c+d x)}{11 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a*(11*B + 10*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a*(11*B + 10*C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + 10*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Se
c[c + d*x]]) + (2*a*C*Sec[c + d*x]^5*Tan[c + d*x])/(11*d*Sqrt[a + a*Sec[c + d*x]]) - (64*(11*B + 10*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (32*(11*B + 10*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*a*
d)

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]

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 3803

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

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 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 4072

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 ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^5(c+d x) \sqrt {a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{11} (11 B+10 C) \int \sec ^5(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{99} (8 (11 B+10 C)) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{231} (16 (11 B+10 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}+\frac {32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {(32 (11 B+10 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 a}\\ &=\frac {16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}-\frac {64 (11 B+10 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {1}{495} (16 (11 B+10 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt {a+a \sec (c+d x)}}-\frac {64 (11 B+10 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}\\ \end {align*}

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Mathematica [A]  time = 6.08, size = 283, normalized size = 1.23 \[ \frac {2 B \tan (c+d x) \left (35 (1-\sec (c+d x))^{9/2}-180 (1-\sec (c+d x))^{7/2}+378 (1-\sec (c+d x))^{5/2}-420 (1-\sec (c+d x))^{3/2}+315 \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (\sec (c+d x)+1)}}{315 d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)}+\frac {2 C \tan (c+d x) \left (-63 (1-\sec (c+d x))^{11/2}+385 (1-\sec (c+d x))^{9/2}-990 (1-\sec (c+d x))^{7/2}+1386 (1-\sec (c+d x))^{5/2}-1155 (1-\sec (c+d x))^{3/2}+693 \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (\sec (c+d x)+1)}}{693 d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(315*Sqrt[1 - Sec[c + d*x]] - 420*(1 - Sec[c + d*x])^(3/2) + 378*(1 - Sec[c + d*x])^(5/2) - 180*(1 - Sec[
c + d*x])^(7/2) + 35*(1 - Sec[c + d*x])^(9/2))*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(315*d*Sqrt[1 - Sec[c
+ d*x]]*(1 + Sec[c + d*x])) + (2*C*(693*Sqrt[1 - Sec[c + d*x]] - 1155*(1 - Sec[c + d*x])^(3/2) + 1386*(1 - Sec
[c + d*x])^(5/2) - 990*(1 - Sec[c + d*x])^(7/2) + 385*(1 - Sec[c + d*x])^(9/2) - 63*(1 - Sec[c + d*x])^(11/2))
*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(693*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x]))

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fricas [A]  time = 0.48, size = 139, normalized size = 0.60 \[ \frac {2 \, {\left (128 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 64 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + 48 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3} + 40 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(128*(11*B + 10*C)*cos(d*x + c)^5 + 64*(11*B + 10*C)*cos(d*x + c)^4 + 48*(11*B + 10*C)*cos(d*x + c)^3 +
40*(11*B + 10*C)*cos(d*x + c)^2 + 35*(11*B + 10*C)*cos(d*x + c) + 315*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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giac [A]  time = 3.60, size = 314, normalized size = 1.37 \[ -\frac {2 \, {\left (3465 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (8085 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5775 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (14322 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 16170 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (13266 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8910 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (4741 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5885 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1177 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 755 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/3465*(3465*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (8085*sqrt(2)*B*a^6*sgn
(cos(d*x + c)) + 5775*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (14322*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 16170*sqrt(2)
*C*a^6*sgn(cos(d*x + c)) - (13266*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 8910*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (47
41*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 5885*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (1177*sqrt(2)*B*a^6*sgn(cos(d*x +
c)) + 755*sqrt(2)*C*a^6*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)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 2.43, size = 160, normalized size = 0.70 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (1408 B \left (\cos ^{5}\left (d x +c \right )\right )+1280 C \left (\cos ^{5}\left (d x +c \right )\right )+704 B \left (\cos ^{4}\left (d x +c \right )\right )+640 C \left (\cos ^{4}\left (d x +c \right )\right )+528 B \left (\cos ^{3}\left (d x +c \right )\right )+480 C \left (\cos ^{3}\left (d x +c \right )\right )+440 B \left (\cos ^{2}\left (d x +c \right )\right )+400 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+350 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(1408*B*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+704*B*cos(d*x+c)^4+640*C*cos(d*x+c)^4+528*B
*cos(d*x+c)^3+480*C*cos(d*x+c)^3+440*B*cos(d*x+c)^2+400*C*cos(d*x+c)^2+385*B*cos(d*x+c)+350*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)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 12.33, size = 626, normalized size = 2.72 \[ -\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\,32{}\mathrm {i}}{9\,d}+\frac {C\,128{}\mathrm {i}}{9\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {C\,256{}\mathrm {i}}{33\,d}+\frac {\left (352\,B+704\,C\right )\,1{}\mathrm {i}}{99\,d}\right )\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 (\frac {B\,32{}\mathrm {i}}{11\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,32{}\mathrm {i}}{11\,d}-\frac {\left (32\,B+64\,C\right )\,1{}\mathrm {i}}{11\,d}\right )-\frac {\left (32\,B+64\,C\right )\,1{}\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 {\left (\frac {B\,32{}\mathrm {i}}{5\,d}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (352\,B+320\,C\right )\,1{}\mathrm {i}}{1155\,d}\right )\,\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}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\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 {\left (352\,B+896\,C\right )\,1{}\mathrm {i}}{693\,d}+\frac {\left (3168\,B+6336\,C\right )\,1{}\mathrm {i}}{693\,d}\right )+\frac {B\,32{}\mathrm {i}}{7\,d}+\frac {\left (3168\,B-6336\,C\right )\,1{}\mathrm {i}}{693\,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 {{\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 (2816\,B+2560\,C\right )\,1{}\mathrm {i}}{3465\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {{\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 (1408\,B+1280\,C\right )\,1{}\mathrm {i}}{3465\,d\,\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 )} \]

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))^(1/2))/cos(c + d*x)^4,x)

[Out]

(((B*32i)/(5*d) - (exp(c*1i + d*x*1i)*(352*B + 320*C)*1i)/(1155*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((a + a/(exp(- c*1i - d*x*1i)/2 +
exp(c*1i + d*x*1i)/2))^(1/2)*((B*32i)/(11*d) + exp(c*1i + d*x*1i)*((B*32i)/(11*d) - ((32*B + 64*C)*1i)/(11*d)
) - ((32*B + 64*C)*1i)/(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)*((C*128i)/(9*d) - (B*32i)/(9*d) + exp(c*1i + d*x*1i)*((C*256i)/(33*d)
+ ((352*B + 704*C)*1i)/(99*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)*((B*32i)/(7*d) - exp(c*1i + d*x*1i)*(((352*B + 896*C)*1i)/(693*d) +
((3168*B + 6336*C)*1i)/(693*d)) + ((3168*B - 6336*C)*1i)/(693*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*
2i) + 1)^3) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(2816*B + 2560
*C)*1i)/(3465*d*(exp(c*1i + d*x*1i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*
x*1i)/2))^(1/2)*(1408*B + 1280*C)*1i)/(3465*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(B + C*sec(c + d*x))*sec(c + d*x)**5, x)

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