3.189 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{c+d \sec (e+f x)} \, dx\)

Optimal. Leaf size=69 \[ \frac {a \tanh ^{-1}(\sin (e+f x))}{d f}-\frac {2 a \sqrt {c-d} \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{d f \sqrt {c+d}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 69, 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 = {3998, 3770, 3831, 2659, 208} \[ \frac {a \tanh ^{-1}(\sin (e+f x))}{d f}-\frac {2 a \sqrt {c-d} \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{d f \sqrt {c+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3770

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

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 3998

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

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Mathematica [A]  time = 0.19, size = 107, normalized size = 1.55 \[ \frac {a \left (\frac {2 (c-d) \tanh ^{-1}\left (\frac {(d-c) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 255, normalized size = 3.70 \[ \left [\frac {a \sqrt {\frac {c - d}{c + d}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{2} + c d + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {c - d}{c + d}} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + a \log \left (\sin \left (f x + e\right ) + 1\right ) - a \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, d f}, -\frac {2 \, a \sqrt {-\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (d \cos \left (f x + e\right ) + c\right )} \sqrt {-\frac {c - d}{c + d}}}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) - a \log \left (\sin \left (f x + e\right ) + 1\right ) + a \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*(-a*1/2/d*ln(abs(tan((f*x+exp(1))/2)-1))+a*1/2/d*ln(abs(ta
n((f*x+exp(1))/2)+1))+(2*a*c-2*a*d)/d*1/2/sqrt(-c^2+d^2)*(atan((c*tan((f*x+exp(1))/2)-d*tan((f*x+exp(1))/2))/s
qrt(-c^2+d^2))+pi*sign(2*c-2*d)*floor((f*x+exp(1))/2/pi+1/2)))

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maple [B]  time = 0.70, size = 135, normalized size = 1.96 \[ -\frac {2 a \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) c}{f d \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 a \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f \sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {a \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f d}+\frac {a \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(sec(f*x+e)*(a+a*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 details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 2.16, size = 195, normalized size = 2.83 \[ \frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,\left (c+d\right )}+\frac {2\,a\,\left (\mathrm {atanh}\left (\frac {d^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-c^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-c^2\,d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (c^2-d^2\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^2-d^2}\,\left (d^2+c\,d\right )}\right )\,\sqrt {c^2-d^2}+c\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\right )}{d\,f\,\left (c+d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(f*(c + d)) + (2*a*(atanh((d^3*sin(e/2 + (f*x)/2) - c^3*sin
(e/2 + (f*x)/2) + c*d^2*sin(e/2 + (f*x)/2) - c^2*d*sin(e/2 + (f*x)/2) + c*sin(e/2 + (f*x)/2)*(c^2 - d^2))/(cos
(e/2 + (f*x)/2)*(c^2 - d^2)^(1/2)*(c*d + d^2)))*(c^2 - d^2)^(1/2) + c*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)
/2))))/(d*f*(c + d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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