3.640 \(\int \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=369 \[ \frac{b \left (59 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{24 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}{24 d}-\frac{\left (33 a^2+16 b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{5 a \left (a^2+4 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{b^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{3 d}+\frac{13 a b \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{12 d} \]

[Out]

(b*(59*a^2 + 16*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x
]])/(24*d*Sqrt[a + b*Sec[c + d*x]]) + (5*a*(a^2 + 4*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c +
d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(8*d*Sqrt[a + b*Sec[c + d*x]]) - ((33*a^2 + 16*b^2)*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(24*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]])
+ ((33*a^2 + 16*b^2)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (13*a*b*Sec[c + d*x]^
(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*d) + (b^2*Sec[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c +
d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 1.34757, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {3842, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{\left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}{24 d}+\frac{b \left (59 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (33 a^2+16 b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{5 a \left (a^2+4 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{b^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{3 d}+\frac{13 a b \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(59*a^2 + 16*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x
]])/(24*d*Sqrt[a + b*Sec[c + d*x]]) + (5*a*(a^2 + 4*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c +
d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(8*d*Sqrt[a + b*Sec[c + d*x]]) - ((33*a^2 + 16*b^2)*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(24*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]])
+ ((33*a^2 + 16*b^2)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (13*a*b*Sec[c + d*x]^
(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*d) + (b^2*Sec[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c +
d*x])/(3*d)

Rule 3842

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[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

Rule 4102

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*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]

Rule 4108

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 4035

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

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3}{2} a \left (2 a^2+b^2\right )+b \left (9 a^2+2 b^2\right ) \sec (c+d x)+\frac{13}{2} a b^2 \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{13 a^2 b^2}{4}+\frac{1}{2} a b \left (12 a^2+19 b^2\right ) \sec (c+d x)+\frac{1}{4} b^2 \left (33 a^2+16 b^2\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{6 b}\\ &=\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\int \frac{-\frac{1}{8} a b^2 \left (33 a^2+16 b^2\right )+\frac{13}{4} a^2 b^3 \sec (c+d x)+\frac{15}{8} a b^2 \left (a^2+4 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{6 b^2}\\ &=\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\int \frac{-\frac{1}{8} a b^2 \left (33 a^2+16 b^2\right )+\frac{13}{4} a^2 b^3 \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{6 b^2}+\frac{1}{16} \left (5 a \left (a^2+4 b^2\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{48} \left (-33 a^2-16 b^2\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{48} \left (b \left (59 a^2+16 b^2\right )\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{\left (5 a \left (a^2+4 b^2\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{16 \sqrt{a+b \sec (c+d x)}}\\ &=\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\left (b \left (59 a^2+16 b^2\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{48 \sqrt{a+b \sec (c+d x)}}+\frac{\left (5 a \left (a^2+4 b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{16 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-33 a^2-16 b^2\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{48 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{5 a \left (a^2+4 b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\left (b \left (59 a^2+16 b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{48 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-33 a^2-16 b^2\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{48 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{b \left (59 a^2+16 b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{24 d \sqrt{a+b \sec (c+d x)}}+\frac{5 a \left (a^2+4 b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{8 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (33 a^2+16 b^2\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{24 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{\left (33 a^2+16 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{13 a b \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{12 d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 6.56889, size = 602, normalized size = 1.63 \[ \frac{(a+b \sec (c+d x))^{5/2} \left (\frac{1}{24} \sec (c+d x) \left (33 a^2 \sin (c+d x)+16 b^2 \sin (c+d x)\right )+\frac{13}{12} a b \tan (c+d x) \sec (c+d x)+\frac{1}{3} b^2 \tan (c+d x) \sec ^2(c+d x)\right )}{d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^2}-\frac{a (a+b \sec (c+d x))^{5/2} \left (\frac{2 i \left (33 a^2+16 b^2\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt{\frac{a-a \cos (c+d x)}{a+b}} \sqrt{\frac{a \cos (c+d x)+a}{a-b}} \left (a \left (2 b \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{a \cos (c+d x)+b}\right ),\frac{b-a}{a+b}\right )+a \Pi \left (1-\frac{a}{b};i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right )-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right )}{b \sqrt{\frac{1}{a-b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{\frac{a^2-a^2 \cos ^2(c+d x)}{a^2}} \left (-a^2+2 (a \cos (c+d x)+b)^2-4 b (a \cos (c+d x)+b)+2 b^2\right )}-\frac{104 a b \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{\sqrt{a \cos (c+d x)+b}}+\frac{2 \left (3 a^2-104 b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{\sqrt{a \cos (c+d x)+b}}\right )}{96 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*Sec[c + d*x])^(5/2)*((-104*a*b*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a +
b)])/Sqrt[b + a*Cos[c + d*x]] + (2*(3*a^2 - 104*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*
x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((2*I)*(33*a^2 + 16*b^2)*Sqrt[(a - a*Cos[c + d*x])/(a + b)]*S
qrt[(a + a*Cos[c + d*x])/(a - b)]*Cos[2*(c + d*x)]*(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b
+ a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*(2*b*EllipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]
]], (-a + b)/(a + b)] + a*EllipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)
/(a + b)]))*Sin[c + d*x])/(Sqrt[(a - b)^(-1)]*b*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[(a^2 - a^2*Cos[c + d*x]^2)/a^2]*
(-a^2 + 2*b^2 - 4*b*(b + a*Cos[c + d*x]) + 2*(b + a*Cos[c + d*x])^2))))/(96*d*(b + a*Cos[c + d*x])^(5/2)*Sec[c
+ d*x]^(5/2)) + ((a + b*Sec[c + d*x])^(5/2)*((Sec[c + d*x]*(33*a^2*Sin[c + d*x] + 16*b^2*Sin[c + d*x]))/24 +
(13*a*b*Sec[c + d*x]*Tan[c + d*x])/12 + (b^2*Sec[c + d*x]^2*Tan[c + d*x])/3))/(d*(b + a*Cos[c + d*x])^2*Sec[c
+ d*x]^(5/2))

