Integrand size = 23, antiderivative size = 203 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {284 a^3 \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {710 a^3 \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {568 a^2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac {2 a^2 \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {284 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d} \] Output:
284/99*a^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+710/693*a^3*sec(d*x+c)^3*ta n(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+46/99*a^3*sec(d*x+c)^4*tan(d*x+c)/d/(a+a *sec(d*x+c))^(1/2)-568/693*a^2*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/11*a^ 2*sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+284/231*a*(a+a*sec(d*x+ c))^(3/2)*tan(d*x+c)/d
Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.39 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^3 \left (1136+568 \sec (c+d x)+426 \sec ^2(c+d x)+355 \sec ^3(c+d x)+224 \sec ^4(c+d x)+63 \sec ^5(c+d x)\right ) \tan (c+d x)}{693 d \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]
Output:
(2*a^3*(1136 + 568*Sec[c + d*x] + 426*Sec[c + d*x]^2 + 355*Sec[c + d*x]^3 + 224*Sec[c + d*x]^4 + 63*Sec[c + d*x]^5)*Tan[c + d*x])/(693*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.22 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4301, 27, 3042, 4504, 3042, 4290, 3042, 4287, 27, 3042, 4489, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^4(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}dx\) |
\(\Big \downarrow \) 4301 |
\(\displaystyle \frac {2}{11} a \int \frac {1}{2} \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a} (23 \sec (c+d x) a+19 a)dx+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} a \int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a} (23 \sec (c+d x) a+19 a)dx+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (23 \csc \left (c+d x+\frac {\pi }{2}\right ) a+19 a\right )dx+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 4287 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} a \left (\frac {355}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}+\frac {1}{11} a \left (\frac {46 a^2 \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}+\frac {355}{9} a \left (\frac {6}{7} \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )\right )\) |
Input:
Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]
Output:
(2*a^2*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (a*( (46*a^2*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + (355 *a*((2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (6* ((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d* x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7))/9))/11
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
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] + Simp[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)]
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] + Simp[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] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[b/(m + n - 1) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(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)]
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*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(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]
Time = 47.62 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {2 a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\) | \(95\) |
Input:
int(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/693/d*a^2*(1136*cos(d*x+c)^5+568*cos(d*x+c)^4+426*cos(d*x+c)^3+355*cos(d *x+c)^2+224*cos(d*x+c)+63)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x +c)*sec(d*x+c)^4
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.60 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 \, {\left (1136 \, a^{2} \cos \left (d x + c\right )^{5} + 568 \, a^{2} \cos \left (d x + c\right )^{4} + 426 \, a^{2} \cos \left (d x + c\right )^{3} + 355 \, a^{2} \cos \left (d x + c\right )^{2} + 224 \, a^{2} \cos \left (d x + c\right ) + 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
2/693*(1136*a^2*cos(d*x + c)^5 + 568*a^2*cos(d*x + c)^4 + 426*a^2*cos(d*x + c)^3 + 355*a^2*cos(d*x + c)^2 + 224*a^2*cos(d*x + c) + 63*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)
Timed out. \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))**(5/2),x)
Output:
Timed out
Timed out. \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
Timed out
Time = 0.53 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.03 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {8 \, {\left (693 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1617 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (3003 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 25 \, {\left (99 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, {\left (2 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{693 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
-8/693*(693*sqrt(2)*a^8*sgn(cos(d*x + c)) - (1617*sqrt(2)*a^8*sgn(cos(d*x + c)) - (3003*sqrt(2)*a^8*sgn(cos(d*x + c)) - 25*(99*sqrt(2)*a^8*sgn(cos(d *x + c)) + 4*(2*sqrt(2)*a^8*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 11* sqrt(2)*a^8*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)/ ((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)
Time = 20.16 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.67 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx =\text {Too large to display} \] Input:
int((a + a/cos(c + d*x))^(5/2)/cos(c + d*x)^4,x)
Output:
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*64i)/ (11*d) + (a^2*exp(c*1i + d*x*1i)*64i)/(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)*((a^2*16i)/d + (a^2*exp(c*1i + d*x*1i)*640i)/(231*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^2*64i)/(9*d) + (a^2*exp(c* 1i + d*x*1i)*2176i)/(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)* ((a^2*80i)/(7*d) - (a^2*exp(c*1i + d*x*1i)*12688i)/(693*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - (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)*2272i)/(693*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)*1136i)/(693*d*(exp(c*1i + d*x*1i) + 1)* (exp(c*2i + d*x*2i) + 1))
\[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) \] Input:
int(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**6,x) + 2*int(sqrt(s ec(c + d*x) + 1)*sec(c + d*x)**5,x) + int(sqrt(sec(c + d*x) + 1)*sec(c + d *x)**4,x))