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

[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))*((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c
^2)^(1/2)))^(1/2)/e/cos(e*x+d)^(1/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)

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Rubi [A]  time = 0.15, 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 = {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 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 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]

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*}

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Mathematica [C]  time = 2.90, size = 506, normalized size = 4.29 \[ \frac {4 \left (\sqrt {a^2-b^2+c^2}+i a-i b+c\right ) (\cos (d+e x)+i \sin (d+e x)) \sqrt {-\frac {i \left (\sqrt {a^2-b^2+c^2}+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )-c\right )}{\left (\sqrt {a^2-b^2+c^2}-i a+i b-c\right ) \left (\tan \left (\frac {1}{2} (d+e x)\right )-i\right )}} \sqrt {-\frac {i \left (\sqrt {a^2-b^2+c^2}+(b-a) \tan \left (\frac {1}{2} (d+e x)\right )+c\right )}{\left (\sqrt {a^2-b^2+c^2}+i a-i b+c\right ) \left (\tan \left (\frac {1}{2} (d+e x)\right )-i\right )}} \sqrt {\frac {\left (\sqrt {a^2-b^2+c^2}-i a+i b+c\right ) (-\cos (d+e x)+i \sin (d+e x))}{\sqrt {a^2-b^2+c^2}+i a-i b+c}} F\left (\sin ^{-1}\left (\sqrt {\frac {\left (-i a+i b+c+\sqrt {a^2-b^2+c^2}\right ) (i \sin (d+e x)-\cos (d+e x))}{i a-i b+c+\sqrt {a^2-b^2+c^2}}}\right )|\frac {b+i \sqrt {a^2-b^2+c^2}}{b-i \sqrt {a^2-b^2+c^2}}\right )}{e \left (a+i \left (\sqrt {a^2-b^2+c^2}+i b+c\right )\right ) \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]

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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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}}{b \cos \left (e x + d\right ) \sec \left (e x + d\right ) + c \cos \left (e x + d\right ) \tan \left (e x + d\right ) + a \cos \left (e x + d\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}}\,{d x} \]

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)

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maple [C]  time = 1.53, size = 714, normalized size = 6.05 \[ -\frac {4 i \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} \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 ) \left (\sqrt {\cos }\left (e x +d \right )\right ) \left (\cos \left (e x +d \right )-1\right )^{2} \left (i \sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-i a \cos \left (e x +d \right )+i \cos \left (e x +d \right ) b -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(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*((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*co
s(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*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))*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*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 \[ \int \frac {1}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}}\,{d x} \]

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

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

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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