\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) [578]

Optimal result

Integrand size = 33, antiderivative size = 86 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}+\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b (a+b) d}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(206\) vs. \(2(86)=172\).

Time = 1.74 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.40 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {\frac {2 (2 A b-3 a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {2 b B \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {4 B \sin (c+d x)}{\sqrt {\cos (c+d x)}}-\frac {2 B \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{2 b d} \] Input:

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

Output:

((2*(2*A*b - 3*a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) - ( 
2*b*B*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d 
*x)/2, 2])/(a + b)))/a + (4*B*Sin[c + d*x])/Sqrt[Cos[c + d*x]] - (2*B*(-2* 
a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcS 
in[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt 
[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/(2*b*d)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3433, 3042, 3479, 27, 3042, 3538, 25, 27, 3042, 3119, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {A \cos (c+d x)+B}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

\(\displaystyle \frac {2 \int \frac {-a B \cos ^2(c+d x)-b B \cos (c+d x)+A b-a B}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-a B \cos ^2(c+d x)-b B \cos (c+d x)+A b-a B}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-a B \sin \left (c+d x+\frac {\pi }{2}\right )^2-b B \sin \left (c+d x+\frac {\pi }{2}\right )+A b-a B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {-\frac {\int -\frac {a (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-B \int \sqrt {\cos (c+d x)}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {a (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-B \int \sqrt {\cos (c+d x)}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx-B \int \sqrt {\cos (c+d x)}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-B \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {(A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}+\frac {2 B \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}\)

Input:

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

Output:

((-2*B*EllipticE[(c + d*x)/2, 2])/d + (2*(A*b - a*B)*EllipticPi[(2*a)/(a + 
 b), (c + d*x)/2, 2])/((a + b)*d))/b + (2*B*Sin[c + d*x])/(b*d*Sqrt[Cos[c 
+ d*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3479

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0]) 
))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(88)=176\).

Time = 2.87 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.47

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 B \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {2 \left (A b -B a \right ) a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{b \left (a^{2}-a b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(298\)

Input:

int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b/sin(1/2* 
d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-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)*sin(1/2*d*x+1/2*c)^2-EllipticE(co 
s(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))-2*(A*b-B*a)/b/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2* 
cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/sin(1/2*d*x+1/2 
*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)