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

Optimal. Leaf size=282 \[ \frac {2 a^3 (299 B+280 C) \tan (c+d x) \sec ^4(c+d x)}{1287 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (4615 B+4184 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (13 B+16 C) \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{143 d}-\frac {8 a^2 (4615 B+4184 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{45045 d}+\frac {4 a (4615 B+4184 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{13 d} \]

[Out]

4/15015*a*(4615*B+4184*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/13*a*C*sec(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*tan
(d*x+c)/d+4/6435*a^3*(4615*B+4184*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9009*a^3*(4615*B+4184*C)*sec(d*x+c)
^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^3*(299*B+280*C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/
2)-8/45045*a^2*(4615*B+4184*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/143*a^2*(13*B+16*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 = 0.84, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4072, 4018, 4016, 3803, 3800, 4001, 3792} \[ \frac {2 a^2 (13 B+16 C) \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{143 d}+\frac {2 a^3 (299 B+280 C) \tan (c+d x) \sec ^4(c+d x)}{1287 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (4615 B+4184 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}-\frac {8 a^2 (4615 B+4184 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{45045 d}+\frac {4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (4615 B+4184 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(4615*B + 4184*C)*Tan[c + d*x])/(6435*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(4615*B + 4184*C)*Sec[c + d*
x]^3*Tan[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(299*B + 280*C)*Sec[c + d*x]^4*Tan[c + d*x])/(12
87*d*Sqrt[a + a*Sec[c + d*x]]) - (8*a^2*(4615*B + 4184*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(45045*d) + (
2*a^2*(13*B + 16*C)*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(143*d) + (4*a*(4615*B + 4184*C)*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(15015*d) + (2*a*C*Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*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 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 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 ^3(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 ^4(c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {2}{13} \int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (13 B+8 C)+\frac {1}{2} a (13 B+16 C) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {4}{143} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (247 B+216 C)+\frac {1}{4} a^2 (299 B+280 C) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {\left (a^2 (4615 B+4184 C)\right ) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{1287}\\ &=\frac {2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {\left (2 a^2 (4615 B+4184 C)\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3003}\\ &=\frac {2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {(4 a (4615 B+4184 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 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a^2 (4615 B+4184 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}+\frac {\left (2 a^2 (4615 B+4184 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{6435}\\ &=\frac {4 a^3 (4615 B+4184 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (4615 B+4184 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (299 B+280 C) \sec ^4(c+d x) \tan (c+d x)}{1287 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a^2 (4615 B+4184 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (13 B+16 C) \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{143 d}+\frac {4 a (4615 B+4184 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a C \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{13 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 131, normalized size = 0.46 \[ \frac {2 a^3 \tan (c+d x) \left (315 (13 B+38 C) \sec ^5(c+d x)+35 (416 B+523 C) \sec ^4(c+d x)+5 (4615 B+4184 C) \sec ^3(c+d x)+6 (4615 B+4184 C) \sec ^2(c+d x)+8 (4615 B+4184 C) \sec (c+d x)+73840 B+3465 C \sec ^6(c+d x)+66944 C\right )}{45045 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(73840*B + 66944*C + 8*(4615*B + 4184*C)*Sec[c + d*x] + 6*(4615*B + 4184*C)*Sec[c + d*x]^2 + 5*(4615*B
+ 4184*C)*Sec[c + d*x]^3 + 35*(416*B + 523*C)*Sec[c + d*x]^4 + 315*(13*B + 38*C)*Sec[c + d*x]^5 + 3465*C*Sec[c
+ d*x]^6)*Tan[c + d*x])/(45045*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.50, size = 177, normalized size = 0.63 \[ \frac {2 \, {\left (16 \, {\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 8 \, {\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 6 \, {\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (416 \, B + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (13 \, B + 38 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \]

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

[Out]

2/45045*(16*(4615*B + 4184*C)*a^2*cos(d*x + c)^6 + 8*(4615*B + 4184*C)*a^2*cos(d*x + c)^5 + 6*(4615*B + 4184*C
)*a^2*cos(d*x + c)^4 + 5*(4615*B + 4184*C)*a^2*cos(d*x + c)^3 + 35*(416*B + 523*C)*a^2*cos(d*x + c)^2 + 315*(1
3*B + 38*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 = 2.39, size = 351, normalized size = 1.24 \[ \frac {8 \, {\left ({\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (1625 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13 \, \sqrt {2} {\left (1625 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 143 \, \sqrt {2} {\left (1625 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 858 \, \sqrt {2} {\left (415 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 362 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6006 \, \sqrt {2} {\left (50 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 30030 \, \sqrt {2} {\left (5 \, B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45045 \, \sqrt {2} {\left (B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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

