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

Optimal. Leaf size=223 \[ \frac {2 a (99 A+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (99 A+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+80 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {4 a (99 A+80 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

4/1155*(99*A+80*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/495*a*(99*A+80*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/
2)+2/693*a*(99*A+80*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a*C*sec(d*x+c)^4*tan(d*x+c)/d/(a+
a*sec(d*x+c))^(1/2)-8/3465*(99*A+80*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/11*C*sec(d*x+c)^4*(a+a*sec(d*x+c)
)^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.52, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4089, 4016, 3803, 3800, 4001, 3792} \[ \frac {2 a (99 A+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (99 A+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+80 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {4 a (99 A+80 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(99*A + 80*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(99*A + 80*C)*Sec[c + d*x]^3*Tan[c +
d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) -
(8*(99*A + 80*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*C*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x
]]*Tan[c + d*x])/(11*d) + (4*(99*A + 80*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 4089

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

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (11 A+8 C)+\frac {1}{2} a C \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{99} (99 A+80 C) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (99 A+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{231} (2 (99 A+80 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (99 A+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (99 A+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {(4 (99 A+80 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 {2 a (99 A+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (99 A+80 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (99 A+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {1}{495} (2 (99 A+80 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {4 a (99 A+80 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (99 A+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (99 A+80 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (99 A+80 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 = 1.19, size = 143, normalized size = 0.64 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (\sec (c+d x)+1)} ((2871 A+3020 C) \cos (c+d x)+13 (99 A+80 C) \cos (2 (c+d x))+1287 A \cos (3 (c+d x))+198 A \cos (4 (c+d x))+198 A \cos (5 (c+d x))+1089 A+1040 C \cos (3 (c+d x))+160 C \cos (4 (c+d x))+160 C \cos (5 (c+d x))+1510 C)}{3465 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1089*A + 1510*C + (2871*A + 3020*C)*Cos[c + d*x] + 13*(99*A + 80*C)*Cos[2*(c + d*x)] + 1287*A*Cos[3*(c + d*x
)] + 1040*C*Cos[3*(c + d*x)] + 198*A*Cos[4*(c + d*x)] + 160*C*Cos[4*(c + d*x)] + 198*A*Cos[5*(c + d*x)] + 160*
C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(3465*d)

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fricas [A]  time = 0.42, size = 133, normalized size = 0.60 \[ \frac {2 \, {\left (16 \, {\left (99 \, A + 80 \, C\right )} \cos \left (d x + c\right )^{5} + 8 \, {\left (99 \, A + 80 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (99 \, A + 80 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 80 \, C\right )} \cos \left (d x + c\right )^{2} + 350 \, C \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(16*(99*A + 80*C)*cos(d*x + c)^5 + 8*(99*A + 80*C)*cos(d*x + c)^4 + 6*(99*A + 80*C)*cos(d*x + c)^3 + 5*
(99*A + 80*C)*cos(d*x + c)^2 + 350*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 = 3.29, size = 314, normalized size = 1.41 \[ -\frac {2 \, {\left (3465 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (10395 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5775 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (15246 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 16170 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (14058 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8910 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (6633 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5885 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (891 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 755 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(3465*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (10395*sqrt(2)*A*a^6*sg
n(cos(d*x + c)) + 5775*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (15246*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 16170*sqrt(2
)*C*a^6*sgn(cos(d*x + c)) - (14058*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 8910*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (6
633*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 5885*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (891*sqrt(2)*A*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 = 1.95, size = 151, normalized size = 0.68 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (1584 A \left (\cos ^{5}\left (d x +c \right )\right )+1280 C \left (\cos ^{5}\left (d x +c \right )\right )+792 A \left (\cos ^{4}\left (d x +c \right )\right )+640 C \left (\cos ^{4}\left (d x +c \right )\right )+594 A \left (\cos ^{3}\left (d x +c \right )\right )+480 C \left (\cos ^{3}\left (d x +c \right )\right )+495 A \left (\cos ^{2}\left (d x +c \right )\right )+400 C \left (\cos ^{2}\left (d x +c \right )\right )+350 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465/d*(-1+cos(d*x+c))*(1584*A*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+792*A*cos(d*x+c)^4+640*C*cos(d*x+c)^4+594*A
*cos(d*x+c)^3+480*C*cos(d*x+c)^3+495*A*cos(d*x+c)^2+400*C*cos(d*x+c)^2+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*(A+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 = 12.20, size = 636, normalized size = 2.85 \[ -\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\,16{}\mathrm {i}}{9\,d}+\frac {C\,256{}\mathrm {i}}{33\,d}+\frac {\left (176\,A+704\,C\right )\,1{}\mathrm {i}}{99\,d}\right )-\frac {A\,16{}\mathrm {i}}{9\,d}+\frac {C\,64{}\mathrm {i}}{9\,d}+\frac {\left (176\,A+704\,C\right )\,1{}\mathrm {i}}{99\,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 (\frac {A\,32{}\mathrm {i}}{11\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,32{}\mathrm {i}}{11\,d}-\frac {\left (32\,A+64\,C\right )\,1{}\mathrm {i}}{11\,d}\right )-\frac {\left (32\,A+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 {A\,16{}\mathrm {i}}{5\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,16{}\mathrm {i}}{5\,d}+\frac {\left (528\,A-320\,C\right )\,1{}\mathrm {i}}{1155\,d}\right )\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 {A\,16{}\mathrm {i}}{7\,d}-\frac {C\,7232{}\mathrm {i}}{693\,d}\right )+\frac {A\,16{}\mathrm {i}}{7\,d}-\frac {C\,64{}\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 {{\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 (3168\,A+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 (1584\,A+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(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^4,x)

[Out]

(((A*16i)/(5*d) + exp(c*1i + d*x*1i)*((A*16i)/(5*d) + ((528*A - 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)*((A*32i)/(11*d) + exp(c*1i + d*x*1i)*((A*32i)/(11*d) - ((32*A
+ 64*C)*1i)/(11*d)) - ((32*A + 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)*(exp(c*1i + d*x*1i)*((C*256i)/(33*d) - (A*16i)/(9*d
) + ((176*A + 704*C)*1i)/(99*d)) - (A*16i)/(9*d) + (C*64i)/(9*d) + ((176*A + 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)*(exp
(c*1i + d*x*1i)*((A*16i)/(7*d) - (C*7232i)/(693*d)) + (A*16i)/(7*d) - (C*64i)/(7*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)*(3168*A + 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)*(1584*A + 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 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)

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