3.721 \(\int \frac{1}{(a+b \cos (c+d x))^2 \sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=208 \[ -\frac{b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac{a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \left (a^2-b^2\right )}-\frac{\left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d (a-b) (a+b)^2} \]

[Out]

(Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a^2 - b^2)*d) + (a*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*(a^2 - b^2)*d) - ((a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)
/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b*(a + b)^2*d) - (b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((
a^2 - b^2)*d*(b + a*Sec[c + d*x]))

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Rubi [A]  time = 0.419767, antiderivative size = 208, normalized size of antiderivative = 1., 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 = {3238, 3844, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac{a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \left (a^2-b^2\right )}-\frac{\left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a^2 - b^2)*d) + (a*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*(a^2 - b^2)*d) - ((a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)
/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b*(a + b)^2*d) - (b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((
a^2 - b^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 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 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))^2 \sqrt{\sec (c+d x)}} \, dx &=\int \frac{\sec ^{\frac{3}{2}}(c+d x)}{(b+a \sec (c+d x))^2} \, dx\\ &=-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac{\int \frac{-\frac{b}{2}-a \sec (c+d x)+\frac{1}{2} b \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{a^2-b^2}\\ &=-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac{\int \frac{-\frac{b^2}{2}-\frac{1}{2} a b \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{b^2 \left (a^2-b^2\right )}-\frac{\left (a^2+b^2\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{\int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}+\frac{a \int \sqrt{\sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}-\frac{\left (\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)} (a+b \cos (c+d x))} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2+b^2\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)}}{(a-b) b (a+b)^2 d}-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )}+\frac{\left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{\sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{\left (a^2-b^2\right ) d}+\frac{a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b \left (a^2-b^2\right ) d}-\frac{\left (a^2+b^2\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)}}{(a-b) b (a+b)^2 d}-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.59814, size = 580, normalized size = 2.79 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{a \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\sin (c+d x)}{a^2-b^2}\right )}{d}+\frac{\frac{\sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}-\frac{8 a \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}-\frac{2 b \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}}{4 d (a-b) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 7.491, size = 713, normalized size = 3.4 \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))^2/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)*(-4/(-2*a*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)*EllipticPi(cos(1/2*d*x+1/2
*c),-2*b/(a-b),2^(1/2))-2*a/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*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*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*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))))/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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \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))^2/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \cos{\left (c + d x \right )}\right )^{2} \sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2} \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))^2/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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