\(\int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [110]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)}}+\frac {2 \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {16, 2716, 2721, 2719} \[ \int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}} \]

[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 2716

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 2719

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

Rule 2721

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

Rubi steps \begin{align*} \text {integral}& = 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] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \left (-\sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x)\right )}{d \sqrt {b \cos (c+d x)}} \]

[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]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(194\) vs. \(2(85)=170\).

Time = 1.90 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.00

method	result	size
default	\(-\frac {2 \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(195\)

[In]

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

[Out]

-2*(-2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^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)*(-2*sin(1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d \cos \left (d x + c\right )} \]

[In]

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

[Out]

(-I*sqrt(2)*sqrt(b)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) + I*sqrt(2)*sqrt(b)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d
*x + c))) + 2*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c))

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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