3.759 \(\int \frac{\sqrt{\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=307 \[ -\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 b \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a^2 d \sqrt{a+b} \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a d \sqrt{a+b} \sqrt{\sec (c+d x)}} \]

[Out]

(2*b*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]]
)], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*Sqrt[a
+ b]*d*Sqrt[Sec[c + d*x]]) + (2*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq
rt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c +
d*x]))/(a - b)])/(a*Sqrt[a + b]*d*Sqrt[Sec[c + d*x]]) - (2*b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*S
qrt[a + b*Cos[c + d*x]])

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Rubi [A]  time = 0.479046, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4222, 2800, 2998, 2816, 2994} \[ -\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 b \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a^2 d \sqrt{a+b} \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{a d \sqrt{a+b} \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]]
)], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*Sqrt[a
+ b]*d*Sqrt[Sec[c + d*x]]) + (2*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq
rt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c +
d*x]))/(a - b)])/(a*Sqrt[a + b]*d*Sqrt[Sec[c + d*x]]) - (2*b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*S
qrt[a + b*Cos[c + d*x]])

Rule 4222

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

Rule 2800

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[(2
*b*Cos[e + f*x])/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[e + f*x]]), x] + Dist[d/(a^2 - b^2), Int[(
b + a*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{a, b, d, e, f}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2998

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 2994

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx &=\left (\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))^{3/2}} \, dx\\ &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{b+a \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{a^2-b^2}\\ &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left ((a-b) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx}{a^2-b^2}+\frac{\left (b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 b \sqrt{\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt{a+b} d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{a \sqrt{a+b} d \sqrt{\sec (c+d x)}}-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.63532, size = 237, normalized size = 0.77 \[ \frac{2 \sqrt{\sec (c+d x)} \left (b (b-a) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right )+2 a (a+b) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{1}{\sec (c+d x)+1}} \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )-2 b (a+b) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{1}{\sec (c+d x)+1}} \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{a d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Sec[c + d*x]]*(-2*b*(a + b)*Cos[(c + d*x)/2]^2*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]
*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] + 2*a*(a + b)*Cos[(c + d*
x)/2]^2*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(
a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] + b*(-a + b)*Cos[c + d*x]*Tan[(c + d*x)/2]))/(a*(a^2 - b^2)*d*Sqrt[a + b
*Cos[c + d*x]])

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Maple [B]  time = 0.636, size = 832, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/(a+b)/(a-b)/a*(1/cos(d*x+c))^(1/2)/(a+b*cos(d*x+c))^(1/2)*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+
b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^
(1/2))*a^2+cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellipt
icF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*
sin(d*x+c)*a*b-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*si
n(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(
a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*sin
(d*x+c)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*
x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b*sin(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+
c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b*sin(d*x+c)-(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(
-(a-b)/(a+b))^(1/2))*b^2*sin(d*x+c)-cos(d*x+c)^2*a*b+b^2*cos(d*x+c)^2+cos(d*x+c)*a*b-cos(d*x+c)*b^2)/sin(d*x+c
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec{\left (c + d x \right )}}}{\left (a + b \cos{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sec(c + d*x))/(a + b*cos(c + d*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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