[Out]

8/45045*(((((4*(2*sqrt(2)*(1625*B*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)*(1625*B*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)*(1625*B*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 - 858*sqrt(2)*(415*B
*a^9*sgn(cos(d*x + c)) + 362*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 6006*sqrt(2)*(50*B*a^9*sgn(cos
(d*x + c)) + 49*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 30030*sqrt(2)*(5*B*a^9*sgn(cos(d*x + c)) +
4*C*a^9*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 45045*sqrt(2)*(B*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 = 1.82, size = 185, normalized size = 0.66 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (73840 B \left (\cos ^{6}\left (d x +c \right )\right )+66944 C \left (\cos ^{6}\left (d x +c \right )\right )+36920 B \left (\cos ^{5}\left (d x +c \right )\right )+33472 C \left (\cos ^{5}\left (d x +c \right )\right )+27690 B \left (\cos ^{4}\left (d x +c \right )\right )+25104 C \left (\cos ^{4}\left (d x +c \right )\right )+23075 B \left (\cos ^{3}\left (d x +c \right )\right )+20920 C \left (\cos ^{3}\left (d x +c \right )\right )+14560 B \left (\cos ^{2}\left (d x +c \right )\right )+18305 C \left (\cos ^{2}\left (d x +c \right )\right )+4095 B \cos \left (d x +c \right )+11970 C \cos \left (d x +c \right )+3465 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{45045 d \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )} \]

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

[Out]

-2/45045/d*(-1+cos(d*x+c))*(73840*B*cos(d*x+c)^6+66944*C*cos(d*x+c)^6+36920*B*cos(d*x+c)^5+33472*C*cos(d*x+c)^
5+27690*B*cos(d*x+c)^4+25104*C*cos(d*x+c)^4+23075*B*cos(d*x+c)^3+20920*C*cos(d*x+c)^3+14560*B*cos(d*x+c)^2+183
05*C*cos(d*x+c)^2+4095*B*cos(d*x+c)+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)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 14.05, size = 988, normalized size = 3.50 \[ \text {result too large to display} \]

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)^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^2*(4*B + 5*C)*32i)/(7*d
) - (B*a^2*16i)/(7*d) + (a^2*(650*B + 811*C)*32i)/(9009*d)) + (C*a^2*32i)/d - (a^2*(5*B + 2*C)*16i)/(7*d)))/((
exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((exp(c*1i + d*x*1i)*((B*a^2*16i)/(5*d) - (a^2*(403*B +
1046*C)*16i)/(15015*d)) + (a^2*(5*B + 2*C)*16i)/(5*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)*(exp(c*1i + d*x*1i)*((B*a^2*16i)/(3*d) + (a^2*(9*B + 10*C)*16i)/(9*d) + (a^2*(13*B - 6*C)*64i
)/(1287*d)) + (a^2*(B + 6*C)*64i)/(9*d) - (a^2*(5*B + 2*C)*16i)/(9*d) + (a^2*(5*B + 16*C)*16i)/(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)*((B*a^2*16i)/(13*d) + (a^2*(3*B + 4*C)*80i)/(13*d) - (a^2*(5*B + 2*C)*16i)/(13*d) - (a^2
*(11*B + 10*C)*16i)/(13*d)) - (B*a^2*16i)/(13*d) - (a^2*(3*B + 4*C)*80i)/(13*d) + (a^2*(5*B + 2*C)*16i)/(13*d)
+ (a^2*(11*B + 10*C)*16i)/(13*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^6) - ((a + a/(exp(- c*1
i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((B*a^2*16i)/(11*d) + (C*a^2*1792i)/(143*d) +
(a^2*(B + 2*C)*80i)/(11*d) - (a^2*(B + C)*160i)/(11*d)) - (C*a^2*128i)/(11*d) + (a^2*(B - 8*C)*16i)/(11*d) +
(a^2*(5*B + 2*C)*16i)/(11*d) - (a^2*(5*B + 9*C)*32i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) +
1)^5) - (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)*(4615*B + 4184*C
)*32i)/(45045*d*(exp(c*1i + d*x*1i) + 1)) - (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)*(4615*B + 4184*C)*16i)/(45045*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 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)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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