3.3.55 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))}{a+b \sec (e+f x)} \, dx\) [255]

Optimal. Leaf size=76 \[ \frac {d \tanh ^{-1}(\sin (e+f x))}{b f}+\frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} f} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4083, 3855, 3916, 2738, 214} \begin {gather*} \frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{b f \sqrt {a-b} \sqrt {a+b}}+\frac {d \tanh ^{-1}(\sin (e+f x))}{b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*ArcTanh[Sin[e + f*x]])/(b*f) + (2*(b*c - a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a
- b]*b*Sqrt[a + b]*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*} \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{a+b \sec (e+f x)} \, dx &=\frac {d \int \sec (e+f x) \, dx}{b}+\frac {(b c-a d) \int \frac {\sec (e+f x)}{a+b \sec (e+f x)} \, dx}{b}\\ &=\frac {d \tanh ^{-1}(\sin (e+f x))}{b f}+\frac {(b c-a d) \int \frac {1}{1+\frac {a \cos (e+f x)}{b}} \, dx}{b^2}\\ &=\frac {d \tanh ^{-1}(\sin (e+f x))}{b f}+\frac {(2 (b c-a d)) \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} (e+f x)\right )\right )}{b^2 f}\\ &=\frac {d \tanh ^{-1}(\sin (e+f x))}{b f}+\frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} f}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 112, normalized size = 1.47 \begin {gather*} \frac {\frac {2 (-b c+a d) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+d \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 )}{b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 92, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {\frac {d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b}-\frac {2 \left (a d -b c \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b}}{f}\)	\(92\)
default	\(\frac {\frac {d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b}-\frac {2 \left (a d -b c \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b}}{f}\)	\(92\)
risch	\(\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a d}{\sqrt {a^{2}-b^{2}}\, f b}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) c}{\sqrt {a^{2}-b^{2}}\, f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a d}{\sqrt {a^{2}-b^{2}}\, f b}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) c}{\sqrt {a^{2}-b^{2}}\, f}+\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{b f}-\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{b f}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Fricas [A]
time = 2.91, size = 321, normalized size = 4.22 \begin {gather*} \left [\frac {{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) - \sqrt {a^{2} - b^{2}} {\left (b c - a d\right )} \log \left (\frac {2 \, a b \cos \left (f x + e\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + b^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} f}, \frac {{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (b c - a d\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (f x + e\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 127, normalized size = 1.67 \begin {gather*} \frac {\frac {d \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{b} - \frac {d \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{b} - \frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (b c - a d\right )}}{\sqrt {-a^{2} + b^{2}} b}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.81, size = 571, normalized size = 7.51 \begin {gather*} \frac {b^2\,c\,\ln \left (\frac {b\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (a^2-b^2\right )}^{3/2}}-\frac {a^2\,c\,\ln \left (\frac {b\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (a^2-b^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 (a^2-b^2\right )}+\frac {c\,\ln \left (\frac {a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{f\,\left (a^2-b^2\right )}-\frac {a\,b\,d\,\ln \left (\frac {b\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,{\left (a^2-b^2\right )}^{3/2}}+\frac {2\,a^2\,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 )}{b\,f\,\left (a^2-b^2\right )}+\frac {a^3\,d\,\ln \left (\frac {b\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{b\,f\,{\left (a^2-b^2\right )}^{3/2}}-\frac {a\,d\,\ln \left (\frac {a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{b\,f\,\left (a^2-b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*c*log((b*sin(e/2 + (f*x)/2) - a*sin(e/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(a^2 - b^2)^(1/2))/cos(e/2 + (f*x
)/2)))/(f*(a^2 - b^2)^(3/2)) - (a^2*c*log((b*sin(e/2 + (f*x)/2) - a*sin(e/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(a
^2 - b^2)^(1/2))/cos(e/2 + (f*x)/2)))/(f*(a^2 - b^2)^(3/2)) - (2*b*d*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/
2)))/(f*(a^2 - b^2)) + (c*log((a*cos(e/2 + (f*x)/2) + b*cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2)*(a^2 - b^2)^(1
/2))/cos(e/2 + (f*x)/2))*((a + b)*(a - b))^(1/2))/(f*(a^2 - b^2)) - (a*b*d*log((b*sin(e/2 + (f*x)/2) - a*sin(e
/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(a^2 - b^2)^(1/2))/cos(e/2 + (f*x)/2)))/(f*(a^2 - b^2)^(3/2)) + (2*a^2*d*at
anh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(b*f*(a^2 - b^2)) + (a^3*d*log((b*sin(e/2 + (f*x)/2) - a*sin(e/2 +
(f*x)/2) + cos(e/2 + (f*x)/2)*(a^2 - b^2)^(1/2))/cos(e/2 + (f*x)/2)))/(b*f*(a^2 - b^2)^(3/2)) - (a*d*log((a*c
os(e/2 + (f*x)/2) + b*cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2)*(a^2 - b^2)^(1/2))/cos(e/2 + (f*x)/2))*((a + b)*
(a - b))^(1/2))/(b*f*(a^2 - b^2))

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