3.831 \(\int \frac{\sqrt{\cos (c+d x)}}{(a+b \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=282 \[ -\frac{3 b \left (-11 a^2 b^2+8 a^4+5 b^4\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (-29 a^2 b^2+8 a^4+15 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (-38 a^2 b^2+35 a^4+15 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2} \]
[Out]
((8*a^4 - 29*a^2*b^2 + 15*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d) - (3*b*(8*a^4 - 11*a^2*b^2 +
5*b^4)*EllipticF[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + (b^2*(35*a^4 - 38*a^2*b^2 + 15*b^4)*EllipticPi[(2
*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + (b^2*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Sqrt[Cos[c
+ d*x]]*(a + b*Sec[c + d*x])^2) + (b^2*(11*a^2 - 5*b^2)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x
]]*(a + b*Sec[c + d*x]))
________________________________________________________________________________________
Rubi [A] time = 0.806788, antiderivative size = 282, normalized size of antiderivative
= 1., 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 = {4264, 3847, 4100, 4106, 3849, 2805, 3787, 3771, 2639, 2641}
\[ -\frac{3 b \left (-11 a^2 b^2+8 a^4+5 b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (-29 a^2 b^2+8 a^4+15 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (-38 a^2 b^2+35 a^4+15 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[Cos[c + d*x]]/(a + b*Sec[c + d*x])^3,x]
[Out]
((8*a^4 - 29*a^2*b^2 + 15*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d) - (3*b*(8*a^4 - 11*a^2*b^2 +
5*b^4)*EllipticF[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + (b^2*(35*a^4 - 38*a^2*b^2 + 15*b^4)*EllipticPi[(2
*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + (b^2*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Sqrt[Cos[c
+ d*x]]*(a + b*Sec[c + d*x])^2) + (b^2*(11*a^2 - 5*b^2)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x
]]*(a + b*Sec[c + d*x]))
Rule 4264
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]
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 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]
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 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 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 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 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]
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (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{1}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx\\ &=\frac{b^2 \sin (c+d x)}{2 a \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{-2 a^2+\frac{5 b^2}{2}+2 a b \sec (c+d x)-\frac{3}{2} b^2 \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \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} \left (8 a^4-29 a^2 b^2+15 b^4\right )-a b \left (4 a^2-b^2\right ) \sec (c+d x)+\frac{1}{4} b^2 \left (11 a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \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 (8 a^4-29 a^2 b^2+15 b^4\right )-\left (a^2 b \left (4 a^2-b^2\right )+\frac{1}{4} b \left (8 a^4-29 a^2 b^2+15 b^4\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 a^4 \left (a^2-b^2\right )^2}+\frac{\left (\left (\frac{1}{4} a^2 b^2 \left (11 a^2-5 b^2\right )+a^2 b^2 \left (4 a^2-b^2\right )+\frac{1}{4} b^2 \left (8 a^4-29 a^2 b^2+15 b^4\right )\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}{2 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}-\frac{\left (3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac{\left (\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}-\frac{\left (3 b \left (8 a^4-11 a^2 b^2+5 b^4\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac{\left (8 a^4-29 a^2 b^2+15 b^4\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac{3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac{b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac{b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.75075, size = 315, normalized size = 1.12 \[ \frac{\frac{-\frac{16 \left (4 a^2 b-b^3\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac{2 \left (-29 a^2 b^2+8 a^4+15 b^4\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-7 a^2 b^2+8 a^4+5 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}+\frac{4 b^2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (a \left (13 a^2-7 b^2\right ) \cos (c+d x)+11 a^2 b-5 b^3\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{16 a^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[Cos[c + d*x]]/(a + b*Sec[c + d*x])^3,x]
[Out]
((4*b^2*Sqrt[Cos[c + d*x]]*(11*a^2*b - 5*b^3 + a*(13*a^2 - 7*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(
b + a*Cos[c + d*x])^2) + ((2*(8*a^4 - 7*a^2*b^2 + 5*b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) -
(16*(4*a^2*b - b^3)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b)
- (2*(8*a^4 - 29*a^2*b^2 + 15*b^4)*(2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] - 2*b*(a + b)*EllipticF[A
rcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(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*a^2*d)
________________________________________________________________________________________
Maple [B] time = 7.339, size = 1957, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(1/2)/(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^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))+a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/a^4*b^4*(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*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*co
s(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)*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*si
n(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)))-12*b^2/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))-8/a^4*b^3*(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*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(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)*Elli
pticE(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
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos{\left (c + d x \right )}}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**(1/2)/(a+b*sec(d*x+c))**3,x)
[Out]
Integral(sqrt(cos(c + d*x))/(a + b*sec(c + d*x))**3, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate(sqrt(cos(d*x + c))/(b*sec(d*x + c) + a)^3, x)