3.799 \(\int \frac{a+b \sec (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=57 \[ \frac{2 a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}-\frac{2 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

[Out]

(-2*b*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d + (2*b*Sin[c + d*x])/(d*Sqrt[Cos[c + d*
x]])

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Rubi [A]  time = 0.066769, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4225, 2748, 2636, 2639, 2641} \[ \frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}-\frac{2 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])/Sqrt[Cos[c + d*x]],x]

[Out]

(-2*b*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d + (2*b*Sin[c + d*x])/(d*Sqrt[Cos[c + d*
x]])

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{a+b \sec (c+d x)}{\sqrt{\cos (c+d x)}} \, dx &=\int \frac{b+a \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=a \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+b \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-b \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.140197, size = 51, normalized size = 0.89 \[ \frac{2 \left (a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{b \sin (c+d x)}{\sqrt{\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*EllipticE[(c + d*x)/2, 2]) + a*EllipticF[(c + d*x)/2, 2] + (b*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/d

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Maple [A]  time = 1.556, size = 148, normalized size = 2.6 \begin{align*} -2\,{\frac{\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}a+\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}b-2\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a+(sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b-2*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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