\(\int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx\) [625]

Optimal result

Integrand size = 23, antiderivative size = 323 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\frac {3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 (a+b)^3 d}-\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]

[Out]

-1/2*b*sin(d*x+c)*sec(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1/4*b*(7*a^2-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))+3/4*b*(3*a^2-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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d+1/4*(8*a^4-5*a^2*b^2+3*b^4)*(
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))*cos(d*x+c)^(1/2)*sec(d*x+
c)^(1/2)/a^3/(a^2-b^2)^2/d-3/4*b*(5*a^4-2*a^2*b^2+b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/(a-b)^2/(a+b)^3/d

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3928, 4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=-\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {b \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}+\frac {\left (8 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 d (a-b)^2 (a+b)^3} \]

[In]

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

[Out]

(3*b*(3*a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a^2 - b^2)^2*d) +
((8*a^4 - 5*a^2*b^2 + 3*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^3*(a^2 - b^
2)^2*d) - (3*b*(5*a^4 - 2*a^2*b^2 + b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(4*a^3*(a - b)^2*(a + b)^3*d) - (b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x])^2) - (b*(7*a^2 - b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*a*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3928

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

Rule 3934

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

Rule 4191

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {-\frac {b}{2}-2 a \sec (c+d x)+\frac {3}{2} b \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )} \\ & = -\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\frac {3}{4} b \left (3 a^2-b^2\right )+a \left (2 a^2+b^2\right ) \sec (c+d x)-\frac {1}{4} b \left (7 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 a \left (a^2-b^2\right )^2} \\ & = -\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\frac {3}{4} a b \left (3 a^2-b^2\right )-\left (\frac {3}{4} b^2 \left (3 a^2-b^2\right )-a^2 \left (2 a^2+b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )^2}-\frac {\left (3 b \left (5 a^4-2 a^2 b^2+b^4\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (3 b \left (3 a^2-b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}+\frac {\left (8 a^4-5 a^2 b^2+3 b^4\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac {\left (3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 (a+b)^3 d}-\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^2 \left (a^2-b^2\right )^2}+\frac {\left (\left (8 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4-5 a^2 b^2+3 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 (a+b)^3 d}-\frac {b \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.80 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {4 b \left (-7 a^2 b+b^3+\left (-9 a^3+3 a b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}}+\frac {2 \cot (c+d x) \left (6 a b \left (3 a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin ^2(c+d x)-6 a b \left (3 a^2-b^2\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-2 a \left (7 a^3-9 a^2 b-a b^2+3 b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+6 \left (5 a^4-2 a^2 b^2+b^4\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{a^3 (a-b)^2 (a+b)^2}}{16 d} \]

[In]

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

[Out]

((4*b*(-7*a^2*b + b^3 + (-9*a^3 + 3*a*b^2)*Cos[c + d*x])*Sin[c + d*x])/(a*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2
*Sqrt[Sec[c + d*x]]) + (2*Cot[c + d*x]*(6*a*b*(3*a^2 - b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x]^2 - 6*a*b*(3*a^2 -
b^2)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*a*(7*a^3 - 9*a^2*b - a*b^2 + 3*b^3)*
EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 6*(5*a^4 - 2*a^2*b^2 + b^4)*EllipticPi[-(b/a
), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(a^3*(a - b)^2*(a + b)^2))/(16*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1935\) vs. \(2(375)=750\).

Time = 28.16 (sec) , antiderivative size = 1936, normalized size of antiderivative = 5.99

method	result	size
default	\(\text {Expression too large to display}\)	\(1936\)

[In]

int(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-
2/a^3*b^3*(1/2*a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*co
s(1/2*d*x+1/2*c)^2-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-3/8/(a+b)/(a^2-b^2)/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2
*c),2^(1/2))*a^2-1/4/(a+b)/(a^2-b^2)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a+7/8/(a+b)/(a^2-b^2)*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9/8
*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2))+9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(
a^2-a*b)*a^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^3*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2-a*b)*a*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic
Pi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+6/a^3*b^2*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos
(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(
1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2
*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+6/a^2*b/(a^2-a*b)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sqrt {\sec {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(sec(d*x+c)**(1/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Integral(sqrt(sec(c + d*x))/(a + b*sec(c + d*x))**3, x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

[In]

int((1/cos(c + d*x))^(1/2)/(a + b/cos(c + d*x))^3,x)

[Out]

int((1/cos(c + d*x))^(1/2)/(a + b/cos(c + d*x))^3, x)