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 \[ \frac {2 \sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]

[Out]

2*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))*EllipticF(sin(1/2*d+1/2*e*x-
1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*sec(e*x+d)^(1/2)*((b+a*cos(e*x+d)+c*sin(
e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)/e/(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.87, 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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fricas [F]  time = 2.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sec \left (e x + d\right )}}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}, x\right ) \]

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

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)

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maple [C]  time = 1.86, size = 722, normalized size = 6.12 \[ -\frac {4 i \EllipticF \left (\sqrt {\frac {\left (i \sin \left (e x +d \right )+\cos \left (e x +d \right )\right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}-c}}, \sqrt {\frac {\left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}-c \right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{\left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}}\right ) \sqrt {\frac {1}{\cos \left (e x +d \right )}}\, \sqrt {\frac {b +a \cos \left (e x +d \right )+c \sin \left (e x +d \right )}{\cos \left (e x +d \right )}}\, \sqrt {\frac {\left (i \sin \left (e x +d \right )+\cos \left (e x +d \right )\right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}-c}}\, \sqrt {-\frac {i \left (\cos \left (e x +d \right ) \sqrt {a^{2}-b^{2}+c^{2}}-a \sin \left (e x +d \right )+b \sin \left (e x +d \right )+c \cos \left (e x +d \right )+\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{\left (i \cos \left (e x +d \right )+\sin \left (e x +d \right )+i\right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}}\, \sqrt {\frac {i \left (a \sin \left (e x +d \right )-b \sin \left (e x +d \right )+\cos \left (e x +d \right ) \sqrt {a^{2}-b^{2}+c^{2}}-c \cos \left (e x +d \right )+\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}{\left (i \cos \left (e x +d \right )+\sin \left (e x +d \right )+i\right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}}\, \left (\cos \left (e x +d \right )+1\right )^{2} \cos \left (e x +d \right ) \left (\cos \left (e x +d \right )-1\right )^{2} \left (i a \cos \left (e x +d \right )-i \cos \left (e x +d \right ) b -i \sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )+i c \sin \left (e x +d \right )+\cos \left (e x +d \right ) \sqrt {a^{2}-b^{2}+c^{2}}-c \cos \left (e x +d \right )+a \sin \left (e x +d \right )-b \sin \left (e x +d \right )\right )}{e \sin \left (e x +d \right )^{4} \left (b +a \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right )} \]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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