\(\int \sqrt {a+b \sec (c+d x)} \, dx\) [324]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 125 \[ \int \sqrt {a+b \sec (c+d x)} \, dx=-\frac {2 \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a}{a+b},\arcsin \left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right ),\frac {a-b}{a+b}\right ) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (1+\sec (c+d x))}{a+b \sec (c+d x)}} (a+b \sec (c+d x))}{\sqrt {a+b} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3865} \[ \int \sqrt {a+b \sec (c+d x)} \, dx=-\frac {2 \cot (c+d x) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (\sec (c+d x)+1)}{a+b \sec (c+d x)}} (a+b \sec (c+d x)) \operatorname {EllipticPi}\left (\frac {a}{a+b},\arcsin \left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right ),\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]

[In]

Int[Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(-2*Cot[c + d*x]*EllipticPi[a/(a + b), ArcSin[Sqrt[a + b]/Sqrt[a + b*Sec[c + d*x]]], (a - b)/(a + b)]*Sqrt[-((
b*(1 - Sec[c + d*x]))/(a + b*Sec[c + d*x]))]*Sqrt[(b*(1 + Sec[c + d*x]))/(a + b*Sec[c + d*x])]*(a + b*Sec[c +
d*x]))/(Sqrt[a + b]*d)

Rule 3865

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*((a + b*Csc[c + d*x])/(d*Rt[a + b, 2]*Cot[
c + d*x]))*Sqrt[b*((1 + Csc[c + d*x])/(a + b*Csc[c + d*x]))]*Sqrt[(-b)*((1 - Csc[c + d*x])/(a + b*Csc[c + d*x]
))]*EllipticPi[a/(a + b), ArcSin[Rt[a + b, 2]/Sqrt[a + b*Csc[c + d*x]]], (a - b)/(a + b)], x] /; FreeQ[{a, b,
c, d}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a}{a+b},\arcsin \left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right ),\frac {a-b}{a+b}\right ) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (1+\sec (c+d x))}{a+b \sec (c+d x)}} (a+b \sec (c+d x))}{\sqrt {a+b} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.21 \[ \int \sqrt {a+b \sec (c+d x)} \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left ((-a+b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+2 a \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sqrt {a+b \sec (c+d x)}}{d (b+a \cos (c+d x))} \]

[In]

Integrate[Sqrt[a + b*Sec[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 6.57 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.46

method result size
default \(\frac {2 \left (\cos \left (d x +c \right )+1\right ) \left (\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a -\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b -2 a \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a +b \sec \left (d x +c \right )}}{d \left (b +a \cos \left (d x +c \right )\right )}\) \(182\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c) + a), x)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a + b*sec(c + d*x)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b/cos(c + d*x))^(1/2),x)

[Out]

int((a + b/cos(c + d*x))^(1/2), x)