3.6.32 \(\int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\) [532]

3.6.32.1 Optimal result
3.6.32.2 Mathematica [A] (verified)
3.6.32.3 Rubi [A] (verified)
3.6.32.4 Maple [B] (verified)
3.6.32.5 Fricas [F]
3.6.32.6 Sympy [F]
3.6.32.7 Maxima [F]
3.6.32.8 Giac [F]
3.6.32.9 Mupad [F(-1)]

3.6.32.1 Optimal result

Integrand size = 23, antiderivative size = 241 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=-\frac {2 a (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]

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)/b^2/d-2/3*(a-b)*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+2/3*(a+b*sec(d*x+c))^(1/2)*tan(d* 
x+c)/d
 
3.6.32.2 Mathematica [A] (verified)

Time = 15.79 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.22 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=-\frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+b \sec (c+d x)} \left (2 a (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b d (b+a \cos (c+d x))}+\frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 a \sin (c+d x)}{3 b}+\frac {2}{3} \tan (c+d x)\right )}{d} \]

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

Time = 0.78 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4322, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4322

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

input
Int[Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]],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))])/(b^2*d) - (2*(a - b)*Sqrt[a + 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 + (2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3* 
d)
 

3.6.32.3.1 Defintions of rubi rules used

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 4322
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] 
+ Simp[m/(m + 1)   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(b + a*Csc 
[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 
 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]
 
3.6.32.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1126\) vs. \(2(215)=430\).

Time = 9.64 (sec) , antiderivative size = 1127, normalized size of antiderivative = 4.68

method result size
default \(\text {Expression too large to display}\) \(1127\)

input
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
-2/3/d/b*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(EllipticF 
(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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*cos(d*x+c)^2+Ellip 
ticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b^2*cos(d*x+c)^2-E 
llipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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^2*cos(d*x+c) 
^2-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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*cos(d* 
x+c)^2+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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* 
cos(d*x+c)+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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)* 
b^2*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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^2*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))* 
(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*cos(d*x+c)+(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ell 
ipticF(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)*a*b+(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF...
 
3.6.32.5 Fricas [F]

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

input
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
 
output
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^2, x)
 
3.6.32.6 Sympy [F]

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

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

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

input
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
 
output
integrate(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^2, x)
 
3.6.32.8 Giac [F]

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

input
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
 
output
integrate(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^2, x)
 
3.6.32.9 Mupad [F(-1)]

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

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