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

Optimal. Leaf size=243 \[ \frac {2 a^2 (99 A+110 B+84 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (429 A+374 B+336 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (429 A+374 B+336 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac {4 a (429 A+374 B+336 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a (11 B+3 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

[Out]

2/1155*(429*A+374*B+336*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/11*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(
d*x+c)/d+2/495*a^2*(429*A+374*B+336*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^2*(99*A+110*B+84*C)*sec(d*x
+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/3465*a*(429*A+374*B+336*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/9
9*a*(11*B+3*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 = 0.69, antiderivative size = 243, 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, 4018, 4016, 3800, 4001, 3792} \[ \frac {2 a^2 (99 A+110 B+84 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (429 A+374 B+336 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (429 A+374 B+336 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac {4 a (429 A+374 B+336 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a (11 B+3 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(429*A + 374*B + 336*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(99*A + 110*B + 84*C)*S
ec[c + d*x]^3*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(429*A + 374*B + 336*C)*Sqrt[a + a*Sec[c +
d*x]]*Tan[c + d*x])/(3465*d) + (2*a*(11*B + 3*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(99*d)
+ (2*(429*A + 374*B + 336*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*d) + (2*C*Sec[c + d*x]^3*(a + a*S
ec[c + d*x])^(3/2)*Tan[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 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 ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (11 A+6 C)+\frac {1}{2} a (11 B+3 C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 a (11 B+3 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {4 \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (33 A+22 B+24 C)+\frac {1}{4} a^2 (99 A+110 B+84 C) \sec (c+d x)\right ) \, dx}{99 a}\\ &=\frac {2 a^2 (99 A+110 B+84 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+3 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{231} (a (429 A+374 B+336 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (99 A+110 B+84 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+3 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 (429 A+374 B+336 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {(2 (429 A+374 B+336 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^2 (99 A+110 B+84 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (429 A+374 B+336 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (11 B+3 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 (429 A+374 B+336 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{495} (a (429 A+374 B+336 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (429 A+374 B+336 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (99 A+110 B+84 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (429 A+374 B+336 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (11 B+3 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 (429 A+374 B+336 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 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 [A]  time = 2.05, size = 185, normalized size = 0.76 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (\sec (c+d x)+1)} ((12441 A+12386 B+12684 C) \cos (c+d x)+(4422 A+4862 B+4368 C) \cos (2 (c+d x))+5577 A \cos (3 (c+d x))+858 A \cos (4 (c+d x))+858 A \cos (5 (c+d x))+3564 A+4862 B \cos (3 (c+d x))+748 B \cos (4 (c+d x))+748 B \cos (5 (c+d x))+4114 B+4368 C \cos (3 (c+d x))+672 C \cos (4 (c+d x))+672 C \cos (5 (c+d x))+4956 C)}{6930 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3564*A + 4114*B + 4956*C + (12441*A + 12386*B + 12684*C)*Cos[c + d*x] + (4422*A + 4862*B + 4368*C)*Cos[2*(
c + d*x)] + 5577*A*Cos[3*(c + d*x)] + 4862*B*Cos[3*(c + d*x)] + 4368*C*Cos[3*(c + d*x)] + 858*A*Cos[4*(c + d*x
)] + 748*B*Cos[4*(c + d*x)] + 672*C*Cos[4*(c + d*x)] + 858*A*Cos[5*(c + d*x)] + 748*B*Cos[5*(c + d*x)] + 672*C
*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(6930*d)

