\(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [624]

Optimal result

Integrand size = 23, antiderivative size = 306 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {\left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a \left (a^2-b^2\right )^2 d}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^2 (a-b)^2 (a+b)^3 d}+\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]

[Out]

1/2*a*sin(d*x+c)*sec(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2+3/4*(a^2+b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/(a^
2-b^2)^2/d/(a+b*sec(d*x+c))-1/4*(5*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/(a^2-b^2)^2/d-1/4*b*(7*a^2-b^2)*(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^2/(a^2-b^2
)^2/d+1/4*(3*a^4+10*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)*EllipticPi(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^2/(a-b)^2/(a+b)^3/d

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 306, 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 = {3929, 4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {3 \left (a^2+b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {a \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac {\left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^2 d (a-b)^2 (a+b)^3} \]

[In]

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

[Out]

-1/4*((5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*(a^2 - b^2)^2*d) - (b*
(7*a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a^2 - b^2)^2*d) + ((3*a
^4 + 10*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^2
*(a - b)^2*(a + b)^3*d) + (a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (3*(a
^2 + b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*(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 3929

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(f*(m + 1)*(a^2 - b^2))), x] - Dist[d^2/((
m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(a*(n - 2) + b*(m + 1)*Csc[e +
f*x] - a*(m + n)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && Lt
Q[1, n, 2] && 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 {a \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 {a}{2}-2 b \sec (c+d x)+\frac {3}{2} a \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 {a \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 {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{4} a \left (5 a^2+b^2\right )+3 a^2 b \sec (c+d x)-\frac {3}{4} a \left (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 {a \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 {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{4} a^2 \left (5 a^2+b^2\right )-\left (-3 a^3 b+\frac {1}{4} a b \left (5 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 a^4+10 a^2 b^2-b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {a \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 {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (b \left (7 a^2-b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (5 a^2+b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 a \left (a^2-b^2\right )^2}+\frac {\left (\left (3 a^4+10 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^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (3 a^4+10 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^2 (a-b)^2 (a+b)^3 d}+\frac {a \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 {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (\left (5 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 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (5 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 \left (a^2-b^2\right )^2 d}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4+10 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^2 (a-b)^2 (a+b)^3 d}+\frac {a \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 {3 \left (a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.71 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {4 a^2 \left (3 b \left (a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}+\frac {4 \cos (c+d x) (b+a \cos (c+d x)) \cot (c+d x) (a+b \sec (c+d x)) \left (5 a^3 b+a b^3-5 a^3 b \sec ^2(c+d x)-a b^3 \sec ^2(c+d x)+a b \left (5 a^2+b^2\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+a \left (3 a^3-5 a^2 b+3 a b^2-b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-3 a^4 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-10 a^2 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+b^4 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{(a-b)^2 b (a+b)^2}}{16 a^2 d (b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1857\) vs. \(2(358)=716\).

Time = 26.06 (sec) , antiderivative size = 1858, normalized size of antiderivative = 6.07

method	result	size
default	\(\text {Expression too large to display}\)	\(1858\)

[In]

int(sec(d*x+c)^(3/2)/(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/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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))+2/a^2*b^2*(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*cos(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)*Ellip
ticF(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)*E
llipticE(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*(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))-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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-4*b/a^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(co
s(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))))/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. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

[In]

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

[Out]

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