3.5.56 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [456]

Optimal. Leaf size=344 \[ \frac {(4 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{4 b d \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{4 b^2 d \sqrt {a+b \sec (c+d x)}}-\frac {(4 A b-3 a B) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{4 b^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d} \]

[Out]

1/4*(4*A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))
^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/b/d/(a+b*sec(d*x+c))^(1/2)-1/4*(4*A*a*b-3*B*a^2-4*B*b^
2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((
b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/b^2/d/(a+b*sec(d*x+c))^(1/2)+1/2*B*sec(d*x+c)^(3/2)*sin(d*x+c)*(
a+b*sec(d*x+c))^(1/2)/b/d-1/4*(4*A*b-3*B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*
d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/b^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/
2)+1/4*(4*A*b-3*B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/b^2/d

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Rubi [A]
time = 0.72, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {4118, 4187, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} -\frac {\left (-3 a^2 B+4 a A b-4 b^2 B\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 )}{4 b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {(4 A b-3 a B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{4 b^2 d}-\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {(4 A b-a B) \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 )}{4 b d \sqrt {a+b \sec (c+d x)}}+\frac {B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(4
*b*d*Sqrt[a + b*Sec[c + d*x]]) - ((4*a*A*b - 3*a^2*B - 4*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[
2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(4*b^2*d*Sqrt[a + b*Sec[c + d*x]]) - ((4*A*b - 3*a*B)*Ellip
ticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(4*b^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Se
c[c + d*x]]) + ((4*A*b - 3*a*B)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*b^2*d) + (B*Sec[c
+ d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*b*d)

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 3941

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 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[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 3944

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d*Sqrt
[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 4118

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

Rule 4120

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 4187

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 4193

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]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx &=\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (\frac {a B}{2}+b B \sec (c+d x)+\frac {1}{2} (4 A b-3 a B) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 b}\\ &=\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}+\frac {\int \frac {-\frac {1}{4} a (4 A b-3 a B)+\frac {1}{2} a b B \sec (c+d x)-\frac {1}{4} \left (4 a A b-3 a^2 B-4 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{2 b^2}\\ &=\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}+\frac {\int \frac {-\frac {1}{4} a (4 A b-3 a B)+\frac {1}{2} a b B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{2 b^2}-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 b^2}\\ &=\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}-\frac {(4 A b-3 a B) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{8 b^2}+\frac {(4 A b-a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 b}-\frac {\left (\left (4 a A b-3 a^2 B-4 b^2 B\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}{8 b^2 \sqrt {a+b \sec (c+d x)}}\\ &=\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}+\frac {\left ((4 A b-a B) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{8 b \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (4 a A b-3 a^2 B-4 b^2 B\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}{8 b^2 \sqrt {a+b \sec (c+d x)}}-\frac {\left ((4 A b-3 a B) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{8 b^2 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\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)}}{4 b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}+\frac {\left ((4 A b-a B) \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}{8 b \sqrt {a+b \sec (c+d x)}}-\frac {\left ((4 A b-3 a B) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{8 b^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {(4 A b-a B) \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)}}{4 b d \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\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)}}{4 b^2 d \sqrt {a+b \sec (c+d x)}}-\frac {(4 A b-3 a B) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{4 b^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {(4 A b-3 a B) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d}+\frac {B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 13.91, size = 451, normalized size = 1.31 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (\frac {8 a B \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{b}+\frac {2 \left (-12 a A b+9 a^2 B+8 b^2 B\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 )}{b^2}+\frac {2 i (-4 A b+3 a B) \sqrt {-\frac {a (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {a (1+\cos (c+d x))}{a-b}} \sqrt {b+a \cos (c+d x)} \csc (c+d x) \left (-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b F\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{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 {-a+b}{a+b}\right )\right )\right )}{a \sqrt {\frac {1}{a-b}} b^3}-\frac {4 a (-4 A b+3 a B) \sin (c+d x)}{b^2}+\frac {8 a B \tan (c+d x)}{b}+\frac {4 (4 A b-3 a B) \tan (c+d x)}{b}+8 B \sec (c+d x) \tan (c+d x)\right )}{16 d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*((8*a*B*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/b + (2*(
-12*a*A*b + 9*a^2*B + 8*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/b
^2 + ((2*I)*(-4*A*b + 3*a*B)*Sqrt[-((a*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(a*(1 + Cos[c + d*x]))/(a - b)]*Sqr
t[b + a*Cos[c + d*x]]*Csc[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)])))/(a*
Sqrt[(a - b)^(-1)]*b^3) - (4*a*(-4*A*b + 3*a*B)*Sin[c + d*x])/b^2 + (8*a*B*Tan[c + d*x])/b + (4*(4*A*b - 3*a*B
)*Tan[c + d*x])/b + 8*B*Sec[c + d*x]*Tan[c + d*x]))/(16*d*Sqrt[a + b*Sec[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 12.74, size = 2737, normalized size = 7.96

method	result	size
default	\(\text {Expression too large to display}\)	\(2737\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/d*(-4*A*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+2*B*cos(d*x+c)^2*sin(d*x+c)*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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-4*A*cos(d*x+c)^2*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*(1/(1+cos(d*x+c)))^(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-3*B*cos(d*x+c)^2*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Ellipt
icE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-3*B*cos(d*x+c)^3*((a-b)/(a+b))^(1
/2)*a^2+4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^2+2*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^2-2*B*((a-b)/(a+b))^(1
/2)*b^2-4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b-3*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b+B*cos(d*x+c)*((a-b)/
(a+b))^(1/2)*a*b+8*A*cos(d*x+c)^2*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^
(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+4*A*sin(d*x+c)*cos(d*
x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2-6*B*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2-4*B*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2
)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2+3*B*sin(d*x+c)*cos(d*x+c)
^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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+6*B*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2+8*B*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x
+c)))^(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))*b^2+4
*A*sin(d*x+c)*cos(d*x+c)^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2-4*B*sin(d*x+c)*cos(d*x+c)^2*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(-(a+b)/(a-b))^(1/2))*b^2+3*B*sin(d*x+c)*cos(d*x+c)^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/
(1+cos(d*x+c)))^(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+6*B*s
in(d*x+c)*cos(d*x+c)^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticPi((-1+c
os(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^2+8*B*sin(d*x+c)*cos(d*x+c)^2*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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))*b^2-8*A*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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*B*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/
(1+cos(d*x+c)))^(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-3*B*s
in(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-8*A*sin(d*x+c)*cos(d*x+c)^2*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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+8*A*sin(d*x+c)*cos(d*x+c)^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*(1/(1+cos(d*x+c)))^(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+4*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b+2*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b+3*B*cos(d*x+c)^2*((a-b)
/(a+b))^(1/2)*a^2-4*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-6*B*cos(d*x+c)^2*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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)*cos(d*x+c)*(1/cos(d*x+c))^(5/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/(b+a*cos(d*
x+c))/b^2/((a-b)/(a+b))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + b/cos(c + d*x))^(1/2),x)

[Out]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + b/cos(c + d*x))^(1/2), x)

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