3.224 \(\int \frac {1}{\sqrt {a-a \sec ^2(c+d x)}} \, dx\)

Optimal. Leaf size=32 \[ \frac {\tan (c+d x) \log (\sin (c+d x))}{d \sqrt {-a \tan ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4121, 3658, 3475} \[ \frac {\tan (c+d x) \log (\sin (c+d x))}{d \sqrt {-a \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a - a*Sec[c + d*x]^2],x]

[Out]

(Log[Sin[c + d*x]]*Tan[c + d*x])/(d*Sqrt[-(a*Tan[c + d*x]^2)])

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 4121

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(b*tan[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a-a \sec ^2(c+d x)}} \, dx &=\int \frac {1}{\sqrt {-a \tan ^2(c+d x)}} \, dx\\ &=\frac {\tan (c+d x) \int \cot (c+d x) \, dx}{\sqrt {-a \tan ^2(c+d x)}}\\ &=\frac {\log (\sin (c+d x)) \tan (c+d x)}{d \sqrt {-a \tan ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 40, normalized size = 1.25 \[ \frac {\tan (c+d x) (\log (\tan (c+d x))+\log (\cos (c+d x)))}{d \sqrt {-a \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a - a*Sec[c + d*x]^2],x]

[Out]

((Log[Cos[c + d*x]] + Log[Tan[c + d*x]])*Tan[c + d*x])/(d*Sqrt[-(a*Tan[c + d*x]^2)])

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fricas [A]  time = 2.87, size = 56, normalized size = 1.75 \[ -\frac {\sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{a d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-a \sec \left (d x + c\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.96, size = 76, normalized size = 2.38 \[ -\frac {\left (-\ln \left (-\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+\ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right )\right ) \sin \left (d x +c \right )}{d \sqrt {-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{2}}}\, \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 37, normalized size = 1.16 \[ -\frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt {-a}} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt {-a}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\sqrt {a-\frac {a}{{\cos \left (c+d\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - a/cos(c + d*x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- a \sec ^{2}{\left (c + d x \right )} + a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(-a*sec(c + d*x)**2 + a), x)

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