\(\int \sqrt {a+b \sec (c+d x)} (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [976]

Optimal result

Integrand size = 50, antiderivative size = 382 \[ \int \sqrt {a+b \sec (c+d x)} \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}-\frac {2 \sqrt {a+b} \left (b^2 (3 B-C)-a b (6 B-C)+3 a^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}-\frac {2 a \sqrt {a+b} (b B-a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b^2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \] Output:

-2/3*(a-b)*(a+b)^(1/2)*(3*B*b+C*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^( 
1/2)/(a+b)^(1/2),((a+b)/(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-2/3*(a+b)^(1/2)*(b^2*(3*B-C)-a*b*(6*B-C)+3*C* 
a^2)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(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 
*a*(a+b)^(1/2)*(B*b-C*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b 
)^(1/2),(a+b)/a,((a+b)/(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+2/3*b^2*C*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1131\) vs. \(2(382)=764\).

Time = 16.53 (sec) , antiderivative size = 1131, normalized size of antiderivative = 2.96 \[ \int \sqrt {a+b \sec (c+d x)} \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(a + b*Sec[c + d*x])^(3/2)*(b*B - a*C + b*C*Sec[c + d*x])*(3*a*b^2*B*Ta 
n[(c + d*x)/2] + 3*b^3*B*Tan[(c + d*x)/2] + a^2*b*C*Tan[(c + d*x)/2] + a*b 
^2*C*Tan[(c + d*x)/2] - 6*a*b^2*B*Tan[(c + d*x)/2]^3 - 2*a^2*b*C*Tan[(c + 
d*x)/2]^3 + 3*a*b^2*B*Tan[(c + d*x)/2]^5 - 3*b^3*B*Tan[(c + d*x)/2]^5 + a^ 
2*b*C*Tan[(c + d*x)/2]^5 - a*b^2*C*Tan[(c + d*x)/2]^5 - 6*a^2*b*B*Elliptic 
Pi[-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)] + 
 6*a^3*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)] - 6*a^2*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)] + 6*a^3*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]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x) 
/2]^2)/(a + b)] + b*(a + b)*(3*b*B + a*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)] - (3* 
a^3*C - 3*a^2*b*(B + C) + b^3*(3*B + C) + a*b^2*(6*B + C))*EllipticF[ArcSi 
n[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Ta 
n[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/...
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.260, Rules used = {3042, 4529, 3042, 4406, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4406

\(\displaystyle \frac {\frac {2}{3} \int \frac {(3 b B+a C) \sec ^2(c+d x) b^4+\left (-3 C a^2+6 b B a+b^2 C\right ) \sec (c+d x) b^3+3 a^2 (b B-a C) b^2}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(3 b B+a C) \sec ^2(c+d x) b^4+\left (-3 C a^2+6 b B a+b^2 C\right ) \sec (c+d x) b^3+3 a^2 (b B-a C) b^2}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {(3 b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+\left (-3 C a^2+6 b B a+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+3 a^2 (b B-a C) b^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {\frac {1}{3} \left (\int \frac {3 a^2 (b B-a C) b^2+\left (b^3 \left (-3 C a^2+6 b B a+b^2 C\right )-b^4 (3 b B+a C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b^4 (a C+3 b B) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\int \frac {3 a^2 (b B-a C) b^2+\left (b^3 \left (-3 C a^2+6 b B a+b^2 C\right )-b^4 (3 b B+a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^4 (a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {\frac {1}{3} \left (3 a^2 b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b^3 \left (3 a^2 C-a b (6 B-C)+b^2 (3 B-C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b^4 (a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (3 a^2 b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^3 \left (3 a^2 C-a b (6 B-C)+b^2 (3 B-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^4 (a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {\frac {1}{3} \left (-b^3 \left (3 a^2 C-a b (6 B-C)+b^2 (3 B-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^4 (a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{3} \left (b^4 (a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} \left (3 a^2 C-a b (6 B-C)+b^2 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {6 a b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{3} \left (-\frac {2 b^2 \sqrt {a+b} \left (3 a^2 C-a b (6 B-C)+b^2 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 b^2 (a-b) \sqrt {a+b} (a C+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {6 a b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2}\)

Input:

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

Output:

(((-2*(a - b)*b^2*Sqrt[a + b]*(3*b*B + a*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))])/d - (2*b^2*Sqr 
t[a + b]*(b^2*(3*B - C) - a*b*(6*B - C) + 3*a^2*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 - (6* 
a*b^2*Sqrt[a + b]*(b*B - a*C)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sq 
rt[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)/3 + (2*b^4*C* 
Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/b^2
 

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 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 4406

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c*m 
 + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m 
 - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b* 
c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], 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 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]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1151\) vs. \(2(347)=694\).

Time = 44.08 (sec) , antiderivative size = 1152, normalized size of antiderivative = 3.02

method	result	size
parts	\(\text {Expression too large to display}\)	\(1152\)
default	\(\text {Expression too large to display}\)	\(1454\)

Input:

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

Output:

-2*a*(B*b-C*a)/d*(2*a*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^( 
1/2))-EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a+EllipticF(-c 
sc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b)*(cos(d*x+c)+1)*(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)*(a+b*s 
ec(d*x+c))^(1/2)/(b+a*cos(d*x+c))-1/2*B*b^2/d/(b+a*cos(d*x+c))*(cos(d*x+c) 
+1)^2*((-(1-cos(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a+((1-cos(d* 
x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*b-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)*EllipticF(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a-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)*EllipticF(-csc(d*x+c)+cot( 
d*x+c),((a-b)/(a+b))^(1/2))*b+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(-csc(d*x+c)+cot(d*x+c),( 
(a-b)/(a+b))^(1/2))*a+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co 
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a 
+b))^(1/2))*b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(a+b*sec(d*x+c))^(1/2)*se 
c(d*x+c)+2/3*C*b/d*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*c 
os(d*x+c)+b)*((cos(d*x+c)^2+2*cos(d*x+c)+1)*(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^2*EllipticE(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*(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/...
 

Fricas [F]

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

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

Output:

integral((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)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

 - int(sqrt(sec(c + d*x)*b + a),x)*a**2*c + int(sqrt(sec(c + d*x)*b + a),x 
)*a*b**2 + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**2,x)*b**2*c + int(sq 
rt(sec(c + d*x)*b + a)*sec(c + d*x),x)*b**3