3.265 \(\int \frac{\sec ^2(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=39 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{b} f} \]

[Out]

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

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Rubi [A]  time = 0.0714095, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4146, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

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

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 0.147321, size = 87, normalized size = 2.23 \[ \frac{\sec (e+f x) \sqrt{a \cos (2 e+2 f x)+a+2 b} \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )}{\sqrt{2} \sqrt{b} f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x
])/(Sqrt[2]*Sqrt[b]*f*Sqrt[a + b*Sec[e + f*x]^2])

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Maple [C]  time = 0.411, size = 379, normalized size = 9.7 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{f\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{1}{ \left ( a+b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) } \left ( i\cos \left ( fx+e \right ) \sqrt{a}\sqrt{b}-i\sqrt{a}\sqrt{b}+a\cos \left ( fx+e \right ) +b \right ) }}\sqrt{-2\,{\frac{i\cos \left ( fx+e \right ) \sqrt{a}\sqrt{b}-i\sqrt{a}\sqrt{b}-a\cos \left ( fx+e \right ) -b}{ \left ( a+b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}} \left ({\it EllipticF} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }\sqrt{{\frac{1}{a+b} \left ( 2\,i\sqrt{a}\sqrt{b}+a-b \right ) }}},\sqrt{-{\frac{1}{ \left ( a+b \right ) ^{2}} \left ( 4\,i{a}^{{\frac{3}{2}}}\sqrt{b}-4\,i\sqrt{a}{b}^{{\frac{3}{2}}}-{a}^{2}+6\,ab-{b}^{2} \right ) }} \right ) -2\,{\it EllipticPi} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }\sqrt{{\frac{2\,i\sqrt{a}\sqrt{b}+a-b}{a+b}}}},{\frac{a+b}{2\,i\sqrt{a}\sqrt{b}+a-b}},{\sqrt{-{\frac{2\,i\sqrt{a}\sqrt{b}-a+b}{a+b}}}{\frac{1}{\sqrt{{\frac{2\,i\sqrt{a}\sqrt{b}+a-b}{a+b}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{1}{a+b} \left ( 2\,i\sqrt{a}\sqrt{b}+a-b \right ) }}}}{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+
a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b
)/(1+cos(f*x+e)))^(1/2)*(EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a
^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))-2*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)
/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)))*sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)/(-1
+cos(f*x+e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.655339, size = 543, normalized size = 13.92 \begin{align*} \left [\frac{\log \left (\frac{{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right )}{4 \, \sqrt{b} f}, \frac{\sqrt{-b} \arctan \left (-\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \,{\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, b f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*
cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)/(sqrt(
b)*f), 1/2*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))/(b*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (e + f x \right )}}{\sqrt{a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(sec(e + f*x)**2/sqrt(a + b*sec(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{2}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^2/sqrt(b*sec(f*x + e)^2 + a), x)