3.2.89 \(\int \frac {a+b \sec (e+f x)}{c+d \sec (e+f x)} \, dx\) [189]

Optimal. Leaf size=67 \[ \frac {a x}{c}+\frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c \sqrt {c-d} \sqrt {c+d} f} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4004, 3916, 2738, 214} \begin {gather*} \frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c f \sqrt {c-d} \sqrt {c+d}}+\frac {a x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x])/(c + d*Sec[e + f*x]),x]

[Out]

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

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 3916

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

Rule 4004

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]

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 68, normalized size = 1.01 \begin {gather*} \frac {a (e+f x)+\frac {2 (-b c+a d) \tanh ^{-1}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{c f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x])/(c + d*Sec[e + f*x]),x]

[Out]

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

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Maple [A]
time = 0.17, size = 73, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (a d -b c \right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{c \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 a \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}}{f}\)	\(73\)
default	\(\frac {-\frac {2 \left (a d -b c \right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{c \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 a \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}}{f}\)	\(73\)
risch	\(\frac {a x}{c}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a d}{\sqrt {c^{2}-d^{2}}\, f c}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b}{\sqrt {c^{2}-d^{2}}\, f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a d}{\sqrt {c^{2}-d^{2}}\, f c}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b}{\sqrt {c^{2}-d^{2}}\, f}\)	\(294\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+b*sec(f*x+e))/(c+d*sec(f*x+e)),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*c^2-4*d^2>0)', see `assume?`
for more de

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Fricas [A]
time = 2.59, size = 258, normalized size = 3.85 \begin {gather*} \left [\frac {2 \, {\left (a c^{2} - a d^{2}\right )} f x - {\left (b c - a d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right )}{2 \, {\left (c^{3} - c d^{2}\right )} f}, \frac {{\left (a c^{2} - a d^{2}\right )} f x + {\left (b c - a d\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right )}{{\left (c^{3} - c d^{2}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (58) = 116\).
time = 0.50, size = 274, normalized size = 4.09 \begin {gather*} \frac {\frac {{\left (\sqrt {-c^{2} + d^{2}} a {\left (c - 2 \, d\right )} {\left | -c + d \right |} + \sqrt {-c^{2} + d^{2}} b c {\left | -c + d \right |} - \sqrt {-c^{2} + d^{2}} a {\left | c \right |} {\left | -c + d \right |} + \sqrt {-c^{2} + d^{2}} b {\left | c \right |} {\left | -c + d \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-\frac {d + \sqrt {{\left (c + d\right )} {\left (c - d\right )} + d^{2}}}{c - d}}}\right )\right )}}{{\left (c^{2} - 2 \, c d + d^{2}\right )} c^{2} + {\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} {\left | c \right |}} + \frac {{\left (a c + b c - 2 \, a d + a {\left | c \right |} - b {\left | c \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-\frac {d - \sqrt {{\left (c + d\right )} {\left (c - d\right )} + d^{2}}}{c - d}}}\right )\right )}}{c^{2} - d {\left | c \right |}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.68, size = 573, normalized size = 8.55 \begin {gather*} \frac {b\,c^2\,\ln \left (\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (c^2-d^2\right )}^{3/2}}-\frac {b\,d^2\,\ln \left (\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (c^2-d^2\right )}^{3/2}}+\frac {2\,a\,c\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,\left (c^2-d^2\right )}-\frac {b\,\ln \left (\frac {c\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}}{f\,\left (c^2-d^2\right )}-\frac {a\,c\,d\,\ln \left (\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (c^2-d^2\right )}^{3/2}}-\frac {2\,a\,d^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{c\,f\,\left (c^2-d^2\right )}+\frac {a\,d^3\,\ln \left (\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{c\,f\,{\left (c^2-d^2\right )}^{3/2}}+\frac {a\,d\,\ln \left (\frac {c\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}}{c\,f\,\left (c^2-d^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x))/(c + d/cos(e + f*x)),x)

[Out]

(b*c^2*log((c*sin(e/2 + (f*x)/2) - d*sin(e/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(c^2 - d^2)^(1/2))/cos(e/2 + (f*x
)/2)))/(f*(c^2 - d^2)^(3/2)) - (b*d^2*log((c*sin(e/2 + (f*x)/2) - d*sin(e/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(c
^2 - d^2)^(1/2))/cos(e/2 + (f*x)/2)))/(f*(c^2 - d^2)^(3/2)) + (2*a*c*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2
)))/(f*(c^2 - d^2)) - (b*log((c*cos(e/2 + (f*x)/2) + d*cos(e/2 + (f*x)/2) - sin(e/2 + (f*x)/2)*(c^2 - d^2)^(1/
2))/cos(e/2 + (f*x)/2))*((c + d)*(c - d))^(1/2))/(f*(c^2 - d^2)) - (a*c*d*log((c*sin(e/2 + (f*x)/2) - d*sin(e/
2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(c^2 - d^2)^(1/2))/cos(e/2 + (f*x)/2)))/(f*(c^2 - d^2)^(3/2)) - (2*a*d^2*ata
n(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(c*f*(c^2 - d^2)) + (a*d^3*log((c*sin(e/2 + (f*x)/2) - d*sin(e/2 + (
f*x)/2) + cos(e/2 + (f*x)/2)*(c^2 - d^2)^(1/2))/cos(e/2 + (f*x)/2)))/(c*f*(c^2 - d^2)^(3/2)) + (a*d*log((c*cos
(e/2 + (f*x)/2) + d*cos(e/2 + (f*x)/2) - sin(e/2 + (f*x)/2)*(c^2 - d^2)^(1/2))/cos(e/2 + (f*x)/2))*((c + d)*(c
- d))^(1/2))/(c*f*(c^2 - d^2))

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