3.5.92 \(\int \frac {1}{a+b \sec (c+d x)} \, dx\) [492]

Optimal. Leaf size=59 \[ \frac {x}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d} \]

[Out]

x/a-2*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3868, 2738, 214} \begin {gather*} \frac {x}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(-1),x]

[Out]

x/a - (2*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)

Rule 214

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a/b)*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 60, normalized size = 1.02 \begin {gather*} \frac {\frac {c}{d}+x+\frac {2 b \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^(-1),x]

[Out]

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

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Maple [A]
time = 0.09, size = 65, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(65\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(65\)
risch	\(\frac {x}{a}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d a}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(2/a*arctan(tan(1/2*d*x+1/2*c))-2*b/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(
1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [A]
time = 2.71, size = 230, normalized size = 3.90 \begin {gather*} \left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d x + \sqrt {a^{2} - b^{2}} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x - \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a + b*sec(c + d*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (50) = 100\).
time = 0.48, size = 218, normalized size = 3.69 \begin {gather*} \frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} {\left (a - 2 \, b\right )} {\left | -a + b \right |} - \sqrt {-a^{2} + b^{2}} {\left | a \right |} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} + \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )} {\left (a - 2 \, b + {\left | a \right |}\right )}}{a^{2} - b {\left | a \right |}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((sqrt(-a^2 + b^2)*(a - 2*b)*abs(-a + b) - sqrt(-a^2 + b^2)*abs(a)*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1
/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b + sqrt((a + b)*(a - b) + b^2))/(a - b))))/((a^2 - 2*a*b + b^2)*a^2
+ (a^2*b - 2*a*b^2 + b^3)*abs(a)) + (pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b -
sqrt((a + b)*(a - b) + b^2))/(a - b))))*(a - 2*b + abs(a))/(a^2 - b*abs(a)))/d

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Mupad [B]
time = 1.09, size = 186, normalized size = 3.15 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-2\,a\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (a\,b-a^2\right )}\right )}{a\,d\,\sqrt {a^2-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cos(c + d*x)),x)

[Out]

(2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d) + (2*b*atanh((2*b^4*sin(c/2 + (d*x)/2) + a^2*sin(c/2 + (
d*x)/2)*(a^2 - b^2) + 2*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 2*a*b^3*sin(c/2 + (d*x)/2) - 2*a*b*sin(c/2 + (d*x
)/2)*(a^2 - b^2))/(a*cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*(a*b - a^2))))/(a*d*(a^2 - b^2)^(1/2))

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