3.450 \(\int \frac{\sqrt{\sec (d+e x)}}{\sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \, dx\)

Optimal. Leaf size=118 \[ \frac{2 \sqrt{\sec (d+e x)} \sqrt{\frac{a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt{a^2+c^2}+b}} \text{EllipticF}\left (\frac{1}{2} \left (-\tan ^{-1}(a,c)+d+e x\right ),\frac{2 \sqrt{a^2+c^2}}{\sqrt{a^2+c^2}+b}\right )}{e \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \]

[Out]

(2*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[Sec[d + e*x]]*Sqrt[(b
+ a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])/(e*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])

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Rubi [A]  time = 0.166285, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3167, 3127, 2661} \[ \frac{2 \sqrt{\sec (d+e x)} \sqrt{\frac{a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt{a^2+c^2}+b}} F\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac{2 \sqrt{a^2+c^2}}{b+\sqrt{a^2+c^2}}\right )}{e \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[d + e*x]]/Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]],x]

[Out]

(2*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[Sec[d + e*x]]*Sqrt[(b
+ a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])/(e*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])

Rule 3167

Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_)
, x_Symbol] :> Dist[(Sec[d + e*x]^n*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n)/(a + b*Sec[d + e*x] + c*Tan[d + e
*x])^n, Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + n, 0] &&
!IntegerQ[n]

Rule 3127

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rule 2661

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

Rubi steps

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

Mathematica [C]  time = 0.91945, size = 339, normalized size = 2.87 \[ \frac{2 \sqrt{\sec (d+e x)} \sec \left (\tan ^{-1}\left (\frac{a}{c}\right )+d+e x\right ) \sqrt{-\frac{c \sqrt{\frac{a^2}{c^2}+1} \left (\sin \left (\tan ^{-1}\left (\frac{a}{c}\right )+d+e x\right )-1\right )}{c \sqrt{\frac{a^2}{c^2}+1}+b}} \sqrt{\frac{c \sqrt{\frac{a^2}{c^2}+1} \left (\sin \left (\tan ^{-1}\left (\frac{a}{c}\right )+d+e x\right )+1\right )}{c \sqrt{\frac{a^2}{c^2}+1}-b}} \sqrt{c \sqrt{\frac{a^2}{c^2}+1} \sin \left (\tan ^{-1}\left (\frac{a}{c}\right )+d+e x\right )+b} \sqrt{a \cos (d+e x)+b+c \sin (d+e x)} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\frac{b+\sqrt{\frac{a^2}{c^2}+1} c \sin \left (d+e x+\tan ^{-1}\left (\frac{a}{c}\right )\right )}{b-\sqrt{\frac{a^2}{c^2}+1} c},\frac{b+\sqrt{\frac{a^2}{c^2}+1} c \sin \left (d+e x+\tan ^{-1}\left (\frac{a}{c}\right )\right )}{b+\sqrt{\frac{a^2}{c^2}+1} c}\right )}{c e \sqrt{\frac{a^2}{c^2}+1} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Sec[d + e*x]]/Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]],x]

[Out]

(2*AppellF1[1/2, 1/2, 1/2, 3/2, (b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(b - Sqrt[1 + a^2/c^2]*c)
, (b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(b + Sqrt[1 + a^2/c^2]*c)]*Sqrt[Sec[d + e*x]]*Sec[d + e
*x + ArcTan[a/c]]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]*Sqrt[-((Sqrt[1 + a^2/c^2]*c*(-1 + Sin[d + e*x + Ar
cTan[a/c]]))/(b + Sqrt[1 + a^2/c^2]*c))]*Sqrt[(Sqrt[1 + a^2/c^2]*c*(1 + Sin[d + e*x + ArcTan[a/c]]))/(-b + Sqr
t[1 + a^2/c^2]*c)]*Sqrt[b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]]])/(Sqrt[1 + a^2/c^2]*c*e*Sqrt[a + b
*Sec[d + e*x] + c*Tan[d + e*x]])

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Maple [C]  time = 0.596, size = 722, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*I/e/(I*a-I*b-(a^2-b^2+c^2)^(1/2)+c)*EllipticF(((I*a-I*b-(a^2-b^2+c^2)^(1/2)+c)/(I*a-I*b+(a^2-b^2+c^2)^(1/2)
-c)*(I*sin(e*x+d)+cos(e*x+d)))^(1/2),((I*a-I*b+(a^2-b^2+c^2)^(1/2)-c)*(I*a-I*b+(a^2-b^2+c^2)^(1/2)+c)/(I*a-I*b
-(a^2-b^2+c^2)^(1/2)+c)/(I*a-I*b-(a^2-b^2+c^2)^(1/2)-c))^(1/2))*(1/cos(e*x+d))^(1/2)*((b+a*cos(e*x+d)+c*sin(e*
x+d))/cos(e*x+d))^(1/2)*((I*a-I*b-(a^2-b^2+c^2)^(1/2)+c)/(I*a-I*b+(a^2-b^2+c^2)^(1/2)-c)*(I*sin(e*x+d)+cos(e*x
+d)))^(1/2)*(-I/(I*a-I*b-(a^2-b^2+c^2)^(1/2)-c)*(cos(e*x+d)*(a^2-b^2+c^2)^(1/2)-a*sin(e*x+d)+b*sin(e*x+d)+c*co
s(e*x+d)+(a^2-b^2+c^2)^(1/2)+c)/(I*cos(e*x+d)+I+sin(e*x+d)))^(1/2)*(I/(I*a-I*b+(a^2-b^2+c^2)^(1/2)-c)*(a*sin(e
*x+d)-b*sin(e*x+d)+cos(e*x+d)*(a^2-b^2+c^2)^(1/2)-c*cos(e*x+d)+(a^2-b^2+c^2)^(1/2)-c)/(I*cos(e*x+d)+I+sin(e*x+
d)))^(1/2)*(cos(e*x+d)+1)^2*cos(e*x+d)*(cos(e*x+d)-1)^2*(I*a*cos(e*x+d)-I*cos(e*x+d)*b-I*(a^2-b^2+c^2)^(1/2)*s
in(e*x+d)+I*c*sin(e*x+d)+cos(e*x+d)*(a^2-b^2+c^2)^(1/2)-c*cos(e*x+d)+a*sin(e*x+d)-b*sin(e*x+d))/sin(e*x+d)^4/(
b+a*cos(e*x+d)+c*sin(e*x+d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sec(e*x + d))/sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\sec \left (e x + d\right )}}{\sqrt{b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sec(e*x + d))/sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(e*x+d)**(1/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))**(1/2),x)

[Out]

Integral(sqrt(sec(d + e*x))/sqrt(a + b*sec(d + e*x) + c*tan(d + e*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sec(e*x + d))/sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a), x)