\(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [470]

Optimal result

Integrand size = 35, antiderivative size = 346 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (3 a^2 A-2 A b^2-a b B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/3*(A*b-B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3*(5*A*a^2*b-A*b^3-2*B*a^3-2*B
*a*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/3*(3*A*a^2-2*A*b^2-B*a*b)*(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)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/3*(6*A*a^2*b-2*A*b^3-3*B*a^3-B*a*b^2
)*(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*s
ec(d*x+c))^(1/2)/a^2/(a^2-b^2)^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4112, 4185, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 a^2 A-a b B-2 A b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-2 a^3 B+5 a^2 A b-2 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^3 B+6 a^2 A b-a b^2 B-2 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]

[In]

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

[Out]

(2*(3*a^2*A - 2*A*b^2 - a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[S
ec[c + d*x]])/(3*a^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(6*a^2*A*b - 2*A*b^3 - 3*a^3*B - a*b^2*B)*El
lipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])
/(a + b)]*Sqrt[Sec[c + d*x]]) - (2*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x])^(3/2)) - (2*(5*a^2*A*b - A*b^3 - 2*a^3*B - 2*a*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2
)^2*d*Sqrt[a + b*Sec[c + d*x]])

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 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 4112

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-d)*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^
(n - 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e
+ f*x])^(n - 1)*Simp[d*(n - 1)*(A*b - a*B) + d*(a*A - b*B)*(m + 1)*Csc[e + f*x] - d*(A*b - a*B)*(m + n + 1)*C
sc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[
m, -1] && LtQ[0, n, 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 4185

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} (-A b+a B)-\frac {3}{2} (a A-b B) \sec (c+d x)+(A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )} \\ & = -\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {1}{4} \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right )+\frac {1}{4} a \left (3 a^2 A+A b^2-4 a b B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2} \\ & = -\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (3 a^2 A-2 A b^2-a b B\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )}+\frac {\left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^2 A-2 A b^2-a b B\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}{3 a^2 \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \\ & = -\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^2 A-2 A b^2-a b B\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}{3 a^2 \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}} \\ & = \frac {2 \left (3 a^2 A-2 A b^2-a b B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.32 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \left (-\frac {\left (\frac {b+a \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (-6 a^2 A b+2 A b^3+3 a^3 B+a b^2 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )-(a-b) \left (3 a^2 A-2 A b^2-a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )\right )}{(a-b)^2}+\frac {a \left (b \left (-5 a^2 A b+A b^3+2 a^3 B+2 a b^2 B\right )+a \left (-6 a^2 A b+2 A b^3+3 a^3 B+a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 a^2 d (a+b \sec (c+d x))^{5/2}} \]

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(5/2)*(-((((b + a*Cos[c + d*x])/(a + b))^(3/2)*((-6*a^2*A*b + 2*A*b^3 + 3
*a^3*B + a*b^2*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - (a - b)*(3*a^2*A - 2*A*b^2 - a*b*B)*EllipticF[(c + d
*x)/2, (2*a)/(a + b)]))/(a - b)^2) + (a*(b*(-5*a^2*A*b + A*b^3 + 2*a^3*B + 2*a*b^2*B) + a*(-6*a^2*A*b + 2*A*b^
3 + 3*a^3*B + a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3*a^2*d*(a + b*Sec[c + d*x])^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(5257\) vs. \(2(376)=752\).

Time = 18.27 (sec) , antiderivative size = 5258, normalized size of antiderivative = 15.20

method	result	size
parts	\(\text {Expression too large to display}\)	\(5258\)
default	\(\text {Expression too large to display}\)	\(5389\)

[In]

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

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.18 (sec) , antiderivative size = 1071, normalized size of antiderivative = 3.10 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/9*(sqrt(2)*(-9*I*A*a^4*b^2 + 6*I*B*a^3*b^3 + 9*I*A*a^2*b^4 - 2*I*B*a*b^5 - 4*I*A*b^6 + (-9*I*A*a^6 + 6*I*B*a
^5*b + 9*I*A*a^4*b^2 - 2*I*B*a^3*b^3 - 4*I*A*a^2*b^4)*cos(d*x + c)^2 - 2*(9*I*A*a^5*b - 6*I*B*a^4*b^2 - 9*I*A*
a^3*b^3 + 2*I*B*a^2*b^4 + 4*I*A*a*b^5)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/2
7*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)*(9*I*A*a^4*b^2 - 6*I*B
*a^3*b^3 - 9*I*A*a^2*b^4 + 2*I*B*a*b^5 + 4*I*A*b^6 + (9*I*A*a^6 - 6*I*B*a^5*b - 9*I*A*a^4*b^2 + 2*I*B*a^3*b^3
+ 4*I*A*a^2*b^4)*cos(d*x + c)^2 - 2*(-9*I*A*a^5*b + 6*I*B*a^4*b^2 + 9*I*A*a^3*b^3 - 2*I*B*a^2*b^4 - 4*I*A*a*b^
5)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*co
s(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*(3*I*B*a^4*b^2 - 6*I*A*a^3*b^3 + I*B*a^2*b^4 + 2*I*A*a*b
^5 + (3*I*B*a^6 - 6*I*A*a^5*b + I*B*a^4*b^2 + 2*I*A*a^3*b^3)*cos(d*x + c)^2 + 2*(3*I*B*a^5*b - 6*I*A*a^4*b^2 +
I*B*a^3*b^3 + 2*I*A*a^2*b^4)*cos(d*x + c))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b -
8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) +
3*I*a*sin(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(-3*I*B*a^4*b^2 + 6*I*A*a^3*b^3 - I*B*a^2*b^4 - 2*I*A*a*b^5 + (-3*I*
B*a^6 + 6*I*A*a^5*b - I*B*a^4*b^2 - 2*I*A*a^3*b^3)*cos(d*x + c)^2 + 2*(-3*I*B*a^5*b + 6*I*A*a^4*b^2 - I*B*a^3*
b^3 - 2*I*A*a^2*b^4)*cos(d*x + c))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^
3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin
(d*x + c) + 2*b)/a)) + 6*((3*B*a^6 - 6*A*a^5*b + B*a^4*b^2 + 2*A*a^3*b^3)*cos(d*x + c)^2 + (2*B*a^5*b - 5*A*a^
4*b^2 + 2*B*a^3*b^3 + A*a^2*b^4)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d
*x + c)))/((a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c)^2 + 2*(a^8*b - 2*a^6*b^3 + a^4*b^5)*d*cos(d*x + c) + (a^
7*b^2 - 2*a^5*b^4 + a^3*b^6)*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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