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

Optimal. Leaf size=239 \[ \frac {2 a (99 A+88 B+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (99 A+88 B+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+88 B+80 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d} \]

[Out]

4/1155*(99*A+88*B+80*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/495*a*(99*A+88*B+80*C)*tan(d*x+c)/d/(a+a*sec(d
*x+c))^(1/2)+2/693*a*(99*A+88*B+80*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a*(11*B+C)*sec(d*x
+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-8/3465*(99*A+88*B+80*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/11*C*s
ec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.56, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4088, 4016, 3803, 3800, 4001, 3792} \[ \frac {2 a (99 A+88 B+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (99 A+88 B+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac {8 (99 A+88 B+80 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(99*A + 88*B + 80*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(99*A + 88*B + 80*C)*Sec[c + d
*x]^3*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt
[a + a*Sec[c + d*x]]) - (8*(99*A + 88*B + 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 + 88*B + 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 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]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+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 (11 B+C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 a (11 B+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+88 B+80 C) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+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+88 B+80 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+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+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {(4 (99 A+88 B+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+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (99 A+88 B+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+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac {1}{495} (2 (99 A+88 B+80 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (99 A+88 B+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+88 B+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 = 1.72, size = 185, normalized size = 0.77 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (\sec (c+d x)+1)} ((2871 A+3322 B+3020 C) \cos (c+d x)+13 (99 A+88 B+80 C) \cos (2 (c+d x))+1287 A \cos (3 (c+d x))+198 A \cos (4 (c+d x))+198 A \cos (5 (c+d x))+1089 A+1144 B \cos (3 (c+d x))+176 B \cos (4 (c+d x))+176 B \cos (5 (c+d x))+968 B+1040 C \cos (3 (c+d x))+160 C \cos (4 (c+d x))+160 C \cos (5 (c+d x))+1510 C)}{3465 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1089*A + 968*B + 1510*C + (2871*A + 3322*B + 3020*C)*Cos[c + d*x] + 13*(99*A + 88*B + 80*C)*Cos[2*(c + d*x)]
+ 1287*A*Cos[3*(c + d*x)] + 1144*B*Cos[3*(c + d*x)] + 1040*C*Cos[3*(c + d*x)] + 198*A*Cos[4*(c + d*x)] + 176*
B*Cos[4*(c + d*x)] + 160*C*Cos[4*(c + d*x)] + 198*A*Cos[5*(c + d*x)] + 176*B*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 = 0.43, size = 151, normalized size = 0.63 \[ \frac {2 \, {\left (16 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{5} + 8 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+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*(16*(99*A + 88*B + 80*C)*cos(d*x + c)^5 + 8*(99*A + 88*B + 80*C)*cos(d*x + c)^4 + 6*(99*A + 88*B + 80*C
)*cos(d*x + c)^3 + 5*(99*A + 88*B + 80*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 = 1.62, size = 410, normalized size = 1.72 \[ -\frac {2 \, {\left (3465 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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 (10395 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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 (15246 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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 (14058 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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 (6633 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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 (891 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 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*(A+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)*A*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^6*sgn(
cos(d*x + c)) - (10395*sqrt(2)*A*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)) - (15246*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 14322*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 1617
0*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (14058*sqrt(2)*A*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)) - (6633*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 4741*sqrt(2)*B*a^6*sgn(c
os(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)) + 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*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 2.04, size = 204, normalized size = 0.85 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (1584 A \left (\cos ^{5}\left (d x +c \right )\right )+1408 B \left (\cos ^{5}\left (d x +c \right )\right )+1280 C \left (\cos ^{5}\left (d x +c \right )\right )+792 A \left (\cos ^{4}\left (d x +c \right )\right )+704 B \left (\cos ^{4}\left (d x +c \right )\right )+640 C \left (\cos ^{4}\left (d x +c \right )\right )+594 A \left (\cos ^{3}\left (d x +c \right )\right )+528 B \left (\cos ^{3}\left (d x +c \right )\right )+480 C \left (\cos ^{3}\left (d x +c \right )\right )+495 A \left (\cos ^{2}\left (d x +c \right )\right )+440 B \left (\cos ^{2}\left (d x +c \right )\right )+400 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+350 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+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))*(1584*A*cos(d*x+c)^5+1408*B*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+792*A*cos(d*x+c)^4+704*
B*cos(d*x+c)^4+640*C*cos(d*x+c)^4+594*A*cos(d*x+c)^3+528*B*cos(d*x+c)^3+480*C*cos(d*x+c)^3+495*A*cos(d*x+c)^2+
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*(A+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 = 12.22, size = 724, normalized size = 3.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((exp(c*1i + d*x*1i)*((A*16i)/(5*d) - ((352*B - 528*A + 320*C)*1i)/(1155*d)) + ((3696*A + 7392*B)*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)*((C*64i)/(9*d) + exp(c*1i + d*x*1i)*
((C*256i)/(33*d) - (A*16i)/(9*d) + ((176*A + 352*B + 704*C)*1i)/(99*d)) - ((176*A + 352*B)*1i)/(99*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)*(((1584*A + 3168*B)*1i)/(693*d) - exp(c*1i + d*x*1i)*(((352*B + 896*C)*1i)/(6
93*d) - (A*16i)/(7*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) - ((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)/(11*d) - ((32*A + 32*B + 64*C)*1i)/(11*d) + ((16*A + 32*B)*1i)/(11*d)) + (A*16i)/(11*d) - (
(32*A + 32*B + 64*C)*1i)/(11*d) + ((16*A + 32*B)*1i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) +
1)^5) - (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 + 2816*B + 2
560*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 + 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 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+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))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)

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