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

Optimal. Leaf size=463 \[ -\frac{2 (a-b) \sqrt{a+b} \left (57 a^2 b^2+6 a^3 b+8 a^4-606 a b^3+135 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{693 b^3 d}+\frac{2 \left (8 a^2+81 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{693 b^2 d}+\frac{2 \left (57 a^2 b^2+8 a^4+135 b^4\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{693 b^2 d}-\frac{2 a (a-b) \sqrt{a+b} \left (51 a^2 b^2+8 a^4+741 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{693 b^4 d}-\frac{8 a \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac{2 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(693*b^4*d) - (2*(a - b)*Sqrt[a + b]*(8*a^4 + 6*a^3*b + 57*a^2*b^2 - 606*a*b^3 + 135*b^4)*Cot[c + d*x]*Ellipt
icF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-
((b*(1 + Sec[c + d*x]))/(a - b))])/(693*b^3*d) + (2*(8*a^4 + 57*a^2*b^2 + 135*b^4)*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(693*b^2*d) + (2*a*(8*a^2 + 67*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(693*b^2*d) + (2*(8*a
^2 + 81*b^2)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*b^2*d) - (8*a*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*
x])/(99*b^2*d) + (2*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

________________________________________________________________________________________

Rubi [A]  time = 1.04078, antiderivative size = 463, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3865, 4082, 4002, 4005, 3832, 4004} \[ \frac{2 \left (8 a^2+81 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{693 b^2 d}+\frac{2 \left (57 a^2 b^2+8 a^4+135 b^4\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{693 b^2 d}-\frac{2 (a-b) \sqrt{a+b} \left (57 a^2 b^2+6 a^3 b+8 a^4-606 a b^3+135 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{693 b^3 d}-\frac{2 a (a-b) \sqrt{a+b} \left (51 a^2 b^2+8 a^4+741 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{693 b^4 d}-\frac{8 a \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac{2 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(693*b^4*d) - (2*(a - b)*Sqrt[a + b]*(8*a^4 + 6*a^3*b + 57*a^2*b^2 - 606*a*b^3 + 135*b^4)*Cot[c + d*x]*Ellipt
icF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-
((b*(1 + Sec[c + d*x]))/(a - b))])/(693*b^3*d) + (2*(8*a^4 + 57*a^2*b^2 + 135*b^4)*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(693*b^2*d) + (2*a*(8*a^2 + 67*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(693*b^2*d) + (2*(8*a
^2 + 81*b^2)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*b^2*d) - (8*a*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*
x])/(99*b^2*d) + (2*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

Rule 3865

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^3*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + n - 1)), x] + Dist[d^3/(b*(m + n
- 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 3)*Simp[a*(n - 3) + b*(m + n - 2)*Csc[e + f*x] - a*(n
- 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3] && (Integer
Q[n] || IntegersQ[2*m, 2*n]) &&  !IGtQ[m, 2]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*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*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (a+\frac{9}{2} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{11 b}\\ &=-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{4 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (-\frac{5 a b}{2}+\frac{1}{4} \left (8 a^2+81 b^2\right ) \sec (c+d x)\right ) \, dx}{99 b^2}\\ &=\frac{2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{8 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac{15}{8} b \left (2 a^2-27 b^2\right )+\frac{5}{8} a \left (8 a^2+67 b^2\right ) \sec (c+d x)\right ) \, dx}{693 b^2}\\ &=\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{16 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (-\frac{15}{8} a b \left (a^2-101 b^2\right )+\frac{15}{16} \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sec (c+d x)\right ) \, dx}{3465 b^2}\\ &=\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{32 \int \frac{\sec (c+d x) \left (\frac{15}{32} b \left (2 a^4+663 a^2 b^2+135 b^4\right )+\frac{15}{32} a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{10395 b^2}\\ &=\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}-\frac{\left ((a-b) \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{693 b^2}+\frac{\left (a \left (8 a^4+51 a^2 b^2+741 b^4\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{693 b^2}\\ &=-\frac{2 a (a-b) \sqrt{a+b} \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{693 b^4 d}-\frac{2 (a-b) \sqrt{a+b} \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{693 b^3 d}+\frac{2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac{2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac{2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac{8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}\\ \end{align*}

