\(\int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [111]

Optimal result

Integrand size = 21, antiderivative size = 67 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \]

[Out]

2/3*b*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*
d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2716, 2721, 2720} \[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[Sec[c + d*x]^2/Sqrt[b*Cos[c + d*x]],x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b*Sin[c + d*x])/(3*d*(b*Cos[c
+ d*x])^(3/2))

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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^2 \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {1}{3} \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {2 b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 \sqrt {b \cos (c+d x)}} \\ & = \frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \left (\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\tan (c+d x)\right )}{3 d \sqrt {b \cos (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^2/Sqrt[b*Cos[c + d*x]],x]

[Out]

(2*(Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + Tan[c + d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(237\) vs. \(2(83)=166\).

Time = 1.79 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.55

method	result	size
default	\(-\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(238\)

[In]

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

[Out]

-2/3*(-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(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)/(2*cos(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/((2*co
s(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 = 103, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/3*(-I*sqrt(2)*sqrt(b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*s
qrt(b)*cos(d*x + c)^2*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)^2)

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {b \cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(sec(c + d*x)**2/sqrt(b*cos(c + d*x)), x)

Maxima [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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