3.620 \(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=304 \[ \frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}-\frac {b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )} \]

[Out]

1/3*(2*a^2-5*b^2)*sin(d*x+c)/a^2/(a^2-b^2)/d/sec(d*x+c)^(1/2)+b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))/se
c(d*x+c)^(1/2)-b*(4*a^2-5*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^3/(a^2-b^2)/d+1/3*(2*a^4+16*a^2*b^2-15*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^4/(a^2-b^2)
/d-b^3*(7*a^2-5*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*a/(a+b),2
^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^4/(a-b)/(a+b)^2/d

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Rubi [A]  time = 0.68, antiderivative size = 304, 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 = {3847, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {\left (16 a^2 b^2+2 a^4-15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac {b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}-\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

-((b*(4*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a^2 - b^2)*d)) + (
(2*a^4 + 16*a^2*b^2 - 15*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b
^2)*d) - (b^3*(7*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])
/(a^4*(a - b)*(a + b)^2*d) + ((2*a^2 - 5*b^2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (b^2*Si
n[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(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 3847

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[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)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && 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 4104

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*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] && LeQ[n, -1]

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 {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\int \frac {-a^2+\frac {5 b^2}{2}+a b \sec (c+d x)-\frac {3}{2} b^2 \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {2 \int \frac {-\frac {3}{4} b \left (4 a^2-5 b^2\right )+\frac {1}{2} a \left (a^2+2 b^2\right ) \sec (c+d x)+\frac {1}{4} b \left (2 a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {2 \int \frac {-\frac {3}{4} a b \left (4 a^2-5 b^2\right )-\left (-\frac {3}{4} b^2 \left (4 a^2-5 b^2\right )-\frac {1}{2} a^2 \left (a^2+2 b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac {\left (b^3 \left (7 a^2-5 b^2\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\left (b \left (4 a^2-5 b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}-\frac {\left (b^3 \left (7 a^2-5 b^2\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}{2 a^4 \left (a^2-b^2\right )}\\ &=-\frac {b^3 \left (7 a^2-5 b^2\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)}}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\left (b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac {b \left (4 a^2-5 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)}}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b^3 \left (7 a^2-5 b^2\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)}}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 6.85, size = 634, normalized size = 2.09 \[ \frac {\frac {2 \left (4 a^3+8 a b^2\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {2 \left (5 b^3-8 a^2 b\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {\left (15 b^3-12 a^2 b\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (2 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)-2 a (a-2 b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{12 a^2 d (a-b) (a+b)}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\sin (2 (c+d x))}{3 a^2}+\frac {b^4 \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {b^3 \sin (c+d x)}{a^3 \left (b^2-a^2\right )}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

((2*(-8*a^2*b + 5*b^3)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(b/a), ArcSin[S
qrt[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)) + (2*(4*a^3 + 8*a*b^2)*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])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((
-12*a^2*b + 15*b^3)*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[Arc
Sin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*EllipticF[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), ArcSin[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[Sqrt[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)))/(12*a^2*(a - b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*((b^3*Sin[c +
d*x])/(a^3*(-a^2 + b^2)) + (b^4*Sin[c + d*x])/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + Sin[2*(c + d*x)]/(3*a^2
)))/d

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fricas [F]  time = 116.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sec \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{4} + 2 \, a b \sec \left (d x + c\right )^{3} + a^{2} \sec \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sec(d*x + c))/(b^2*sec(d*x + c)^4 + 2*a*b*sec(d*x + c)^3 + a^2*sec(d*x + c)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 11.49, size = 1064, normalized size = 3.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3/a^2*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-sin(1/2*d*x+1/2
*c)^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)-4*(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)*(EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(a^2+2*a*b+3*b^2)/a^4*(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))+8*b^3/a^3/(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^4*b^4*(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+si
n(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*(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))+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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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