\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) [20]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \]

[Out]

2/3*(A+3*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))*cos(d*x+c)^(
1/2)*(b*sec(d*x+c))^(1/2)/b^2/d+2/3*A*tan(d*x+c)/d/(b*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4130, 3856, 2720} \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \]

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^2*d) + (2*A*Tan[c + d*x])
/(3*d*(b*Sec[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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sec ^2(c+d x) \left (2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \sin (2 (c+d x))\right )}{3 d (b \sec (c+d x))^{3/2}} \]

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(3/2),x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.91 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.91

method	result	size
parts	\(-\frac {2 A \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )-\sin \left (d x +c \right )\right )}{3 d \sqrt {b \sec \left (d x +c \right )}\, b}-\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )}{d b \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(218\)
default	\(\frac {\frac {2 i A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{3}+2 i C \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )}{3}+2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sec \left (d x +c \right )+\frac {2 A \sin \left (d x +c \right )}{3}}{d \sqrt {b \sec \left (d x +c \right )}\, b}\)	\(264\)

[In]

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*A/d/(b*sec(d*x+c))^(1/2)/b*(I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-co
t(d*x+c)+csc(d*x+c)),I)+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+c)+
csc(d*x+c)),I)*sec(d*x+c)-sin(d*x+c))-2*I*C/d*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)
/b/(b*sec(d*x+c))^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \, A \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, b^{2} d} \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(3/2), x)

Maxima [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(3/2), x)

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(3/2),x)

[Out]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(3/2), x)