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

Optimal. Leaf size=93 \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d))
+ (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.111831, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3812, 3808, 206} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d))
+ (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])

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)]

Rule 3808

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{2 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.208703, size = 102, normalized size = 1.1 \[ \frac{\tan (c+d x) \left (\frac{2 \sqrt{1-\sec (c+d x)}}{\sqrt{\sec (c+d x)}}+\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.168, size = 100, normalized size = 1.1 \begin{align*}{\frac{1}{ad\sin \left ( dx+c \right ) } \left ( \arctan \left ({\frac{\sin \left ( dx+c \right ) }{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -2\,\cos \left ( dx+c \right ) +2 \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.26689, size = 140, normalized size = 1.51 \begin{align*} -\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 \, \sqrt{a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 2.00059, size = 764, normalized size = 8.22 \begin{align*} \left [\frac{\frac{\sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}} + 4 \, \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{\sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*(a*cos(d*x + c) + a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq
rt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a) + 4
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), (sqrt(2)*(
a*cos(d*x + c) + a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x
+ c))/sin(d*x + c)) + 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x
+ c) + a*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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