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fricas [A]  time = 0.43, size = 157, normalized size = 0.65 \[ \frac {2 \, {\left (8 \, {\left (429 \, A + 374 \, B + 336 \, C\right )} a \cos \left (d x + c\right )^{5} + 4 \, {\left (429 \, A + 374 \, B + 336 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (429 \, A + 374 \, B + 336 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 21 \, C\right )} a \cos \left (d x + c\right ) + 315 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(8*(429*A + 374*B + 336*C)*a*cos(d*x + c)^5 + 4*(429*A + 374*B + 336*C)*a*cos(d*x + c)^4 + 3*(429*A + 3
74*B + 336*C)*a*cos(d*x + c)^3 + 5*(99*A + 187*B + 168*C)*a*cos(d*x + c)^2 + 35*(11*B + 21*C)*a*cos(d*x + c) +
315*C*a)*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 = 2.28, size = 410, normalized size = 1.69 \[ -\frac {4 \, {\left (3465 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt {2} C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (11550 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 9240 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 6930 \, \sqrt {2} C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (17094 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 14784 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15246 \, \sqrt {2} C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (14652 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13662 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 11088 \, \sqrt {2} C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (6897 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5687 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5313 \, \sqrt {2} C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 2 \, {\left (627 \, \sqrt {2} A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 517 \, \sqrt {2} B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 483 \, \sqrt {2} C a^{7} \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)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-4/3465*(3465*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 3465*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^7*sgn(
cos(d*x + c)) - (11550*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 9240*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 6930*sqrt(2)*C
*a^7*sgn(cos(d*x + c)) - (17094*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 14784*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 1524
6*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (14652*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 13662*sqrt(2)*B*a^7*sgn(cos(d*x +
c)) + 11088*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (6897*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 5687*sqrt(2)*B*a^7*sgn(
cos(d*x + c)) + 5313*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - 2*(627*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 517*sqrt(2)*B*
a^7*sgn(cos(d*x + c)) + 483*sqrt(2)*C*a^7*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*t
an(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 = 1.99, size = 205, normalized size = 0.84 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3432 A \left (\cos ^{5}\left (d x +c \right )\right )+2992 B \left (\cos ^{5}\left (d x +c \right )\right )+2688 C \left (\cos ^{5}\left (d x +c \right )\right )+1716 A \left (\cos ^{4}\left (d x +c \right )\right )+1496 B \left (\cos ^{4}\left (d x +c \right )\right )+1344 C \left (\cos ^{4}\left (d x +c \right )\right )+1287 A \left (\cos ^{3}\left (d x +c \right )\right )+1122 B \left (\cos ^{3}\left (d x +c \right )\right )+1008 C \left (\cos ^{3}\left (d x +c \right )\right )+495 A \left (\cos ^{2}\left (d x +c \right )\right )+935 B \left (\cos ^{2}\left (d x +c \right )\right )+840 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+735 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(3432*A*cos(d*x+c)^5+2992*B*cos(d*x+c)^5+2688*C*cos(d*x+c)^5+1716*A*cos(d*x+c)^4+149
6*B*cos(d*x+c)^4+1344*C*cos(d*x+c)^4+1287*A*cos(d*x+c)^3+1122*B*cos(d*x+c)^3+1008*C*cos(d*x+c)^3+495*A*cos(d*x
+c)^2+935*B*cos(d*x+c)^2+840*C*cos(d*x+c)^2+385*B*cos(d*x+c)+735*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

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

[Out]

Timed out

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mupad [B]  time = 13.53, size = 852, normalized size = 3.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^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)*(exp(c*1i + d*x*1i)*((a*(11*B + 39*C)*32i)/(693
*d) - (a*(3*A + 2*B)*8i)/(7*d) + (a*(A + 4*B + 12*C)*8i)/(7*d)) + (A*a*8i)/(7*d) - (a*(3*A + 6*B + 4*C)*8i)/(7
*d) - (a*(B - C)*32i)/(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)*((a*(3*A + 2*B)*8i)/(11*d) - (A*a*8i)/(11*d) + (a*(
5*A + 6*B + 4*C)*8i)/(11*d) - (a*(7*A + 8*B + 12*C)*8i)/(11*d)) + (a*(3*A + 2*B)*8i)/(11*d) - (A*a*8i)/(11*d)
+ (a*(5*A + 6*B + 4*C)*8i)/(11*d) - (a*(7*A + 8*B + 12*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*(3*A +
2*B)*8i)/(5*d) + (a*(33*A + 11*B - 42*C)*16i)/(1155*d)) + (a*(A + 3*B + 2*C)*16i)/(5*d) - (A*a*8i)/(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)*((A*a*8i)/(9*d) - exp(c*1i + d*x*1i)*((a*(3*A + 2*B)*8i)/(9*d) + (a*(A - 8*C)*8i)/(9*d) - (C*a*64i)/(99
*d) - (a*(2*A + 3*B + 6*C)*16i)/(9*d)) + (C*a*64i)/(9*d) - (a*(2*A + 3*B + 2*C)*16i)/(9*d) + (a*(3*A + 2*B + 8
*C)*8i)/(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)*((A*a*8i)/(3*d) - (a*exp(c*1i + d*x*1i)*(429*A + 374*B + 336*C)*8i)/(3465*d)))/((exp(
c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1
i + d*x*1i)/2))^(1/2)*(429*A + 374*B + 336*C)*16i)/(3465*d*(exp(c*1i + d*x*1i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3, x)

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