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

Optimal. Leaf size=62 \[ \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3812, 3807, 215} \[ \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

Antiderivative was successfully verified.

[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 215

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 3807

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 3812

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*Csc
[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*} \int \frac {1}{\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)}}-\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} \operatorname {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} \sinh ^{-1}\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*}

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Mathematica [A]  time = 0.23, size = 90, normalized size = 1.45 \[ \frac {2 \sin (c+d x) \sqrt {-((\sec (c+d x)-1) \sec (c+d x))}+\sqrt {2} \tan (c+d x) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )}{d \sqrt {-\tan ^2(c+d x)}} \]

Warning: Unable to verify antiderivative.

[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])

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fricas [B]  time = 0.70, size = 144, normalized size = 2.32 \[ \frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 1.50, size = 98, normalized size = 1.58 \[ -\frac {\left (-\arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 \cos \left (d x +c \right )-2\right ) \sqrt {\frac {1+\cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{d \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.09, size = 101, normalized size = 1.63 \[ -\frac {\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}+1}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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