3.727 \(\int \frac{1}{(a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)}} \, dx\)
Optimal. Leaf size=317 \[ \frac{b^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac{b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{3 \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b d \left (a^2-b^2\right )^2}+\frac{\left (5 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}-\frac{\left (10 a^2 b^2+3 a^4-b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b d (a-b)^2 (a+b)^3} \]
[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) + (3*(a^
2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b*(a^2 - b^2)^2*d) - ((3*a^4 + 10
*a^2*b^2 - b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a*(a - b)^
2*b*(a + b)^3*d) + (b^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Sec[c + d*x])^2) - (b*(7*a^
2 - b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*a*(a^2 - b^2)^2*d*(b + a*Sec[c + d*x]))
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Rubi [A] time = 0.751586, antiderivative size = 317, 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 = {3238, 3845, 4100, 4106, 3849, 2805, 3787, 3771, 2639, 2641}
\[ \frac{b^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac{b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}+\frac{3 \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b d \left (a^2-b^2\right )^2}+\frac{\left (5 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}-\frac{\left (10 a^2 b^2+3 a^4-b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a b d (a-b)^2 (a+b)^3} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + b*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]),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) + (3*(a^
2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b*(a^2 - b^2)^2*d) - ((3*a^4 + 10
*a^2*b^2 - b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a*(a - b)^
2*b*(a + b)^3*d) + (b^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Sec[c + d*x])^2) - (b*(7*a^
2 - b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*a*(a^2 - b^2)^2*d*(b + a*Sec[c + d*x]))
Rule 3238
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] && !IntegerQ[m] && IntegersQ[n, p]
Rule 3845
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))
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{1}{(a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)}} \, dx &=\int \frac{\sec ^{\frac{5}{2}}(c+d x)}{(b+a \sec (c+d x))^3} \, dx\\ &=\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac{\int \frac{-\frac{b^2}{2}-2 a b \sec (c+d x)+\frac{1}{2} \left (4 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{b \left (7 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} b^2 \left (5 a^2+b^2\right )+a b \left (2 a^2+b^2\right ) \sec (c+d x)-\frac{1}{4} b^2 \left (7 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}\\ &=\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{b \left (7 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} b^3 \left (5 a^2+b^2\right )-\left (-a b^2 \left (2 a^2+b^2\right )+\frac{1}{4} a b^2 \left (5 a^2+b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 a b^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)}{b+a \sec (c+d x)} \, dx}{8 a b \left (a^2-b^2\right )^2}\\ &=\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{b \left (7 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\left (3 \left (a^2+b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 b \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)} (a+b \cos (c+d x))} \, dx}{8 a b \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 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a (a-b)^2 b (a+b)^3 d}+\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{b \left (7 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac{\left (3 \left (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 b \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{3 \left (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 b \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 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a (a-b)^2 b (a+b)^3 d}+\frac{b^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac{b \left (7 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 6.61226, size = 401, normalized size = 1.26 \[ -\frac{\frac{4 b \sin (c+d x) \left (b \left (5 a^2+b^2\right ) \cos (c+d x)+7 a^3-a b^2\right )}{a \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} (a+b \cos (c+d x))^2}+\frac{2 \cot (c+d x) \left (-2 b \left (-7 a^2 b+5 a^3+a b^2+b^3\right ) \sqrt{-\tan ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 a b \left (5 a^2+b^2\right ) \sqrt{-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+20 a^2 b^2 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-5 a^3 b \sec ^{\frac{3}{2}}(c+d x)+5 a^3 b \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)+6 a^4 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-a b^3 \sec ^{\frac{3}{2}}(c+d x)+a b^3 \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)-2 b^4 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a^2 b (a-b)^2 (a+b)^2}}{16 d} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + b*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]),x]
[Out]
-((4*b*(7*a^3 - a*b^2 + b*(5*a^2 + b^2)*Cos[c + d*x])*Sin[c + d*x])/(a*(a^2 - b^2)^2*(a + b*Cos[c + d*x])^2*Sq
rt[Sec[c + d*x]]) + (2*Cot[c + d*x]*(-5*a^3*b*Sec[c + d*x]^(3/2) - a*b^3*Sec[c + d*x]^(3/2) + 5*a^3*b*Cos[2*(c
+ d*x)]*Sec[c + d*x]^(3/2) + a*b^3*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + 2*a*b*(5*a^2 + b^2)*EllipticE[ArcSin
[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*b*(5*a^3 - 7*a^2*b + a*b^2 + b^3)*EllipticF[ArcSin[Sqrt[Se
c[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 6*a^4*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[
c + d*x]^2] + 20*a^2*b^2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*b^4*Ell
ipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(a^2*(a - b)^2*b*(a + b)^2))/(16*d)
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Maple [B] time = 11.668, size = 1736, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b*(-1/a*b^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*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(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)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))-1/2*b/(a^2-b^2)/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*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))+1/2*b/(a^2-b^2)/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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(
a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^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*b/(a-b),2^(1/2)))-2*a/b*
(-1/2/a*b^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*b*cos(1/2*d*x
+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^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*b*cos(1/2*d*x+1/2*c)^2+a-b)-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))+
1/4/(a+b)/(a^2-b^2)/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^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(co
s(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(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*b^3/a^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*b/(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*b^3/a^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))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b*(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*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^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*b/(a-b),2^(1/2))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^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*b/(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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/((b*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)
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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(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [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(1/(a+b*cos(d*x+c))**3/sec(d*x+c)**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate(1/((b*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)