3.107 \(\int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=203 \[ \frac {46 a^3 \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {710 a^3 \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {284 a^3 \tan (c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}-\frac {568 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{693 d}+\frac {284 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d} \]

[Out]

284/231*a*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+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*tan(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

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Rubi [A]  time = 0.37, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3814, 4016, 3803, 3800, 4001, 3792} \[ \frac {2 a^2 \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}+\frac {46 a^3 \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {710 a^3 \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}-\frac {568 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{693 d}+\frac {284 a^3 \tan (c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {284 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(284*a^3*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) + (710*a^3*Sec[c + d*x]^3*Tan[c + d*x])/(693*d*Sqrt[a +
a*Sec[c + d*x]]) + (46*a^3*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Sec[c + d*x]]) - (568*a^2*Sqrt[a + a
*Sec[c + d*x]]*Tan[c + d*x])/(693*d) + (2*a^2*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (
284*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*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 3814

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] + Dist[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] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

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]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac {2 a^2 \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{11} (2 a) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {19 a}{2}+\frac {23}{2} a \sec (c+d x)\right ) \, dx\\ &=\frac {46 a^3 \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{99} \left (355 a^2\right ) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\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 {2 a^2 \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{231} \left (710 a^2\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\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 {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}+\frac {1}{231} (284 a) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\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}+\frac {1}{99} \left (142 a^2\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, 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}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 80, normalized size = 0.39 \[ \frac {2 a^3 \tan (c+d x) \left (63 \sec ^5(c+d x)+224 \sec ^4(c+d x)+355 \sec ^3(c+d x)+426 \sec ^2(c+d x)+568 \sec (c+d x)+1136\right )}{693 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(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])])

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fricas [A]  time = 0.60, size = 121, normalized size = 0.60 \[ \frac {2 \, {\left (1136 \, a^{2} \cos \left (d x + c\right )^{5} + 568 \, a^{2} \cos \left (d x + c\right )^{4} + 426 \, a^{2} \cos \left (d x + c\right )^{3} + 355 \, a^{2} \cos \left (d x + c\right )^{2} + 224 \, a^{2} \cos \left (d x + c\right ) + 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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 + 22
4*a^2*cos(d*x + c) + 63*a^2)*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 = 6.22, size = 209, normalized size = 1.03 \[ -\frac {8 \, {\left (693 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1617 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (3003 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 25 \, {\left (99 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, {\left (2 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{693 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-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 - 1
1*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)

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maple [A]  time = 0.96, size = 105, normalized size = 0.52 \[ -\frac {2 \left (1136 \left (\cos ^{6}\left (d x +c \right )\right )-568 \left (\cos ^{5}\left (d x +c \right )\right )-142 \left (\cos ^{4}\left (d x +c \right )\right )-71 \left (\cos ^{3}\left (d x +c \right )\right )-131 \left (\cos ^{2}\left (d x +c \right )\right )-161 \cos \left (d x +c \right )-63\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{693 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x)

[Out]

-2/693/d*(1136*cos(d*x+c)^6-568*cos(d*x+c)^5-142*cos(d*x+c)^4-71*cos(d*x+c)^3-131*cos(d*x+c)^2-161*cos(d*x+c)-
63)*(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)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 9.32, size = 542, normalized size = 2.67 \[ \frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,64{}\mathrm {i}}{11\,d}+\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,64{}\mathrm {i}}{11\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,16{}\mathrm {i}}{d}+\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,640{}\mathrm {i}}{231\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,64{}\mathrm {i}}{9\,d}+\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,2176{}\mathrm {i}}{99\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,80{}\mathrm {i}}{7\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,12688{}\mathrm {i}}{693\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,2272{}\mathrm {i}}{693\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,1136{}\mathrm {i}}{693\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/2)/cos(c + d*x)^4,x)

[Out]

((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*ex
p(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))

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

[Out]

Timed out

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