3.449 \(\int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx\)
Optimal. Leaf size=118 \[ \frac {2 \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}} \]
[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))*EllipticE(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))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/e/se
c(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)
________________________________________________________________________________________
Rubi [A] time = 0.14, 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, 3119, 2653} \[ \frac {2 \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]]/Sqrt[Sec[d + e*x]],x]
[Out]
(2*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[a + b*Sec[d + e*x] +
c*Tan[d + e*x]])/(e*Sqrt[Sec[d + e*x]]*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 3119
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*C
os[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])], Int[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] /; FreeQ[{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 {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx &=\frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)} \, dx}{\sqrt {\sec (d+e x)} \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}\\ &=\frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \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 {\sec (d+e x)} \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}\\ &=\frac {2 E\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 {a+b \sec (d+e x)+c \tan (d+e x)}}{e \sqrt {\sec (d+e x)} \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 6.25, size = 1580, normalized size = 13.39 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]]/Sqrt[Sec[d + e*x]],x]
[Out]
(2*a*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(c*e*Sqrt[Sec[d + e*x]]) + (2*b*AppellF1[1/2, 1/2, 1/2, 3/2, -
((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -(
(b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Se
c[d + e*x + ArcTan[a/c]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(
b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 +
c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[a + b*Se
c[d + e*x] + c*Tan[d + e*x]])/(Sqrt[1 + a^2/c^2]*c*e*Sqrt[Sec[d + e*x]]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*
x]]) + (a^2*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sq
rt[1 + c^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqr
t[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt
[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c^2)/a^2])]*Sqrt[b
+ a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt[(a^2 + c^2)/a^2]*
Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - A
rcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[b + a*Sqrt[1 + c^2/a^2]
*Cos[d + e*x - ArcTan[c/a]]])*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(c*e*Sqrt[Sec[d + e*x]]*Sqrt[b + a*Co
s[d + e*x] + c*Sin[d + e*x]]) + (c*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*
x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x
- ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1
+ c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^
2 + c^2)/a^2])]*Sqrt[b + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a
*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c
^2/a^2]*Cos[d + e*x - ArcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[
b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]])*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(e*Sqrt[Sec[d
+ e*x]]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]])
________________________________________________________________________________________
fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}{\sqrt {\sec \left (e x + d\right )}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/sec(e*x+d)^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)/sqrt(sec(e*x + d)), x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}{\sqrt {\sec \left (e x + d\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/sec(e*x+d)^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)/sqrt(sec(e*x + d)), x)
________________________________________________________________________________________
maple [C] time = 1.89, size = 12462, normalized size = 105.61 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/sec(e*x+d)^(1/2),x)
[Out]
result too large to display
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}{\sqrt {\sec \left (e x + d\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/sec(e*x+d)^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)/sqrt(sec(e*x + d)), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}}}{\sqrt {\frac {1}{\cos \left (d+e\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + c*tan(d + e*x) + b/cos(d + e*x))^(1/2)/(1/cos(d + e*x))^(1/2),x)
[Out]
int((a + c*tan(d + e*x) + b/cos(d + e*x))^(1/2)/(1/cos(d + e*x))^(1/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}}}{\sqrt {\sec {\left (d + e x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(e*x+d)+c*tan(e*x+d))**(1/2)/sec(e*x+d)**(1/2),x)
[Out]
Integral(sqrt(a + b*sec(d + e*x) + c*tan(d + e*x))/sqrt(sec(d + e*x)), x)
________________________________________________________________________________________