3.582 \(\int \frac{a+b \tan (e+f x)}{\sqrt{d \sec (e+f x)}} \, dx\)
Optimal. Leaf size=58 \[ \frac{2 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b}{f \sqrt{d \sec (e+f x)}} \]
[Out]
(-2*b)/(f*Sqrt[d*Sec[e + f*x]]) + (2*a*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]])
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Rubi [A] time = 0.0492027, antiderivative size = 58, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{3486, 3771, 2639} \[ \frac{2 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b}{f \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])/Sqrt[d*Sec[e + f*x]],x]
[Out]
(-2*b)/(f*Sqrt[d*Sec[e + f*x]]) + (2*a*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]])
Rule 3486
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3771
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{a+b \tan (e+f x)}{\sqrt{d \sec (e+f x)}} \, dx &=-\frac{2 b}{f \sqrt{d \sec (e+f x)}}+a \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx\\ &=-\frac{2 b}{f \sqrt{d \sec (e+f x)}}+\frac{a \int \sqrt{\cos (e+f x)} \, dx}{\sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 b}{f \sqrt{d \sec (e+f x)}}+\frac{2 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.27413, size = 54, normalized size = 0.93 \[ \frac{2 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )-2 b \sqrt{\cos (e+f x)}}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[e + f*x])/Sqrt[d*Sec[e + f*x]],x]
[Out]
(-2*b*Sqrt[Cos[e + f*x]] + 2*a*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]])
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Maple [C] time = 0.258, size = 916, normalized size = 15.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/2),x)
[Out]
-1/2/f*(cos(f*x+e)-1)*(4*I*sin(f*x+e)*cos(f*x+e)^2*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))
^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a-4*I*sin(f*x+e)*cos(f*x+e)^2*El
lipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(1/2)*a+8*I*sin(f*x+e)*cos(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1
))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a-8*I*sin(f*x+e)*cos(f*x+e)*El
lipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(1/2)*a+4*I*a*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-4*I*a*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x
+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*
x+e)-4*sin(f*x+e)*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b-4*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(1/2)*a+b*ln(-2*(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f*
x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*cos(f*x+e)*sin(f*x+e)-b*cos(f*x+e)*ln(-(2*(-cos(f*x+e)/(cos(f*x
+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)
*sin(f*x+e)-4*sin(f*x+e)*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b+4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*a)/sin(f*x+e)^3/cos(f*x+e)/(d/cos(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \tan \left (f x + e\right ) + a}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)/sqrt(d*sec(f*x + e)), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sec \left (f x + e\right )}{\left (b \tan \left (f x + e\right ) + a\right )}}{d \sec \left (f x + e\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)/(d*sec(f*x + e)), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (e + f x \right )}}{\sqrt{d \sec{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(1/2),x)
[Out]
Integral((a + b*tan(e + f*x))/sqrt(d*sec(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \tan \left (f x + e\right ) + a}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)/sqrt(d*sec(f*x + e)), x)