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(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/b^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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])

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 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]

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*}

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Mathematica [B]  time = 0.12, 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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fricas [B]  time = 0.60, size = 215, normalized size = 5.51 \[ \left [\frac {\log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right )}{4 \, \sqrt {b} f}, \frac {\sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, b f}\right ] \]

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

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)

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maple [C]  time = 1.60, size = 379, normalized size = 9.72 \[ -\frac {\sqrt {2}\, \sqrt {\frac {i \sqrt {a}\, \sqrt {b}\, \cos \left (f x +e \right )-i \sqrt {a}\, \sqrt {b}+a \cos \left (f x +e \right )+b}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \sqrt {-\frac {2 \left (i \sqrt {a}\, \sqrt {b}\, \cos \left (f x +e \right )-i \sqrt {a}\, \sqrt {b}-a \cos \left (f x +e \right )-b \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \left (\EllipticF \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}}{\sin \left (f x +e \right )}, \sqrt {-\frac {4 i a^{\frac {3}{2}} \sqrt {b}-4 i \sqrt {a}\, b^{\frac {3}{2}}-a^{2}+6 a b -b^{2}}{\left (a +b \right )^{2}}}\right )-2 \EllipticPi \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}}{\sin \left (f x +e \right )}, \frac {a +b}{2 i \sqrt {a}\, \sqrt {b}+a -b}, \frac {\sqrt {-\frac {2 i \sqrt {a}\, \sqrt {b}-a +b}{a +b}}}{\sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}}\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )}{f \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}} \]

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^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*
(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))/((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)

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maxima [A]  time = 0.33, size = 23, normalized size = 0.59 \[ \frac {\operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b} f} \]

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]

arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt(b)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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