3.8.34 \(\int \sqrt {a+b \sec (c+d x)} (a^2-b^2 \sec ^2(c+d x)) \, dx\) [734]

3.8.34.1 Optimal result

Integrand size = 32, antiderivative size = 353 \[ \int \sqrt {a+b \sec (c+d x)} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {2 a (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}+\frac {2 \sqrt {a+b} \left (3 a^2+a b-b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}-\frac {2 a^2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]

output

2/3*a*(a-b)*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+2/3*(3*a^2+a*b-b^2)*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*a^2*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))*(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*( 
a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
 

3.8.34.2 Mathematica [A] (verified)

Time = 14.58 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.38 \[ \int \sqrt {a+b \sec (c+d x)} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (a^2-b^2 \sec ^2(c+d x)\right ) \left (2 a b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 \left (3 a^3-3 a^2 b+a b^2+b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+12 a^3 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a b \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 d (b+a \cos (c+d x)) \left (a^2-2 b^2+a^2 \cos (2 c+2 d x)\right )}+\frac {\cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (a^2-b^2 \sec ^2(c+d x)\right ) \left (-\frac {4}{3} a b \sin (c+d x)-\frac {4}{3} b^2 \tan (c+d x)\right )}{d \left (a^2-2 b^2+a^2 \cos (2 c+2 d x)\right )} \]

input

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

output

(4*Cos[(c + d*x)/2]^2*Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(a^2 - b^2*S 
ec[c + d*x]^2)*(2*a*b*(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*(3*a^3 - 3*a^2*b + a*b^2 + b^3)*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)] + 12*a^3*S 
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 
 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b 
)] + a*b*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x 
)/2]))/(3*d*(b + a*Cos[c + d*x])*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])) + ( 
Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(a^2 - b^2*Sec[c + d*x]^2)*((-4*a* 
b*Sin[c + d*x])/3 - (4*b^2*Tan[c + d*x])/3))/(d*(a^2 - 2*b^2 + a^2*Cos[2*c 
 + 2*d*x]))
 

3.8.34.3 Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4530, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4530

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4406

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

input

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

output

((2*a*(a - b)*Sqrt[a + b]*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*Sqrt[a + b]*(3*a^2 + a*b 
- b^2)*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^2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(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)/3 - (2*b^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)
 

3.8.34.3.1 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 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 4530

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

3.8.34.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1309\) vs. \(2(318)=636\).

Time = 9.86 (sec) , antiderivative size = 1310, normalized size of antiderivative = 3.71

method	result	size
parts	\(\text {Expression too large to display}\)	\(1310\)
default	\(\text {Expression too large to display}\)	\(1917\)

input

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

output

2*a^2/d*(cos(d*x+c)+1)*(-2*a*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a 
+b))^(1/2))+EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a-Ellipti 
cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b)*(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)*(a+b*sec(d*x+c))^ 
(1/2)/(b+a*cos(d*x+c))+2/3*b/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(co 
s(d*x+c)+1)*(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)* 
a*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^2*cos(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)*a^2*cos(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)*a*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)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(1/2)*a*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^2*cos(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)*a^2*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x...
 

3.8.34.5 Fricas [F]

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

input

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

output

integral(-(b^2*sec(d*x + c)^2 - a^2)*sqrt(b*sec(d*x + c) + a), x)
 

3.8.34.6 Sympy [F]

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

input

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

output

Integral((a - b*sec(c + d*x))*(a + b*sec(c + d*x))**(3/2), x)
 

3.8.34.7 Maxima [F]

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

input

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

output

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

3.8.34.8 Giac [F]

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

input

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

output

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

3.8.34.9 Mupad [F(-1)]

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

input

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

output

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