\(\int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx\) [269]

Optimal result

Integrand size = 23, antiderivative size = 62 \[ \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arcsinh}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}} \]

[Out]

-arcsinh(tan(d*x+c)/(1+sec(d*x+c)))*2^(1/2)/d+2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(1+sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3897, 3892, 221} \[ \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx=\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-\frac {\sqrt {2} \text {arcsinh}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

[In]

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

[Out]

-((Sqrt[2]*ArcSinh[Tan[c + d*x]/(1 + Sec[c + d*x])])/d) + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 + Sec[
c + d*x]])

Rule 221

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

Rule 3892

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

Rule 3897

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

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.45 \[ \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx=\frac {2 \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)+\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {-\tan ^2(c+d x)}} \]

[In]

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

[Out]

(2*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sin[c + d*x] + Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1
- Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[-Tan[c + d*x]^2])

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.53

method	result	size
default	\(-\frac {\sqrt {1+\sec \left (d x +c \right )}\, \left (\sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}\)	\(95\)

[In]

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

[Out]

-1/d*(1+sec(d*x+c))^(1/2)/sec(d*x+c)^(1/2)*(2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+
c)+1))^(1/2))*(-1/(cos(d*x+c)+1))^(1/2)+2*cot(d*x+c)-2*csc(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (56) = 112\).

Time = 0.30 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.32 \[ \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx=\frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/2*((sqrt(2)*cos(d*x + c) + sqrt(2))*log(-(2*sqrt(2)*sqrt((cos(d*x + c) + 1)/cos(d*x + c))*sqrt(cos(d*x + c))
*sin(d*x + c) + cos(d*x + c)^2 - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*sqrt((cos(d*x
+ c) + 1)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.63 \[ \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx=-\frac {\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(
cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 4*sqrt(2)*sin(1/2*d*x + 1/2*c)
)/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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