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

Optimal. Leaf size=273 \[ \frac {2 a^3 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (10439 A+8368 C) \tan (c+d x)}{6435 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d}-\frac {4 a^2 (10439 A+8368 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{45045 d}+\frac {2 a (10439 A+8368 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac {10 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \]

[Out]

2/15015*a*(10439*A+8368*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+10/143*a*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*
tan(d*x+c)/d+2/13*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/6435*a^3*(10439*A+8368*C)*tan(d*x+c)/d/
(a+a*sec(d*x+c))^(1/2)+2/9009*a^3*(2717*A+2224*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/45045*a^2
*(10439*A+8368*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/1287*a^2*(143*A+136*C)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(
1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.86, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4089, 4018, 4016, 3800, 4001, 3792} \[ \frac {2 a^3 (2717 A+2224 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (143 A+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d}+\frac {2 a^3 (10439 A+8368 C) \tan (c+d x)}{6435 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^2 (10439 A+8368 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{45045 d}+\frac {2 a (10439 A+8368 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac {10 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(10439*A + 8368*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(2717*A + 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 + 8368*C)*Sqrt[a + a*Sec[c + d*x]]*Tan
[c + d*x])/(45045*d) + (2*a^2*(143*A + 136*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(1287*d) +
(2*a*(10439*A + 8368*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(15015*d) + (10*a*C*Sec[c + d*x]^3*(a + a*Se
c[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 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 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 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 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 ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+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 {5}{2} a C \sec (c+d x)\right ) \, dx}{13 a}\\ &=\frac {10 a 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+96 C)+\frac {1}{4} a^2 (143 A+136 C) \sec (c+d x)\right ) \, dx}{143 a}\\ &=\frac {2 a^2 (143 A+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {10 a 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 {15}{8} a^3 (143 A+112 C)+\frac {1}{8} a^3 (2717 A+2224 C) \sec (c+d x)\right ) \, dx}{1287 a}\\ &=\frac {2 a^3 (2717 A+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+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {10 a 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+8368 C)\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3003}\\ &=\frac {2 a^3 (2717 A+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+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+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {10 a 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+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+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+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+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+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {10 a 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+8368 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{6435}\\ &=\frac {2 a^3 (10439 A+8368 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2717 A+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+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+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+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {10 a 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*}

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Mathematica [A]  time = 2.01, size = 169, normalized size = 0.62 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \sqrt {a (\sec (c+d x)+1)} (1120 (286 A+347 C) \cos (c+d x)+14 (32747 A+30334 C) \cos (2 (c+d x))+141570 A \cos (3 (c+d x))+156585 A \cos (4 (c+d x))+20878 A \cos (5 (c+d x))+20878 A \cos (6 (c+d x))+322751 A+125520 C \cos (3 (c+d x))+125520 C \cos (4 (c+d x))+16736 C \cos (5 (c+d x))+16736 C \cos (6 (c+d x))+343612 C)}{180180 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(322751*A + 343612*C + 1120*(286*A + 347*C)*Cos[c + d*x] + 14*(32747*A + 30334*C)*Cos[2*(c + d*x)] + 1415
70*A*Cos[3*(c + d*x)] + 125520*C*Cos[3*(c + d*x)] + 156585*A*Cos[4*(c + d*x)] + 125520*C*Cos[4*(c + d*x)] + 20
878*A*Cos[5*(c + d*x)] + 16736*C*Cos[5*(c + d*x)] + 20878*A*Cos[6*(c + d*x)] + 16736*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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fricas [A]  time = 0.44, size = 171, normalized size = 0.63 \[ \frac {2 \, {\left (8 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \, {\left (10439 \, A + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (1859 \, A + 2092 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (143 \, A + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 11970 \, C a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \]

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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/45045*(8*(10439*A + 8368*C)*a^2*cos(d*x + c)^6 + 4*(10439*A + 8368*C)*a^2*cos(d*x + c)^5 + 3*(10439*A + 8368
*C)*a^2*cos(d*x + c)^4 + 10*(1859*A + 2092*C)*a^2*cos(d*x + c)^3 + 35*(143*A + 523*C)*a^2*cos(d*x + c)^2 + 119
70*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)

