3.93 \(\int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {2 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3792} \[ \frac {2 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3792

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

Rubi steps

\begin {align*} \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx &=\frac {2 a \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 29, normalized size = 1.12 \[ \frac {2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/d

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fricas [A]  time = 0.75, size = 41, normalized size = 1.58 \[ \frac {2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)

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giac [B]  time = 7.08, size = 62, normalized size = 2.38 \[ -\frac {2 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*
c)^2 - a)*d)

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maple [A]  time = 0.90, size = 42, normalized size = 1.62 \[ -\frac {2 \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))/sin(d*x+c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.19, size = 41, normalized size = 1.58 \[ \frac {2\,\sin \left (c+d\,x\right )\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}}{d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*sin(c + d*x)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2))/(d*(cos(c + d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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