\(\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\) [858]

Optimal result

Integrand size = 25, antiderivative size = 67 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4349, 3943, 2742, 2740} \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

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

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[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 4349

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

Rubi steps \begin{align*} \text {integral}& = \left (\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 \\ & = \frac {\sqrt {b+a \cos (c+d x)} \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\ & = \frac {\sqrt {\frac {b+a \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\ & = \frac {2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.96 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 i \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.01

method	result	size
default	\(\frac {2 \operatorname {EllipticF}\left (\sqrt {\frac {a -b}{a +b}}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), \sqrt {-\frac {a +b}{a -b}}\right ) \sqrt {a +b \sec \left (d x +c \right )}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \sqrt {\cos \left (d x +c \right )}}{d \sqrt {\frac {a -b}{a +b}}\, \left (b +a \cos \left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(135\)

[In]

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

[Out]

2/d/((a-b)/(a+b))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*(a+b*sec(
d*x+c))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)
+1))^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.18 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\frac {-i \, \sqrt {2} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + i \, \sqrt {2} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )}{a d} \]

[In]

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

[Out]

(-I*sqrt(2)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x
+ c) + 3*I*a*sin(d*x + c) + 2*b)/a) + I*sqrt(2)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9
*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a))/(a*d)

Sympy [F]

\[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \sec {\left (c + d x \right )}} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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