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*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])])

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Rubi [A]  time = 0.146648, 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, 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 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 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 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]

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*}

Mathematica [F]  time = 21.2875, size = 0, normalized size = 0. \[ \int \sqrt{\cos (d+e x)} \sqrt{a+b \sec (d+e x)+c \tan (d+e x)} \, dx \]

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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Maple [C]  time = 0.444, size = 12460, normalized size = 105.6 \begin{align*} \text{output too large to display} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)