Mathematica [A]  time = 16.8228, size = 615, normalized size = 1.33 \[ \frac{\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac{2 a \left (51 a^2 b^2+8 a^4+741 b^4\right ) \sin (c+d x)}{693 b^3}+\frac{2}{693} \sec ^3(c+d x) \left (113 a^2 \sin (c+d x)+81 b^2 \sin (c+d x)\right )+\frac{2 \sec ^2(c+d x) \left (3 a^3 \sin (c+d x)+229 a b^2 \sin (c+d x)\right )}{693 b}+\frac{2 \sec (c+d x) \left (205 a^2 b^2 \sin (c+d x)-4 a^4 \sin (c+d x)+135 b^4 \sin (c+d x)\right )}{693 b^2}+\frac{46}{99} a b \tan (c+d x) \sec ^3(c+d x)+\frac{2}{11} b^2 \tan (c+d x) \sec ^4(c+d x)\right )}{d (a \cos (c+d x)+b)^2}-\frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (-2 b \left (51 a^3 b^2+663 a^2 b^3+2 a^4 b+8 a^5+741 a b^4+135 b^5\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+a \left (51 a^2 b^2+8 a^4+741 b^4\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 a \left (51 a^3 b^2+51 a^2 b^3+8 a^4 b+8 a^5+741 a b^4+741 b^5\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{693 b^3 d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*a*(8*a^5 + 8*a^4*b + 51*a^3*b^2 + 51*a
^2*b^3 + 741*a*b^4 + 741*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^5 + 2*a^4*b + 51*a^3*b^2 + 663*a
^2*b^3 + 741*a*b^4 + 135*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Cos[c +
d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(693*b^3*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c
+ d*x)/2]^2]*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*a*(8*a^4 + 51*a^2*b^2 + 741*
b^4)*Sin[c + d*x])/(693*b^3) + (2*Sec[c + d*x]^3*(113*a^2*Sin[c + d*x] + 81*b^2*Sin[c + d*x]))/693 + (2*Sec[c
+ d*x]^2*(3*a^3*Sin[c + d*x] + 229*a*b^2*Sin[c + d*x]))/(693*b) + (2*Sec[c + d*x]*(-4*a^4*Sin[c + d*x] + 205*a
^2*b^2*Sin[c + d*x] + 135*b^4*Sin[c + d*x]))/(693*b^2) + (46*a*b*Sec[c + d*x]^3*Tan[c + d*x])/99 + (2*b^2*Sec[
c + d*x]^4*Tan[c + d*x])/11))/(d*(b + a*Cos[c + d*x])^2)

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Maple [B]  time = 1.365, size = 2806, normalized size = 6.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/693/d/b^3*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-8*sin(d*x+c)*cos(d*x+c)^
6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/
sin(d*x+c),((a-b)/(a+b))^(1/2))*a^6+135*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^6-63*b^6-4*cos
(d*x+c)^7*a^5*b+51*cos(d*x+c)^7*a^4*b^2+205*cos(d*x+c)^7*a^3*b^3+741*cos(d*x+c)^7*a^2*b^4+135*cos(d*x+c)^7*a*b
^5+8*cos(d*x+c)^6*a^5*b-52*cos(d*x+c)^6*a^4*b^2+51*cos(d*x+c)^6*a^3*b^3-307*cos(d*x+c)^6*a^2*b^4+741*cos(d*x+c
)^6*a*b^5-4*cos(d*x+c)^5*a^5*b-140*cos(d*x+c)^5*a^3*b^3-566*cos(d*x+c)^5*a*b^5+cos(d*x+c)^4*a^4*b^2-160*cos(d*
x+c)^4*a^2*b^4-116*cos(d*x+c)^3*a^3*b^3-86*cos(d*x+c)^3*a*b^5-274*cos(d*x+c)^2*a^2*b^4-224*cos(d*x+c)*a*b^5+8*
cos(d*x+c)^7*a^6-8*cos(d*x+c)^6*a^6+135*cos(d*x+c)^6*b^6-54*cos(d*x+c)^4*b^6-18*cos(d*x+c)^2*b^6+135*sin(d*x+c
)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1
+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^6-8*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(
1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^6-5
1*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El
lipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b^2-51*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+
b))^(1/2))*a^3*b^3-741*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^4-741*sin(d*x+c)*cos(d*x+c)
^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))
/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^5+8*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b
+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b+2*sin(d*x
+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((
-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b^2+51*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a^3*b^3+663*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+
1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^4+741*sin(d*x+c)*cos(d*x+c)^6*(cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+
c),((a-b)/(a+b))^(1/2))*a*b^5-8*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*
x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b-51*sin(d*x+c)*cos(
d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d
*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b^2-51*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^
3-741*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2
)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^4-741*sin(d*x+c)*cos(d*x+c)^6*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b
)/(a+b))^(1/2))*a*b^5+8*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(c
os(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b+2*sin(d*x+c)*cos(d*x+c)^5*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b^2+51*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b
+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^3+663*sin
(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti
cF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^4+741*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^
(1/2))*a*b^5-8*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)
+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b)/(b+a*cos(d*x+c))/cos(d*x+c)^5/sin(
d*x+c)^5

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \sec \left (d x + c\right )^{6} + 2 \, a b \sec \left (d x + c\right )^{5} + a^{2} \sec \left (d x + c\right )^{4}\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*sec(d*x + c)^6 + 2*a*b*sec(d*x + c)^5 + a^2*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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