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

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

[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

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Rubi [A]  time = 0.76, 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 = {4072, 4018, 4016, 3800, 4001, 3792} \[ \frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}+\frac {2 a^3 (803 B+710 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^2 (803 B+710 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a (803 B+710 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

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 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 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 ^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*}

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Mathematica [B]  time = 6.17, size = 487, normalized size = 2.05 \[ \frac {2 B \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{9 d (\sec (c+d x)+1)^2}+\frac {38 B \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{63 d (\sec (c+d x)+1)^3}+\frac {146 B \tan (c+d x) \sec ^2(c+d x) (a (\sec (c+d x)+1))^{5/2}}{105 d (\sec (c+d x)+1)^3}+\frac {584 B \tan (c+d x) \sec (c+d x) (a (\sec (c+d x)+1))^{5/2}}{315 d (\sec (c+d x)+1)^3}+\frac {1168 B \tan (c+d x) (a (\sec (c+d x)+1))^{5/2}}{315 d (\sec (c+d x)+1)^3}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) (a (\sec (c+d x)+1))^{5/2}}{11 d (\sec (c+d x)+1)^2}+\frac {46 C \tan (c+d x) \sec ^4(c+d x) (a (\sec (c+d x)+1))^{5/2}}{99 d (\sec (c+d x)+1)^3}+\frac {710 C \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3}+\frac {284 C \tan (c+d x) \sec ^2(c+d x) (a (\sec (c+d x)+1))^{5/2}}{231 d (\sec (c+d x)+1)^3}+\frac {1136 C \tan (c+d x) \sec (c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3}+\frac {2272 C \tan (c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3} \]

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]

(1168*B*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(315*d*(1 + Sec[c + d*x])^3) + (2272*C*(a*(1 + Sec[c + d*x]
))^(5/2)*Tan[c + d*x])/(693*d*(1 + Sec[c + d*x])^3) + (584*B*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c +
d*x])/(315*d*(1 + Sec[c + d*x])^3) + (1136*C*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(693*d*(
1 + Sec[c + d*x])^3) + (146*B*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(105*d*(1 + Sec[c + d*
x])^3) + (284*C*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(231*d*(1 + Sec[c + d*x])^3) + (38*B
*Sec[c + d*x]^3*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(63*d*(1 + Sec[c + d*x])^3) + (710*C*Sec[c + d*x]^3
*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(693*d*(1 + Sec[c + d*x])^3) + (46*C*Sec[c + d*x]^4*(a*(1 + Sec[c
+ d*x]))^(5/2)*Tan[c + d*x])/(99*d*(1 + Sec[c + d*x])^3) + (2*B*Sec[c + d*x]^3*(a*(1 + Sec[c + d*x]))^(5/2)*Ta
n[c + d*x])/(9*d*(1 + Sec[c + d*x])^2) + (2*C*Sec[c + d*x]^4*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(11*d*
(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.49, size = 157, normalized size = 0.66 \[ \frac {2 \, {\left (8 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

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)

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giac [A]  time = 3.99, size = 306, normalized size = 1.29 \[ \frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 231 \, \sqrt {2} {\left (69 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 65 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (9 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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)

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maple [A]  time = 1.78, size = 163, normalized size = 0.69 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (6424 B \left (\cos ^{5}\left (d x +c \right )\right )+5680 C \left (\cos ^{5}\left (d x +c \right )\right )+3212 B \left (\cos ^{4}\left (d x +c \right )\right )+2840 C \left (\cos ^{4}\left (d x +c \right )\right )+2409 B \left (\cos ^{3}\left (d x +c \right )\right )+2130 C \left (\cos ^{3}\left (d x +c \right )\right )+1430 B \left (\cos ^{2}\left (d x +c \right )\right )+1775 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

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)

[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

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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)^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

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mupad [B]  time = 15.40, size = 855, normalized size = 3.61 \[ \text {result too large to display} \]

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))

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

[Out]

Timed out

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