3.455 \(\int \frac{1}{\sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.151947, 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 = {3163, 3127, 2661} \[ \frac{2 \sqrt{\frac{a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt{a^2+c^2}+b}} F\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac{2 \sqrt{a^2+c^2}}{b+\sqrt{a^2+c^2}}\right )}{e \sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[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[(b + a*Cos[d + e*x] +
c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])/(e*Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])

Rule 3163

Int[cos[(d_.) + (e_.)*(x_)]^(n_)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(n_),
x_Symbol] :> Dist[(Cos[d + e*x]^n*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n)/(b + a*Cos[d + e*x] + c*Sin[d + e*
x])^n, Int[(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] &&  !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{1}{\sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \, dx &=\frac{\sqrt{b+a \cos (d+e x)+c \sin (d+e x)} \int \frac{1}{\sqrt{b+a \cos (d+e x)+c \sin (d+e x)}} \, dx}{\sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}}\\ &=\frac{\sqrt{\frac{b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt{a^2+c^2}}} \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{\cos (d+e x)} \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{\frac{b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt{a^2+c^2}}}}{e \sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}}\\ \end{align*}

Mathematica [C]  time = 2.94404, size = 506, normalized size = 4.29 \[ \frac{4 \left (\sqrt{a^2-b^2+c^2}+i a-i b+c\right ) (\cos (d+e x)+i \sin (d+e x)) \sqrt{-\frac{i \left (\sqrt{a^2-b^2+c^2}+(a-b) \tan \left (\frac{1}{2} (d+e x)\right )-c\right )}{\left (\sqrt{a^2-b^2+c^2}-i a+i b-c\right ) \left (\tan \left (\frac{1}{2} (d+e x)\right )-i\right )}} \sqrt{-\frac{i \left (\sqrt{a^2-b^2+c^2}+(b-a) \tan \left (\frac{1}{2} (d+e x)\right )+c\right )}{\left (\sqrt{a^2-b^2+c^2}+i a-i b+c\right ) \left (\tan \left (\frac{1}{2} (d+e x)\right )-i\right )}} \sqrt{\frac{\left (\sqrt{a^2-b^2+c^2}-i a+i b+c\right ) (-\cos (d+e x)+i \sin (d+e x))}{\sqrt{a^2-b^2+c^2}+i a-i b+c}} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{\left (\sqrt{a^2-b^2+c^2}-i a+i b+c\right ) (-\cos (d+e x)+i \sin (d+e x))}{\sqrt{a^2-b^2+c^2}+i a-i b+c}}\right ),\frac{b+i \sqrt{a^2-b^2+c^2}}{b-i \sqrt{a^2-b^2+c^2}}\right )}{e \left (a+i \left (\sqrt{a^2-b^2+c^2}+i b+c\right )\right ) \sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]]),x]

[Out]

(4*(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])*EllipticF[ArcSin[Sqrt[(((-I)*a + I*b + c + Sqrt[a^2 - b^2 + c^2])*(
-Cos[d + e*x] + I*Sin[d + e*x]))/(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])]], (b + I*Sqrt[a^2 - b^2 + c^2])/(b -
I*Sqrt[a^2 - b^2 + c^2])]*Sqrt[(((-I)*a + I*b + c + Sqrt[a^2 - b^2 + c^2])*(-Cos[d + e*x] + I*Sin[d + e*x]))/
(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])]*(Cos[d + e*x] + I*Sin[d + e*x])*Sqrt[((-I)*(-c + Sqrt[a^2 - b^2 + c^2
] + (a - b)*Tan[(d + e*x)/2]))/(((-I)*a + I*b - c + Sqrt[a^2 - b^2 + c^2])*(-I + Tan[(d + e*x)/2]))]*Sqrt[((-I
)*(c + Sqrt[a^2 - b^2 + c^2] + (-a + b)*Tan[(d + e*x)/2]))/((I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])*(-I + Tan[
(d + e*x)/2]))])/((a + I*(I*b + c + Sqrt[a^2 - b^2 + c^2]))*e*Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*T
an[d + e*x]])

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Maple [C]  time = 0.415, size = 714, 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(1/cos(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)*((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*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)*(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*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))*cos(e*x+d)^(1/2)*(cos(e*x+d)-1)^2*(I*(a^2-b^2+c^2)^(1/2)*sin(e*x+d)-I*a*cos(e*x+d)+I*cos(e*x+d)*b-I*c*si
n(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{1}{\sqrt{b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt{\cos \left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \sec{\left (d + e x \right )} + c \tan{\left (d + e x \right )}} \sqrt{\cos{\left (d + e x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + b*sec(d + e*x) + c*tan(d + e*x))*sqrt(cos(d + e*x))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt{\cos \left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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