3.9.40 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [840]

3.9.40.1 Optimal result
3.9.40.2 Mathematica [A] (warning: unable to verify)
3.9.40.3 Rubi [A] (verified)
3.9.40.4 Maple [B] (verified)
3.9.40.5 Fricas [F]
3.9.40.6 Sympy [F]
3.9.40.7 Maxima [F]
3.9.40.8 Giac [F]
3.9.40.9 Mupad [F(-1)]

3.9.40.1 Optimal result

Integrand size = 34, antiderivative size = 210 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} C \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 d}+\frac {2 \sqrt {a+b} (B-C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d} \]

output
-2*(a-b)*C*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^2/d+2*(B-C)*cot(d*x+c)*EllipticF((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/d
 
3.9.40.2 Mathematica [A] (warning: unable to verify)

Time = 13.04 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.49 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \sqrt {\sec (c+d x)} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) C \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (B+C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+C \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \sec (c+d x)}}+\frac {2 C (b+a \cos (c+d x)) \tan (c+d x)}{b d \sqrt {a+b \sec (c+d x)}} \]

input
Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/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)*C* 
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*(B + C)*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[ArcSin[Tan[(c + d*x)/2]], (a - 
b)/(a + b)] + C*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[( 
c + d*x)/2]))/(b*d*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[a + b*Sec[c + d*x]]) + (2 
*C*(b + a*Cos[c + d*x])*Tan[c + d*x])/(b*d*Sqrt[a + b*Sec[c + d*x]])
 
3.9.40.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3042, 4546, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

input
Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Sqrt[a + b*Sec[c + d*x]],x]
 
output
(-2*(a - b)*Sqrt[a + b]*C*Cot[c + d*x]*EllipticE[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^2*d) + (2*Sqrt[a + b]*(B - C) 
*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)
 

3.9.40.3.1 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 4546
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   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, C}, x] && NeQ[a^2 - b^2, 0]
 
3.9.40.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(930\) vs. \(2(192)=384\).

Time = 10.12 (sec) , antiderivative size = 931, normalized size of antiderivative = 4.43

method result size
parts \(\text {Expression too large to display}\) \(931\)
default \(\text {Expression too large to display}\) \(1082\)

input
int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
-2*B/d*(cos(d*x+c)+1)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos 
(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b) 
)^(1/2))*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))+2*C/d/b*((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)*Elliptic 
E(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*cos(d*x+c)^2+(cos(d*x+c)/(c 
os(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt 
icE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b*cos(d*x+c)^2-EllipticF(co 
t(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*(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)* 
EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*cos(d*x+c)+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 
)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b*cos(d*x+c)-2*Elli 
pticF(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)+(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a+(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti 
cE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b-(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)*EllipticF(cot(d...
 
3.9.40.5 Fricas [F]

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

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

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

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

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

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

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

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

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

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