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

Optimal. Leaf size=80 \[ \frac {\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[Out]

EllipticF(sin(f*x+e),(a/(a+b))^(1/2))*(1-a*sin(f*x+e)^2/(a+b))^(1/2)/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b
-a*sin(f*x+e)^2))^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 103, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4148, 6722, 1974, 421, 419} \[ \frac {\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{f \sqrt {\cos ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b + a*Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(
f*Sqrt[Cos[e + f*x]^2]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x]^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

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

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

Rule 4148

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

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 69, normalized size = 0.86 \[ \frac {\sec (e+f x) \sqrt {\frac {a \cos (2 (e+f x))+a+2 b}{a+b}} F\left (e+f x\left |\frac {a}{a+b}\right .\right )}{\sqrt {2} f \sqrt {a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*EllipticF[e + f*x, a/(a + b)]*Sec[e + f*x])/(Sqrt[2]*f*Sqrt[a +
b*Sec[e + f*x]^2])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (f x + e\right )}{\sqrt {b \sec \left (f x + e\right )^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.55, size = 269, normalized size = 3.36 \[ \frac {\left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, \sqrt {\frac {i \sqrt {a}\, \sqrt {b}\, \cos \left (f x +e \right )-i \sqrt {a}\, \sqrt {b}+a \cos \left (f x +e \right )+b}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \sqrt {-\frac {2 \left (i \sqrt {a}\, \sqrt {b}\, \cos \left (f x +e \right )-i \sqrt {a}\, \sqrt {b}-a \cos \left (f x +e \right )-b \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \EllipticF \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}}{\sin \left (f x +e \right )}, \sqrt {-\frac {4 i a^{\frac {3}{2}} \sqrt {b}-4 i \sqrt {a}\, b^{\frac {3}{2}}-a^{2}+6 a b -b^{2}}{\left (a +b \right )^{2}}}\right )}{f \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a}\, \sqrt {b}+a -b}{a +b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*sin(f*x+e)^2*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)*Ellipt
icF((-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))/((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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )}{\sqrt {b \sec \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(cos(e + f*x)*(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 \[ \int \frac {\sec {\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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