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

Optimal. Leaf size=306 \[ \frac {3 \left (a^2+b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {a \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac {\left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d (a-b)^2 (a+b)^3} \]

[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

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Rubi [A]  time = 0.65, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3844, 4100, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {3 \left (a^2+b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {a \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac {\left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}+\frac {\left (10 a^2 b^2+3 a^4-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*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 2639

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

Rule 2641

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

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 3771

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 3787

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 3844

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 3849

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 4100

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 4106

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*} \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=\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)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |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)} F\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 (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |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*}

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Mathematica [B]  time = 6.77, size = 719, normalized size = 2.35 \[ \frac {\sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)^3 \left (\frac {\left (5 a^2+b^2\right ) \sin (c+d x)}{4 a \left (a^2-b^2\right )^2}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {b^3 \sin (c+d x)-7 a^2 b \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}\right )}{d (a+b \sec (c+d x))^3}-\frac {\sec ^3(c+d x) (a \cos (c+d x)+b)^3 \left (\frac {2 \left (-a^2-5 b^2\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {\left (5 a^2+b^2\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (2 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)-2 a (a-2 b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {48 b \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{\left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}\right )}{16 d (a-b)^2 (a+b)^2 (a+b \sec (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/16*((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(-a^2 - 5*b^2)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c +
d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2
]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (48*b*Cos[c + d*x]^2*EllipticPi[-(b/a), ArcSin
[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/((b + a*Cos[c + d*x])*(1
- Cos[c + d*x]^2)) + ((5*a^2 + b^2)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*
a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*Elli
pticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), A
rcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[S
qrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])
*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))))/((a - b)^2*(a + b)^2*d*(a + b*Sec[c + d*x])^3
) + ((b + a*Cos[c + d*x])^3*Sec[c + d*x]^(7/2)*(((5*a^2 + b^2)*Sin[c + d*x])/(4*a*(a^2 - b^2)^2) + (b^2*Sin[c
+ d*x])/(2*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (-7*a^2*b*Sin[c + d*x] + b^3*Sin[c + d*x])/(4*a*(a^2 - b^2)
^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^3)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 13.65, size = 1858, normalized size = 6.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*b^2/a^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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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