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

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

[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))*cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e
*x+d))^(1/2)/e/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(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, 3119, 2653} \[ \frac {2 \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[d + e*x]]*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[(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 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 \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx &=\frac {\left (\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}\right ) \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)} \, dx}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}\\ &=\frac {\left (\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}\right ) \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 {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}\\ &=\frac {2 \sqrt {\cos (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 ) \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{e \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}\\ \end {align*}

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Mathematica [F]  time = 21.22, size = 0, normalized size = 0.00 \[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 = 1.02, size = 12459, normalized size = 105.58 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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