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/(d*sec(f*x+e))^(1/2)+2*a*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticE(sin(1/2*e+1/2*f*x),2
^(1/2))/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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]

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*}

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Mathematica [A]  time = 0.29, 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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fricas [F]  time = 1.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}{d \sec \left (f x + e\right )}, x\right ) \]

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

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)

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maple [C]  time = 0.92, size = 916, normalized size = 15.79 \[ \text {result too large to display} \]

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*(-1+cos(f*x+e))*(4*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+cos(f*x
+e))^2)^(1/2)*cos(f*x+e)^2*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a-4*I*(1/(1+cos(f*x+e)))^(1/2)
*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*sin(f*x+e)*EllipticE(I*(-
1+cos(f*x+e))/sin(f*x+e),I)*a+8*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+c
os(f*x+e))^2)^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a-8*I*(1/(1+cos(f*x+e)))^(
1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticE(I*
(-1+cos(f*x+e))/sin(f*x+e),I)*a+4*I*a*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/
(1+cos(f*x+e))^2)^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-4*I*a*(1/(1+cos(f*x+e)))^(1/2)*(c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)
*sin(f*x+e)-4*cos(f*x+e)^3*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*a-4*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(1+co
s(f*x+e))^2)^(1/2)*b-4*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*b+b*ln(-2*(2*(-cos(f*x+e)/(1
+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^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)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^
2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*sin(f*x+e)+4*cos(f*x+e)*(-cos(f*x+e)/(1
+cos(f*x+e))^2)^(1/2)*a)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/cos(f*x+e)/sin(f*x+e)^3/(d/cos(f*x+e))^(1/2)

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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