3.706 \(\int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=52 \[ x-\frac {4 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}} \]

[Out]

x-4*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4042, 3919, 3831, 2659, 208} \[ x-\frac {4 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^2,x]

[Out]

x - (4*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d)

Rule 208

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

Rule 2659

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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4042

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=-\int \frac {-a+b \sec (c+d x)}{a+b \sec (c+d x)} \, dx\\ &=x-(2 b) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx\\ &=x-2 \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx\\ &=x-\frac {4 \operatorname {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 )}{d}\\ &=x-\frac {4 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 56, normalized size = 1.08 \[ \frac {4 b \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {c}{d}+x \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^2,x]

[Out]

c/d + x + (4*b*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d)

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fricas [A]  time = 0.52, size = 226, normalized size = 4.35 \[ \left [\frac {{\left (a^{2} - b^{2}\right )} d x + \sqrt {a^{2} - b^{2}} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{{\left (a^{2} - b^{2}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x - 2 \, \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{2} - b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((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^2 -
b^2)*d), ((a^2 - b^2)*d*x - 2*sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*si
n(d*x + c))))/((a^2 - b^2)*d)]

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giac [B]  time = 0.24, size = 235, normalized size = 4.52 \[ -\frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} {\left (a + b\right )} {\left | a \right |} {\left | -a + b \right |} - {\left (a^{2} - 3 \, a b\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} - \frac {{\left (a^{2} - 3 \, a b + a {\left | a \right |} + b {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{a^{2} - b {\left | a \right |}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.80, size = 61, normalized size = 1.17 \[ -\frac {4 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d*b/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+2/d*arctan(tan(1/2*d*x+1/2*c)
)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.98, size = 182, normalized size = 3.50 \[ x-\frac {4\,b\,\mathrm {atanh}\left (\frac {8\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (a\,\left (a^2-b^2\right )+b\,\left (a^2-b^2\right )+4\,a\,b^2-2\,a^2\,b-2\,b^3\right )}\right )}{d\,\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - (4*b*atanh((8*b^4*sin(c/2 + (d*x)/2) + a^2*sin(c/2 + (d*x)/2)*(a^2 - b^2) + 5*b^2*sin(c/2 + (d*x)/2)*(a^2
- b^2) - 8*a*b^3*sin(c/2 + (d*x)/2) - 2*a*b*sin(c/2 + (d*x)/2)*(a^2 - b^2))/(cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1
/2)*(a*(a^2 - b^2) + b*(a^2 - b^2) + 4*a*b^2 - 2*a^2*b - 2*b^3))))/(d*(a^2 - b^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a - b*sec(c + d*x))/(a + b*sec(c + d*x)), x)

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