\(\int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx\) [248]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\frac {b \text {arctanh}(\sin (e+f x))}{d f}-\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} d \sqrt {c+d} f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4083, 3855, 3916, 2738, 214} \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\frac {b \text {arctanh}(\sin (e+f x))}{d f}-\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{d f \sqrt {c-d} \sqrt {c+d}} \]

[In]

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

[Out]

(b*ArcTanh[Sin[e + f*x]])/(d*f) - (2*(b*c - a*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c
- d]*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 3855

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

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 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.47 \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\frac {\frac {2 (b c-a d) \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+b \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{d f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{d}-\frac {2 \left (-a d +b c \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{d \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\)	\(92\)
default	\(\frac {\frac {b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{d}-\frac {2 \left (-a d +b c \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{d \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\)	\(92\)
risch	\(\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a}{\sqrt {c^{2}-d^{2}}\, f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b c}{\sqrt {c^{2}-d^{2}}\, f d}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a}{\sqrt {c^{2}-d^{2}}\, f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b c}{\sqrt {c^{2}-d^{2}}\, f d}-\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{d f}+\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{d f}\)	\(331\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 316, normalized size of antiderivative = 4.16 \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\left [-\frac {{\left (b c - a d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - {\left (b c^{2} - b d^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (b c^{2} - b d^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (c^{2} d - d^{3}\right )} f}, -\frac {2 \, {\left (b c - a d\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (b c^{2} - b d^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (b c^{2} - b d^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (c^{2} d - d^{3}\right )} f}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(f*x+e)*(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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.67 \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\frac {\frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{d} - \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{d} + \frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} {\left (b c - a d\right )}}{\sqrt {-c^{2} + d^{2}} d}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.54 (sec) , antiderivative size = 573, normalized size of antiderivative = 7.54 \[ \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{c+d \sec (e+f x)} \, dx=\frac {a\,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 {a\,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\,b\,d\,\mathrm {atanh}\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 {a\,\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 {b\,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\,b\,c^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{d\,f\,\left (c^2-d^2\right )}-\frac {b\,c^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 )}{d\,f\,{\left (c^2-d^2\right )}^{3/2}}+\frac {b\,c\,\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 )}}{d\,f\,\left (c^2-d^2\right )} \]

[In]

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

[Out]

(a*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)) - (a*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*b*d*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/
2)))/(f*(c^2 - d^2)) - (a*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)) + (b*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*b*c^2*at
anh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(d*f*(c^2 - d^2)) - (b*c^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)))/(d*f*(c^2 - d^2)^(3/2)) + (b*c*log((c*c
os(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))/(d*f*(c^2 - d^2))