3.612 \(\int \frac{1}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx\)

Optimal. Leaf size=135 \[ -\frac{2 b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a^2 d}+\frac{2 b^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.207321, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3852, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{2 b^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}-\frac{2 b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3852

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[b^2/(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 - b*Csc[e + f*x])/Sqrt[d*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, 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}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx &=\frac{\int \frac{a-b \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2}+\frac{b^2 \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{a^2}\\ &=\frac{\int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{a}-\frac{b \int \sqrt{\sec (c+d x)} \, dx}{a^2}+\frac{\left (b^2 \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}{a^2}\\ &=\frac{2 b^2 \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^2 (a+b) d}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{a}-\frac{\left (b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{a^2}\\ &=\frac{2 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a d}-\frac{2 b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 b^2 \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^2 (a+b) d}\\ \end{align*}

Mathematica [A]  time = 6.23237, size = 178, normalized size = 1.32 \[ \frac{\cot (c+d x) \left (2 a \sqrt{-\tan ^2(c+d x)} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+2 b \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+a \sec ^{\frac{7}{2}}(c+d x)-a \sec ^{\frac{3}{2}}(c+d x)+a \cos (2 (c+d x)) \sec ^{\frac{7}{2}}(c+d x)-a \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)-2 a \sqrt{-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]*(-(a*Sec[c + d*x]^(3/2)) - a*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + a*Sec[c + d*x]^(7/2) + a*Cos[
2*(c + d*x)]*Sec[c + d*x]^(7/2) - 2*a*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 2*a*El
lipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 2*b*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*
x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(a^2*d)

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Maple [A]  time = 1.415, size = 226, normalized size = 1.7 \begin{align*} 2\,{\frac{\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}}{ \left ( a-b \right ){a}^{2}\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) ab-{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{2}+{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}-{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) ab+{b}^{2}{\it EllipticPi} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,2\,{\frac{a}{a-b}},\sqrt{2} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+EllipticE(cos(
1/2*d*x+1/2*c),2^(1/2))*a^2-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+b^2*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a
-b),2^(1/2)))/a^2/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \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*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(d*x + c) + a)*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*sec(d*x+c))/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 \sec{\left (c + d x \right )}\right ) \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*sec(d*x+c))/sec(d*x+c)**(1/2),x)

[Out]

Integral(1/((a + b*sec(c + d*x))*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 \sec \left (d x + c\right ) + a\right )} \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*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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