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giac [A]  time = 2.29, size = 351, normalized size = 1.29 \[ \frac {8 \, {\left ({\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 143 \, \sqrt {2} {\left (1859 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1716 \, \sqrt {2} {\left (228 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 181 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6006 \, \sqrt {2} {\left (57 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60060 \, \sqrt {2} {\left (3 \, A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45045 \, \sqrt {2} {\left (A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/45045*(((((4*(2*sqrt(2)*(1859*A*a^9*sgn(cos(d*x + c)) + 1483*C*a^9*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2
- 13*sqrt(2)*(1859*A*a^9*sgn(cos(d*x + c)) + 1483*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 143*sqrt
(2)*(1859*A*a^9*sgn(cos(d*x + c)) + 1483*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 1716*sqrt(2)*(228*
A*a^9*sgn(cos(d*x + c)) + 181*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 6006*sqrt(2)*(57*A*a^9*sgn(co
s(d*x + c)) + 49*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 60060*sqrt(2)*(3*A*a^9*sgn(cos(d*x + c)) +
2*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 45045*sqrt(2)*(A*a^9*sgn(cos(d*x + c)) + C*a^9*sgn(cos(d
*x + c))))*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)

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maple [A]  time = 1.98, size = 176, normalized size = 0.64 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (83512 A \left (\cos ^{6}\left (d x +c \right )\right )+66944 C \left (\cos ^{6}\left (d x +c \right )\right )+41756 A \left (\cos ^{5}\left (d x +c \right )\right )+33472 C \left (\cos ^{5}\left (d x +c \right )\right )+31317 A \left (\cos ^{4}\left (d x +c \right )\right )+25104 C \left (\cos ^{4}\left (d x +c \right )\right )+18590 A \left (\cos ^{3}\left (d x +c \right )\right )+20920 C \left (\cos ^{3}\left (d x +c \right )\right )+5005 A \left (\cos ^{2}\left (d x +c \right )\right )+18305 C \left (\cos ^{2}\left (d x +c \right )\right )+11970 C \cos \left (d x +c \right )+3465 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{45045 d \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )} \]

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+C*sec(d*x+c)^2),x)

[Out]

-2/45045/d*(-1+cos(d*x+c))*(83512*A*cos(d*x+c)^6+66944*C*cos(d*x+c)^6+41756*A*cos(d*x+c)^5+33472*C*cos(d*x+c)^
5+31317*A*cos(d*x+c)^4+25104*C*cos(d*x+c)^4+18590*A*cos(d*x+c)^3+20920*C*cos(d*x+c)^3+5005*A*cos(d*x+c)^2+1830
5*C*cos(d*x+c)^2+11970*C*cos(d*x+c)+3465*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^6/sin(d*x+c)*a^2

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

[Out]

Timed out

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mupad [B]  time = 12.78, size = 1039, normalized size = 3.81 \[ \text {result too large to display} \]

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))^(5/2))/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*8i)/(3*d) - (a^2*exp(c*1i + d*x*1i)*(10
439*A + 8368*C)*8i)/(45045*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)*(exp(c*1i + d*x*1i)*((A*a^2*8i)/d + (a^2*(286*A - 523*C)*32i)/(15015*d)
) - (A*a^2*8i)/(5*d) + (a^2*(2*A + C)*32i)/(5*d)))/((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)*(exp(c*1i + d*x*1i)*((A*a^2*48i)/(13*d) - (a^2*(3*
A + C)*32i)/(13*d) + (a^2*(13*A + 20*C)*16i)/(13*d) - (a^2*(A + C)*160i)/(13*d)) - (A*a^2*48i)/(13*d) + (a^2*(
3*A + C)*32i)/(13*d) - (a^2*(13*A + 20*C)*16i)/(13*d) + (a^2*(A + C)*160i)/(13*d)))/((exp(c*1i + d*x*1i) + 1)*
(exp(c*2i + d*x*2i) + 1)^6) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1
i)*((A*a^2*40i)/(11*d) - (a^2*(A - 16*C)*8i)/(11*d) + (C*a^2*128i)/(143*d) - (a^2*(3*A + 4*C)*40i)/(11*d) + (a
^2*(11*A + 20*C)*8i)/(11*d)) - (A*a^2*8i)/(11*d) - (C*a^2*128i)/(11*d) + (a^2*(11*A + 4*C)*8i)/(11*d) - (a^2*(
5*A + 12*C)*24i)/(11*d) + (a^2*(5*A - 16*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)*((a^2*(A + 4*C)*40i)/(7*
d) - (A*a^2*40i)/(7*d) + (a^2*(143*A + 811*C)*32i)/(9009*d)) + (A*a^2*8i)/(7*d) - (a^2*(A - 7*C)*32i)/(7*d) -
(a^2*(9*A + 4*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)*((A*a^2*8i)/(9*d) - exp(c*1i + d*x*1i)*((A*a^2*16i)/(3*d) - (a^2*(A + 2*
C)*80i)/(9*d) + (C*a^2*128i)/(429*d)) + (C*a^2*128i)/(3*d) - (a^2*(5*A + 2*C)*16i)/(9*d) + (a^2*(5*A + 32*C)*8
i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - (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)*(10439*A + 8368*C)*16i)/(45045*d*(exp(c*1i + d*x*1i) + 1))

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

[Out]

Timed out

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