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

Optimal. Leaf size=65 \[ \frac{2 \sin (c+d x)}{d \sqrt{b \cos (c+d x)}}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0444571, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {16, 2636, 2640, 2639} \[ \frac{2 \sin (c+d x)}{d \sqrt{b \cos (c+d x)}}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.0434025, size = 47, normalized size = 0.72 \[ \frac{2 \left (\sin (c+d x)-\sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 2.019, size = 165, normalized size = 2.5 \begin{align*} -2\,{\frac{\sqrt{-2\,b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b} \left ( \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ) }{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)*((sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-b*(2*sin(1/2*d
*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/(b*(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{\sec \left (d x + c\right )}{\sqrt{b \cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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