3.520 \(\int \frac {a+b \sec (d+e x)}{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \, dx\)

Optimal. Leaf size=92 \[ -\frac {a^2 \tan (d+e x)}{b e \left (a^2 \sec (d+e x)+a b\right )}-\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b^2 e}+\frac {a x}{b^2} \]

[Out]

a*x/b^2-2*arctan((a-b)^(1/2)*tan(1/2*e*x+1/2*d)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^2/e-a^2*tan(e*x+d)/b/e/
(a*b+a^2*sec(e*x+d))

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Rubi [A]  time = 0.30, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4172, 3923, 3919, 3831, 2659, 205} \[ -\frac {a^2 \tan (d+e x)}{b e \left (a^2 \sec (d+e x)+a b\right )}-\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b^2 e}+\frac {a x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/b^2 - (2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/(b^2*e) - (a^2*Tan[
d + e*x])/(b*e*(a*b + a^2*Sec[d + e*x]))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4172

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {a+b \sec (d+e x)}{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \, dx &=\left (4 a^2\right ) \int \frac {a+b \sec (d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^2} \, dx\\ &=-\frac {a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}+\frac {\int \frac {4 a^3 \left (a^2-b^2\right )+4 a^2 b \left (a^2-b^2\right ) \sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=\frac {a x}{b^2}-\frac {a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac {\left (2 a \left (a^2-b^2\right )\right ) \int \frac {\sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{b^2}\\ &=\frac {a x}{b^2}-\frac {a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac {\left (a^2-b^2\right ) \int \frac {1}{1+\frac {b \cos (d+e x)}{a}} \, dx}{a b^2}\\ &=\frac {a x}{b^2}-\frac {a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac {\left (2 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {b}{a}+\left (1-\frac {b}{a}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{a b^2 e}\\ &=\frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b^2 e}-\frac {a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 97, normalized size = 1.05 \[ \frac {2 \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {b^2-a^2}}\right )+\frac {a (a d+a e x-b \sin (d+e x)+b (d+e x) \cos (d+e x))}{a+b \cos (d+e x)}}{b^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 279, normalized size = 3.03 \[ \left [\frac {2 \, a b e x \cos \left (e x + d\right ) + 2 \, a^{2} e x - 2 \, a b \sin \left (e x + d\right ) + \sqrt {-a^{2} + b^{2}} {\left (b \cos \left (e x + d\right ) + a\right )} \log \left (\frac {2 \, a b \cos \left (e x + d\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right )}{2 \, {\left (b^{3} e \cos \left (e x + d\right ) + a b^{2} e\right )}}, \frac {a b e x \cos \left (e x + d\right ) + a^{2} e x - a b \sin \left (e x + d\right ) - \sqrt {a^{2} - b^{2}} {\left (b \cos \left (e x + d\right ) + a\right )} \arctan \left (-\frac {a \cos \left (e x + d\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (e x + d\right )}\right )}{b^{3} e \cos \left (e x + d\right ) + a b^{2} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 145, normalized size = 1.58 \[ {\left (\frac {{\left (x e + d\right )} a}{b^{2}} - \frac {2 \, a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{{\left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + a + b\right )} b} - \frac {2 \, {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{b^{2}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)*a/b^2 - 2*a*tan(1/2*x*e + 1/2*d)/((a*tan(1/2*x*e + 1/2*d)^2 - b*tan(1/2*x*e + 1/2*d)^2 + a + b)*b)
- 2*(pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/2*d)
)/sqrt(a^2 - b^2)))*sqrt(a^2 - b^2)/b^2)*e^(-1)

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maple [A]  time = 0.29, size = 163, normalized size = 1.77 \[ -\frac {2 \tan \left (\frac {d}{2}+\frac {e x}{2}\right ) a}{e b \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+a +b \right )}-\frac {2 \arctan \left (\frac {\tan \left (\frac {d}{2}+\frac {e x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a^{2}}{e \,b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d}{2}+\frac {e x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{e \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 a \arctan \left (\tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/e/b*tan(1/2*d+1/2*e*x)*a/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)-2/e/b^2/((a+b)*(a-b))^(1/2)*ar
ctan(tan(1/2*d+1/2*e*x)*(a-b)/((a+b)*(a-b))^(1/2))*a^2+2/e/((a+b)*(a-b))^(1/2)*arctan(tan(1/2*d+1/2*e*x)*(a-b)
/((a+b)*(a-b))^(1/2))+2/e*a/b^2*arctan(tan(1/2*d+1/2*e*x))

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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+b*sec(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 3.02, size = 444, normalized size = 4.83 \[ \frac {2\,\mathrm {atanh}\left (\frac {64\,a^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^4-128\,a^3\,b+128\,a\,b^3-64\,b^4}-\frac {192\,a^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a\,b^2-128\,a^3-64\,b^3+\frac {64\,a^4}{b}}+\frac {192\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a\,b-64\,b^2-\frac {128\,a^3}{b}+\frac {64\,a^4}{b^2}}-\frac {64\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a\,b-64\,b^2-\frac {128\,a^3}{b}+\frac {64\,a^4}{b^2}}\right )\,\sqrt {b^2-a^2}}{b^2\,e}-\frac {2\,a\,\mathrm {atan}\left (\frac {64\,a^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{64\,a\,b-64\,a^2-\frac {64\,a^3}{b}+\frac {64\,a^4}{b^2}}+\frac {64\,a^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{64\,a\,b^2-64\,a^2\,b-64\,a^3+\frac {64\,a^4}{b}}-\frac {64\,a^4\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{64\,a^4-64\,a^3\,b-64\,a^2\,b^2+64\,a\,b^3}-\frac {64\,a\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{64\,a\,b-64\,a^2-\frac {64\,a^3}{b}+\frac {64\,a^4}{b^2}}\right )}{b^2\,e}-\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{b\,e\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh((64*a^3*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2))/(128*a*b^3 - 128*a^3*b + 64*a^4 - 64*b^4) - (192*a^2*ta
n(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2))/(128*a*b^2 - 128*a^3 - 64*b^3 + (64*a^4)/b) + (192*a*tan(d/2 + (e*x)/2)*(b
^2 - a^2)^(1/2))/(128*a*b - 64*b^2 - (128*a^3)/b + (64*a^4)/b^2) - (64*b*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2))
/(128*a*b - 64*b^2 - (128*a^3)/b + (64*a^4)/b^2))*(b^2 - a^2)^(1/2))/(b^2*e) - (2*a*atan((64*a^2*tan(d/2 + (e*
x)/2))/(64*a*b - 64*a^2 - (64*a^3)/b + (64*a^4)/b^2) + (64*a^3*tan(d/2 + (e*x)/2))/(64*a*b^2 - 64*a^2*b - 64*a
^3 + (64*a^4)/b) - (64*a^4*tan(d/2 + (e*x)/2))/(64*a*b^3 - 64*a^3*b + 64*a^4 - 64*a^2*b^2) - (64*a*b*tan(d/2 +
(e*x)/2))/(64*a*b - 64*a^2 - (64*a^3)/b + (64*a^4)/b^2)))/(b^2*e) - (2*a*tan(d/2 + (e*x)/2))/(b*e*(a + b + ta
n(d/2 + (e*x)/2)^2*(a - b)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(d + e*x))/(a*sec(d + e*x) + b)**2, x)

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