3.9.33 \(\int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\) [833]

Optimal. Leaf size=246 \[ -\frac {\left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a \left (a^2-b^2\right )^2 d}-\frac {b \left (7 a^2-b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 (a-b)^2 (a+b)^3 d}+\frac {a \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \]

[Out]

-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))/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
))/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))/a^2/(a-b)^2/(a+b)^3/d+1/2*a*sin(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2/cos(d
*x+c)^(1/2)+3/4*(a^2+b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.45, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4349, 3929, 4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {b \left (7 a^2-b^2\right ) 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 ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {a \sin (c+d x)}{2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\\ &=\frac {a \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 (\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 {\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 \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (3 a^4+10 a^2 b^2-b^4\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 (b \left (7 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (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 \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 a \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 (a-b)^2 (a+b)^3 d}+\frac {a \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}-\frac {\left (b \left (7 a^2-b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (5 a^2+b^2\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 ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a \left (a^2-b^2\right )^2 d}-\frac {b \left (7 a^2-b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 (a-b)^2 (a+b)^3 d}+\frac {a \sin (c+d x)}{2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {3 \left (a^2+b^2\right ) \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 11.34, size = 272, normalized size = 1.11 \begin {gather*} \frac {\frac {4 \sqrt {\cos (c+d x)} \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 (b+a \cos (c+d x))^2}-\frac {-\frac {2 \left (a^2+5 b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+24 b \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )+\frac {2 \left (5 a^2+b^2\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*Sqrt[Cos[c + d*x]]*(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*(b + a*Co
s[c + d*x])^2) - ((-2*(a^2 + 5*b^2)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 24*b*(2*EllipticF[(c
+ d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + (2*(5*a^2 + b^2)*(-2*a*b*EllipticE[A
rcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*Ellipti
cPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2)
)/(16*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1857\) vs. \(2(310)=620\).
time = 0.61, size = 1858, normalized size = 7.55

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(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)*(-4*b/a^2*(1/b*a^2/(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*cos(1/2*d*x+1/2*c)^2*a-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(c
os(1/2*d*x+1/2*c),2^(1/2))+1/2/b*a/(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/b*a/(a^2-b^2)*(si
n(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)))-2/a/(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^2*b^2*(1/2/b*a^2/(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*cos(1/2*d*x+1/2*c)^2*a-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*cos(1/2*d*x+1/2*c)^2*a-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)*Ellipt
icF(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*(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))))/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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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