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

Optimal. Leaf size=235 \[ -\frac{2 (2 A b-a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{a^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A+a b B-2 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.719584, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {2955, 4030, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A+a b B-2 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{2 (2 A b-a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 4030

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*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)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

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

Mathematica [C]  time = 15.1463, size = 365, normalized size = 1.55 \[ \frac{2 (a \cos (c+d x)+b) (A+B \sec (c+d x)) \left (-a b (a B-A b) \sin (c+d x)+\frac{\left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i a (a+b) (a (A+B)-2 A b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )+\left (a^2 A+a b B-2 A b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)+i (a+b) \left (a^2 A+a b B-2 A b^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{\sec ^{\frac{3}{2}}(c+d x)}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} (A \cos (c+d x)+B)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x])*(-(a*b*(-(A*b) + a*B)*Sin[c + d*x]) + ((Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^(3/2)*(I*(a + b)*(a^2*A - 2*A*b^2 + a*b*B)*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Se
c[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - I*a*(a + b)*(-2*A*b + a*(A + B))*El
lipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c +
d*x)/2]^2)/(a + b)] + (a^2*A - 2*A*b^2 + a*b*B)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/
2]))/Sec[c + d*x]^(3/2)))/(a^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(3/2
))

________________________________________________________________________________________

Maple [B]  time = 0.407, size = 889, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1+cos(d*x+c))*(cos(d*x+c)+1)^2*(A*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*((a-b)/(a+b))^(1/2)*a^2+A*cos(d*
x+c)^2*((a-b)/(a+b))^(1/2)*a*b*(1/(cos(d*x+c)+1))^(1/2)+A*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(
1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2-2*A*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*sin(d*x+c)*b^2-A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*a^2-2*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c)
,(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*a*b-A*cos(d*x+c)*((a-b)/(a+b
))^(1/2)*a^2*(1/(cos(d*x+c)+1))^(1/2)+2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2*(1/(cos(d*x+c)+1))^(1/2)+B*(1/(a+
b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-
b))^(1/2))*sin(d*x+c)*a*b+B*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))
^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2-B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b*(1/(cos(d*x+c
)+1))^(1/2)-A*((a-b)/(a+b))^(1/2)*a*b*(1/(cos(d*x+c)+1))^(1/2)-2*A*((a-b)/(a+b))^(1/2)*b^2*(1/(cos(d*x+c)+1))^
(1/2)+B*((a-b)/(a+b))^(1/2)*a*b*(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)
*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)/a^2/(b+a*cos(d*x+c))/(a-b)/sin(d*x+c)^3

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \sec \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((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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