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Maple [C]  time = 0.364, size = 2295, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(5/2),x)

[Out]

-1/24/d/((a-b)/(a+b))^(1/2)*(33*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3+33*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b
-34*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2-44*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^
(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*
a*b^2+33*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*Elli
pticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-16*cos(d*x+c)^4*sin(d*x+c)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))
^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+120*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c
)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/(
(a-b)/(a+b))^(1/2))*a*b^2+26*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d
*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-44*cos(d*
x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+33*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+
c),(-(a+b)/(a-b))^(1/2))*a^2*b-16*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(
cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+120*
cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticPi((-
1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b^2+26*cos(d*x+c)^4*sin(d*x+
c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(
a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b+18*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d
*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b
))^(1/2))*a^3-33*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1
/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3-33*cos(d*x+c)^3*((a-b)/
(a+b))^(1/2)*a^3+16*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*b^3-8*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^3+26*cos(d*x+c)^
4*((a-b)/(a+b))^(1/2)*a^2*b+16*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^2+18*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^
2-59*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b-8*b^3*((a-b)/(a+b))^(1/2)+16*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(-(a+b)/(a-b))^(1/2))*b^3+30*cos(d*x+c)^4*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1
/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(
1/2))*a^3+18*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*
EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3-33*cos(d*x+c)^3*sin(d*x+c)*
(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b
))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3+16*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)
+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1
/2))*b^3+30*cos(d*x+c)^3*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*E
llipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a^3)*((b+a*cos(d*x
+c))/cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(3/2)/(b+a*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)

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Maxima [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 )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2), x)

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Fricas [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)^(3/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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)**(3/2)*(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 )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(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)^(3/2), x)