\(\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx\) [185]

Optimal result

Integrand size = 29, antiderivative size = 61 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {c} f} \]

[Out]

2*arctan(a^(1/2)*c^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2))*a^(1/2)/f/c^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4019, 209} \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {c} f} \]

[In]

Int[Sqrt[a + a*Sec[e + f*x]]/Sqrt[c + d*Sec[e + f*x]],x]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[c]*Tan[e + f*x])/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])])/(Sqrt[c
]*f)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4019

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(2 a) \text {Subst}\left (\int \frac {1}{1+a c x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f} \\ & = \frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {c} f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {d+c \cos (e+f x)}}\right ) \sqrt {d+c \cos (e+f x)} \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (1+\sec (e+f x))}}{\sqrt {c} f \sqrt {c+d \sec (e+f x)}} \]

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]/Sqrt[c + d*Sec[e + f*x]],x]

[Out]

(Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[c]*Sin[(e + f*x)/2])/Sqrt[d + c*Cos[e + f*x]]]*Sqrt[d + c*Cos[e + f*x]]*Sec[(e +
f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])/(Sqrt[c]*f*Sqrt[c + d*Sec[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(165\) vs. \(2(49)=98\).

Time = 2.90 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.72

method	result	size
default	\(-\frac {2 \sqrt {2}\, \sqrt {-\left (c -d \right )^{4} c}\, \arctan \left (\frac {\left (c -d \right )^{2} c \sqrt {2}\, \sin \left (f x +e \right )}{\sqrt {-\left (c -d \right )^{4} c}\, \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {c +d \sec \left (f x +e \right )}\, \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \cos \left (f x +e \right )}{f \left (c^{2}-2 c d +d^{2}\right ) c \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}}\)	\(166\)

[In]

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

[Out]

-2/f*2^(1/2)*(-(c-d)^4*c)^(1/2)/(c^2-2*c*d+d^2)/c*arctan((c-d)^2*c*2^(1/2)/(-(c-d)^4*c)^(1/2)*sin(f*x+e)/(cos(
f*x+e)+1)/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2))*(c+d*sec(f*x+e))^(1/2)*(a*(sec(f*x+e)+1))^(1/2)*cos(f*x+
e)/(cos(f*x+e)+1)/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 206, normalized size of antiderivative = 3.38 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\left [\frac {\sqrt {-\frac {a}{c}} \log \left (-\frac {2 \, c \sqrt {-\frac {a}{c}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c \cos \left (f x + e\right )^{2} + a c - a d - {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{f}, -\frac {2 \, \sqrt {\frac {a}{c}} \arctan \left (\frac {\sqrt {\frac {a}{c}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a \sin \left (f x + e\right )}\right )}{f}\right ] \]

[In]

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

[Out]

[sqrt(-a/c)*log(-(2*c*sqrt(-a/c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e
))*cos(f*x + e)*sin(f*x + e) - 2*a*c*cos(f*x + e)^2 + a*c - a*d - (a*c + a*d)*cos(f*x + e))/(cos(f*x + e) + 1)
)/f, -2*sqrt(a/c)*arctan(sqrt(a/c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x +
e))*cos(f*x + e)/(a*sin(f*x + e)))/f]

Sympy [F]

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

[In]

integrate((a+a*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))/sqrt(c + d*sec(e + f*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for mor
e details)Is

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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