3.4.72 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [372]

3.4.72.1 Optimal result
3.4.72.2 Mathematica [A] (warning: unable to verify)
3.4.72.3 Rubi [A] (verified)
3.4.72.4 Maple [B] (verified)
3.4.72.5 Fricas [F]
3.4.72.6 Sympy [F]
3.4.72.7 Maxima [F]
3.4.72.8 Giac [F]
3.4.72.9 Mupad [F(-1)]

3.4.72.1 Optimal result

Integrand size = 33, antiderivative size = 261 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 A b-2 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 d}-\frac {2 \sqrt {a+b} (3 A b-(2 a+b) B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b d} \]

output
-2/3*(a-b)*(3*A*b-2*B*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b) 
^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b 
*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d-2/3*(3*A*b-(2*a+b)*B)*cot(d*x+c)*Ellipt 
icF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b 
*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d+2/3*B*( 
a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 
3.4.72.2 Mathematica [A] (warning: unable to verify)

Time = 19.31 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) (-3 A b+2 a B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (3 A b+(-2 a+b) B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-(3 A b-2 a B) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \sec (c+d x)}}+\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (\frac {2 (3 A b-2 a B) \sin (c+d x)}{3 b^2}+\frac {2 B \tan (c+d x)}{3 b}\right )}{d \sqrt {a+b \sec (c+d x)}} \]

input
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x 
]
 
output
(2*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-3 
*A*b + 2*a*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d* 
x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - 
 b)/(a + b)] + 2*b*(3*A*b + (-2*a + b)*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d 
*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - (3*A*b - 2*a*B)*Cos[c + d*x]*(b 
 + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^2*d*Sqrt[Sec 
[(c + d*x)/2]^2]*Sqrt[a + b*Sec[c + d*x]]) + ((b + a*Cos[c + d*x])*Sec[c + 
 d*x]*((2*(3*A*b - 2*a*B)*Sin[c + d*x])/(3*b^2) + (2*B*Tan[c + d*x])/(3*b) 
))/(d*Sqrt[a + b*Sec[c + d*x]])
 
3.4.72.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4498, 27, 3042, 4493, 3042, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

\(\displaystyle \frac {2 \int \frac {\sec (c+d x) (b B+(3 A b-2 a B) \sec (c+d x))}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec (c+d x) (b B+(3 A b-2 a B) \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b B+(3 A b-2 a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {(3 A b-2 a B) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(3 A b-B (2 a+b)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(3 A b-2 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(3 A b-B (2 a+b)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {(3 A b-2 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} (3 A b-B (2 a+b)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {-\frac {2 (a-b) \sqrt {a+b} (3 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 \sqrt {a+b} (3 A b-B (2 a+b)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\)

input
Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]
 
output
((-2*(a - b)*Sqrt[a + b]*(3*A*b - 2*a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqr 
t[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + 
d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*Sqrt 
[a + b]*(3*A*b - (2*a + b)*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec 
[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + 
 b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/(3*b) + (2*B*Sqrt[a + 
 b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d)
 

3.4.72.3.1 Defintions of rubi rules used

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 4319
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4492
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B)   Int[Csc[e 
 + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B   Int[Csc[e + f*x]*((1 + 
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} 
, x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
 

rule 4498
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 
3.4.72.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1944\) vs. \(2(235)=470\).

Time = 22.29 (sec) , antiderivative size = 1945, normalized size of antiderivative = 7.45

method result size
parts \(\text {Expression too large to display}\) \(1945\)
default \(\text {Expression too large to display}\) \(1951\)

input
int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
2*A/d/b*(EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*co 
s(d*x+c)^2+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b* 
cos(d*x+c)^2-EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* 
b*cos(d*x+c)^2+2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos 
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 
/2)*a*cos(d*x+c)+2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ 
(1/2)*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))* 
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*b*cos(d*x+c)+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))* 
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*a+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b-E 
llipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c 
)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b+a*sin(d*x+c) 
*cos(d*x+c)+sin(d*x+c)*b)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x 
+c)+1)-2/3*B/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+...
 
3.4.72.5 Fricas [F]

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

input
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="fricas")
 
output
integral((B*sec(d*x + c)^3 + A*sec(d*x + c)^2)/sqrt(b*sec(d*x + c) + a), x 
)
 
3.4.72.6 Sympy [F]

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

input
integrate(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)
 
output
Integral((A + B*sec(c + d*x))*sec(c + d*x)**2/sqrt(a + b*sec(c + d*x)), x)
 
3.4.72.7 Maxima [F]

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

input
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="maxima")
 
output
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*sec(d*x + c) + a), x)
 
3.4.72.8 Giac [F]

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

input
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="giac")
 
output
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*sec(d*x + c) + a), x)
 
3.4.72.9 Mupad [F(-1)]

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

input
int((A + B/cos(c + d*x))/(cos(c + d*x)^2*(a + b/cos(c + d*x))^(1/2)),x)
 
output
int((A + B/cos(c + d*x))/(cos(c + d*x)^2*(a + b/cos(c + d*x))^(1/2)), x)