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

Optimal. Leaf size=65 \[ \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c-d} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {d} f \sqrt {c-d}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3967, 208} \[ \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c-d} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {d} f \sqrt {c-d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[c - d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[c - d]*Sqrt[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 3967

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

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 98, normalized size = 1.51 \[ \frac {\sqrt {2} \sqrt {\cos (e+f x)} \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c-d} \sqrt {\cos (e+f x)}}\right )}{\sqrt {d} f \sqrt {c-d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sin[(e + f*x)/2])/(Sqrt[c - d]*Sqrt[Cos[e + f*x]])]*Sqrt[Cos[e + f*x]]*Sec[(
e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])/(Sqrt[c - d]*Sqrt[d]*f)

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fricas [B]  time = 0.69, size = 357, normalized size = 5.49 \[ \left [\frac {\sqrt {\frac {a}{c d - d^{2}}} \log \left (-\frac {{\left (a c^{2} - 8 \, a c d + 8 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} + a d^{2} + {\left (a c^{2} - 2 \, a c d\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (c^{2} d - 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c d^{2} - d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a}{c d - d^{2}}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (6 \, a c d - 7 \, a d^{2}\right )} \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{3} + {\left (c^{2} - 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} - {\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )}\right )}{2 \, f}, -\frac {\sqrt {-\frac {a}{c d - d^{2}}} \arctan \left (\frac {2 \, {\left (c d - d^{2}\right )} \sqrt {-\frac {a}{c d - d^{2}}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{{\left (a c - 2 \, a d\right )} \cos \left (f x + e\right )^{2} + a d + {\left (a c - a d\right )} \cos \left (f x + e\right )}\right )}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*sqrt(a/(c*d - d^2))*log(-((a*c^2 - 8*a*c*d + 8*a*d^2)*cos(f*x + e)^3 + a*d^2 + (a*c^2 - 2*a*c*d)*cos(f*x
+ e)^2 + 4*((c^2*d - 3*c*d^2 + 2*d^3)*cos(f*x + e)^2 + (c*d^2 - d^3)*cos(f*x + e))*sqrt(a/(c*d - d^2))*sqrt((a
*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + (6*a*c*d - 7*a*d^2)*cos(f*x + e))/(c^2*cos(f*x + e)^3 + (c^2 -
2*c*d)*cos(f*x + e)^2 + d^2 - (2*c*d - d^2)*cos(f*x + e)))/f, -sqrt(-a/(c*d - d^2))*arctan(2*(c*d - d^2)*sqrt
(-a/(c*d - d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/((a*c - 2*a*d)*cos(f*x + e)
^2 + a*d + (a*c - a*d)*cos(f*x + e)))/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))^(1/2)/(c-d*sec(f*x+e)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)2*a*sqrt(-a)*sign(cos(f*x+exp(1)))*atan(1/2*(c*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+e
xp(1))))^2+d*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+a*c-3*a*d)/sqrt(2)/sqrt(-d^
2+c*d)/a)/sqrt(-d^2+c*d)/a/f

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maple [B]  time = 1.66, size = 414, normalized size = 6.37 \[ -\frac {\left (\ln \left (-\frac {2 \left (\sqrt {-\frac {2 d}{c +d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, c \sin \left (f x +e \right )+\sqrt {-\frac {2 d}{c +d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, d \sin \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{c \cos \left (f x +e \right )+d \cos \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c -d}\right )-\ln \left (-\frac {2 \left (\sqrt {-\frac {2 d}{c +d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, c \sin \left (f x +e \right )+\sqrt {-\frac {2 d}{c +d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-c \sin \left (f x +e \right )-d \sin \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c -d}\right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{f \sqrt {-\frac {2 d}{c +d}}\, \sqrt {\left (c +d \right ) \left (c -d \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-d*sec(f*x+e)),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\sqrt {a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )}{d \sec \left (f x + e\right ) - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-d*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{d-c\,\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)/(cos(e + f*x)*(c - d/cos(e + f*x))),x)

[Out]

-int((a + a/cos(e + f*x))^(1/2)/(d - c*cos(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sec {\left (e + f x \right )}}{c - d \sec {\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**(1/2)/(c-d*sec(f*x+e)),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*sec(e + f*x)/(c - d*sec(e + f*x)), x)

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