3.3.48 \(\int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\) [248]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3897, 3893, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 3893

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 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*} \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 \text {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*}

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Mathematica [A]
time = 0.20, size = 102, normalized size = 1.10 \begin {gather*} \frac {\left (\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+\frac {2 \sqrt {1-\sec (c+d x)}}{\sqrt {\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

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 = 0.12, size = 102, normalized size = 1.10

method	result	size
default	\(-\frac {\left (-\arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 \cos \left (d x +c \right )-2\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, a}\)	\(102\)

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,method=_RETURNVERBOSE)

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

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Maxima [A]
time = 0.55, size = 104, normalized size = 1.12 \begin {gather*} -\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 {gather*}

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 = 3.19, size = 281, normalized size = 3.02 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sqrt {\sec {\left (c + d x \right )}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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