\(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\) [1080]

Optimal result

Integrand size = 23, antiderivative size = 48 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/3*(3*A+C)*(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))/d+2/3*C*sin(
d*x+c)/d/cos(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4151, 3091, 2720} \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[(A + C*Sec[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]

[Out]

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

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 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 4151

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[b^2, Int
[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e + f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {C+A \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {1}{3} (-3 A-C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left ((3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {C \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \]

[In]

Integrate[(A + C*Sec[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(265\) vs. \(2(68)=136\).

Time = 1.46 (sec) , antiderivative size = 266, normalized size of antiderivative = 5.54

method	result	size
default	\(-\frac {2 \left (-2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (3 A +C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\)	\(266\)

[In]

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

[Out]

-2/3*(-2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*EllipticF(cos(1/2*d*x+1/2*c),2^(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)*(3*A+C)*sin(1/2*d*x+1/2*c)^2+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(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)*sin(1/2*d*x+1/2*c)^2)^(
1/2)/(-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)^(3/2)/sin(1/2*d*x+1/2*c)/
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 = 105, normalized size of antiderivative = 2.19 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-3 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.61 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,A\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*A*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(
c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))