\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) [68]

Optimal result

Integrand size = 33, antiderivative size = 117 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \] Output:

2*B*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b/d/cos(d*x+c)^(1/2)/(b*sec(d*x+ 
c))^(1/2)+2/3*(A+3*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.50 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.22 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (6 i B \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-2 i (A+3 C) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )+\sqrt {1+e^{2 i (c+d x)}} (-3 i B+A \sin (c+d x))\right )}{3 b d \sqrt {1+e^{2 i (c+d x)}} \sqrt {b \sec (c+d x)}} \] Input:

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

Output:

(2*((6*I)*B*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - (2*I 
)*(A + 3*C)*E^(I*(c + d*x))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c 
+ d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*((-3*I)*B + A*Sin[c + d*x])))/(3* 
b*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4535, 3042, 4258, 3042, 3119, 4533, 3042, 4258, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4535

\(\displaystyle \int \frac {C \sec ^2(c+d x)+A}{(b \sec (c+d x))^{3/2}}dx+\frac {B \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {B \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\)

\(\Big \downarrow \) 4258

\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {B \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {B \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {(A+3 C) \int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+3 C) \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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 3120

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 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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 4533

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] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.79 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.92

method	result	size
parts	\(\frac {A \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {b \sec \left (d x +c \right )}\, b}+\frac {2 B \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{b d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}-\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d b \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(342\)
default	\(-\frac {2 \left (i B \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-3 \cos \left (d x +c \right )-6-3 \sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) A \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i B \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (3 \cos \left (d x +c \right )+6+3 \sec \left (d x +c \right )\right )+i C \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (3 \cos \left (d x +c \right )+6+3 \sec \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )-1\right ) A -3 B \sin \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b}\)	\(349\)

Input:

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

Output:

A/d*(-2/3*I*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I 
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1+sec(d*x+c))+2/3*sin(d*x+c))/(b*sec( 
d*x+c))^(1/2)/b+2*B/b/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)*(I*(1/(cos(d*x 
+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-co 
t(d*x+c)),I)*(-cos(d*x+c)-2-sec(d*x+c))+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x 
+c)+2+sec(d*x+c))+sin(d*x+c))-2*I*C/d*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I 
*(csc(d*x+c)-cot(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 complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.37 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \, A \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, b^{2} d} \] Input:

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

Output:

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)) + 3*I*sqrt(2)*B*sqrt(b)*weierstrassZeta(-4, 0, weierstrassP 
Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*B*sqrt(b)*wei 
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c))))/(b^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(b)*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d 
*x))/sec(c + d*x),x)*b + int(sqrt(sec(c + d*x)),x)*c))/b**2