3.10.77 \(\int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [977]

3.10.77.1 Optimal result

Integrand size = 50, antiderivative size = 316 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} C \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b \sqrt {a+b} (B-C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 \sqrt {a+b} (b B-a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d} \]

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)/d+2*b*(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)/d-2*(B*b-C*a)*cot(d*x+c)*EllipticPi((a+b*se 
c(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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)/d
 

3.10.77.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(965\) vs. \(2(316)=632\).

Time = 23.90 (sec) , antiderivative size = 965, normalized size of antiderivative = 3.05 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 b C \cos (c+d x) \sqrt {a+b \sec (c+d x)} (b B-a C+b C \sec (c+d x)) \sin (c+d x)}{d (b C+b B \cos (c+d x)-a C \cos (c+d x))}+\frac {2 \sqrt {a+b \sec (c+d x)} (b B-a C+b C \sec (c+d x)) \left (a b C \tan \left (\frac {1}{2} (c+d x)\right )+b^2 C \tan \left (\frac {1}{2} (c+d x)\right )-2 a b C \tan ^3\left (\frac {1}{2} (c+d x)\right )+a b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-b^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-2 a b B \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 a^2 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 a b B \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 a^2 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) C E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (-a b B+a^2 C+b^2 (B+C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{d \sqrt {b+a \cos (c+d x)} (b C+b B \cos (c+d x)-a C \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}} \]

input

Integrate[(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2)/Sqrt 
[a + b*Sec[c + d*x]],x]
 

output

(2*b*C*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(b*B - a*C + b*C*Sec[c + d*x] 
)*Sin[c + d*x])/(d*(b*C + b*B*Cos[c + d*x] - a*C*Cos[c + d*x])) + (2*Sqrt[ 
a + b*Sec[c + d*x]]*(b*B - a*C + b*C*Sec[c + d*x])*(a*b*C*Tan[(c + d*x)/2] 
 + b^2*C*Tan[(c + d*x)/2] - 2*a*b*C*Tan[(c + d*x)/2]^3 + a*b*C*Tan[(c + d* 
x)/2]^5 - b^2*C*Tan[(c + d*x)/2]^5 - 2*a*b*B*EllipticPi[-1, ArcSin[Tan[(c 
+ d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a* 
Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*C*EllipticPi[- 
1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2] 
*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a 
*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d* 
x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + 
b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*C*EllipticPi[-1, ArcSin[Tan[(c + d* 
x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*S 
qrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(a 
+ b)*C*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[( 
c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 
 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (-(a*b*B) + a^2*C + b^2*(B + C))*Ellip 
ticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^ 
2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c 
+ d*x)/2]^2)/(a + b)]))/(d*Sqrt[b + a*Cos[c + d*x]]*(b*C + b*B*Cos[c + ...
 

3.10.77.3 Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4529, 3042, 4404, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4529

\(\displaystyle \frac {\int \sqrt {a+b \sec (c+d x)} \left (C \sec (c+d x) b^3+(b B-a C) b^2\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4404

\(\displaystyle \frac {a b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\sec (c+d x) \left (C \sec (c+d x) b^4+a C b^3+(b B-a C) b^3\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (C \csc \left (c+d x+\frac {\pi }{2}\right ) b^4+a C b^3+(b B-a C) b^3\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (C \csc \left (c+d x+\frac {\pi }{2}\right ) b^4+a C b^3+(b B-a C) b^3\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b^2}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {b^4 (B-C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b^4 C \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^4 (B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^4 C \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b^2}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {b^4 C \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^3 \sqrt {a+b} (B-C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b^2}\)

\(\Big \downarrow \) 4492

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

input

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

output

((-2*(a - b)*b^2*Sqrt[a + b]*C*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Se 
c[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))])/d + (2*b^3*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))])/d - (2*b^2*Sqrt[a + b]*(b*B - a*C)*Cot[c + d*x]*Ellipt 
icPi[(a + b)/a, 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))])/d)/b^2
 

3.10.77.3.1 Defintions of rubi rules used

rule 3042

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

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4404

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ 
.) + (c_)), x_Symbol] :> Simp[a*c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] 
+ Int[Csc[e + f*x]*((b*c + a*d + b*d*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]] 
), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4529

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

3.10.77.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1090\) vs. \(2(289)=578\).

Time = 14.48 (sec) , antiderivative size = 1091, normalized size of antiderivative = 3.45

method	result	size
parts	\(\text {Expression too large to display}\)	\(1091\)
default	\(\text {Expression too large to display}\)	\(2139\)

input

int((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/ 
2),x,method=_RETURNVERBOSE)
 

output

2*a*(B*b-C*a)/d*(1+cos(d*x+c))*(EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ 
b))^(1/2))-2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2)))*(1/ 
(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^( 
1/2)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))-2*B*b^2/d*(1+cos(d*x+c))*(1/( 
a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(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*b/d*(EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ 
b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+ 
cos(d*x+c)))^(1/2)*b*cos(d*x+c)^2-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/( 
a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/( 
1+cos(d*x+c)))^(1/2)*a*cos(d*x+c)^2-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b) 
/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)) 
/(1+cos(d*x+c)))^(1/2)*b*cos(d*x+c)^2+2*EllipticF(cot(d*x+c)-csc(d*x+c),(( 
a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x 
+c))/(1+cos(d*x+c)))^(1/2)*b*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c), 
((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d 
*x+c))/(1+cos(d*x+c)))^(1/2)*a*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c 
),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos 
(d*x+c))/(1+cos(d*x+c)))^(1/2)*b*cos(d*x+c)+(cos(d*x+c)/(1+cos(d*x+c)))^(1 
/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d*x+c...
 

3.10.77.5 Fricas [F]

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

input

integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c 
))^(1/2),x, algorithm="fricas")
 

output

integral((C*b*sec(d*x + c) - C*a + B*b)*sqrt(b*sec(d*x + c) + a), x)
 

3.10.77.6 Sympy [F]

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

input

integrate((B*a*b-C*a**2+b**2*B*sec(d*x+c)+b**2*C*sec(d*x+c)**2)/(a+b*sec(d 
*x+c))**(1/2),x)
 

output

-Integral(-B*b*sqrt(a + b*sec(c + d*x)), x) - Integral(C*a*sqrt(a + b*sec( 
c + d*x)), x) - Integral(-C*b*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)
 

3.10.77.7 Maxima [F]

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

input

integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c 
))^(1/2),x, algorithm="maxima")
 

output

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)/sqrt 
(b*sec(d*x + c) + a), x)
 

3.10.77.8 Giac [F]

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

input

integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c 
))^(1/2),x, algorithm="giac")
 

output

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)/sqrt 
(b*sec(d*x + c) + a), x)
 

3.10.77.9 Mupad [F(-1)]

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

input

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

output

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