3.4.32 \(\int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [332]

3.4.32.1 Optimal result

Integrand size = 23, antiderivative size = 310 \[ \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}+\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d} \]

output

-2*(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*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*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)/a/d
 

3.4.32.2 Mathematica [A] (warning: unable to verify)

Time = 12.55 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.23 \[ \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {-\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \left (2 (a+b) \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-4 b \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+4 b \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {3}{2} (c+d x)\right )-a \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )+2 b \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )\right )+2 (b+a \cos (c+d x)) \tan (c+d x)}{b d \sqrt {a+b \sec (c+d x)}} \]

input

Integrate[Tan[c + d*x]^2/Sqrt[a + b*Sec[c + d*x]],x]
 

output

(-(Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^4]*Sqrt[Sec[c + d 
*x]]*Sqrt[1 + Sec[c + d*x]]*(2*(a + b)*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)] 
- 4*b*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 4*b*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*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Si 
n[(3*(c + d*x))/2] - a*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x) 
/2] + 2*b*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x)/2])) + 2*(b 
+ a*Cos[c + d*x])*Tan[c + d*x])/(b*d*Sqrt[a + b*Sec[c + d*x]])
 

3.4.32.3 Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4382, 3042, 4547, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4382

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4547

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

input

Int[Tan[c + d*x]^2/Sqrt[a + b*Sec[c + d*x]],x]
 

output

(-2*(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)]*S 
qrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*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) + (2*Sqrt[a + b]*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))])/(a*d)
 

3.4.32.3.1 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 4382

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

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 4547

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]* 
(b_.) + (a_)], x_Symbol] :> Int[(A - 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, C}, x] && NeQ[a^2 - b^2, 0]
 

3.4.32.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1068\) vs. \(2(283)=566\).

Time = 10.92 (sec) , antiderivative size = 1069, normalized size of antiderivative = 3.45

method	result	size
default	\(\text {Expression too large to display}\)	\(1069\)

input

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

output

2/d/b*(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)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(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)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(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)/(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*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*cos(d*x+c)^2+4*(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)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b 
)/(a+b))^(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)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(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)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(c 
os(d*x+c)+1))^(1/2)*b*cos(d*x+c)-4*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*cos(d*x+c)+2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c) 
+1))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*b+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))/(c...
 

3.4.32.5 Fricas [F]

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

input

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

output

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

3.4.32.6 Sympy [F]

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

input

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

output

Integral(tan(c + d*x)**2/sqrt(a + b*sec(c + d*x)), x)
 

3.4.32.7 Maxima [F]

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

input

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

output

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

3.4.32.8 Giac [F]

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

input

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

output

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

3.4.32.9 Mupad [F(-1)]

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

input

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

output

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