\(\int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx\) [378]

Optimal result

Integrand size = 23, antiderivative size = 54 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{f} \] Output:

-a^(1/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f+(a+b*sec(f*x+e)^2)^(1 
/2)/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(119\) vs. \(2(54)=108\).

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.20 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\frac {\left (-2 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right ) \cos (e+f x)+\sqrt {2} \sqrt {b} \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{b}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{\sqrt {2} \sqrt {b} f \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{b}}} \] Input:

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x],x]
 

Output:

((-2*Sqrt[a]*ArcSinh[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]]*Cos[e + f*x] + Sqrt[2 
]*Sqrt[b]*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/b])*Sqrt[a + b*Sec[e + f*x]^ 
2])/(Sqrt[2]*Sqrt[b]*f*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/b])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4627, 243, 60, 73, 221}

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.

\(\displaystyle \int \tan (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x) \sqrt {a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\int \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {a \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+2 \sqrt {a+b \sec ^2(e+f x)}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {2 a \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sec ^2(e+f x)}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \sqrt {a+b \sec ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}\)

Input:

Int[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x],x]
 

Output:

(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + b*Sec 
[e + f*x]^2])/(2*f)
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}\)	\(58\)
default	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}\)	\(58\)

Input:

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

Output:

1/f*((a+b*sec(f*x+e)^2)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sec(f*x+e)^2) 
^(1/2))/sec(f*x+e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (46) = 92\).

Time = 0.18 (sec) , antiderivative size = 312, normalized size of antiderivative = 5.78 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\left [\frac {\sqrt {a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, f}, \frac {\sqrt {-a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) + 4 \, \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f}\right ] \] Input:

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

Output:

[1/8*(sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160* 
a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x 
 + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x 
+ e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + 8*sqrt((a*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2))/f, 1/4*(sqrt(-a)*arctan(1/4*(8*a^2*cos 
(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 
+ b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^ 
2)) + 4*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/f]
 

Sympy [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx \] Input:

integrate((a+b*sec(f*x+e)**2)**(1/2)*tan(f*x+e),x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*tan(e + f*x), x)
 

Maxima [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \tan \left (f x + e\right ) \,d x } \] Input:

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

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*tan(f*x + e), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (46) = 92\).

Time = 0.30 (sec) , antiderivative size = 377, normalized size of antiderivative = 6.98 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\frac {2 \, {\left (\frac {a \arctan \left (-\frac {\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} + \sqrt {a + b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left ({\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} b + \sqrt {a + b} b\right )}}{{\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )}^{2} - 2 \, {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} \sqrt {a + b} + a - 3 \, b}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{f} \] Input:

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

Output:

2*(a*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x 
+ 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*t 
an(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))/sqrt(-a))/sqrt(-a) + 2*((sqr 
t(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/ 
2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 
 + a + b))*b + sqrt(a + b)*b)/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt( 
a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/ 
2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - 2*(sqrt(a + b)*tan(1/2*f 
*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 
 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a 
+ b) + a - 3*b))*sgn(cos(f*x + e))/f
 

Mupad [B] (verification not implemented)

Time = 16.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x) \, dx=\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{f}-\frac {\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sqrt {a}}\right )}{f} \] Input:

int(tan(e + f*x)*(a + b/cos(e + f*x)^2)^(1/2),x)
 

Output:

(a + b/cos(e + f*x)^2)^(1/2)/f - (a^(1/2)*atanh((a + b/cos(e + f*x)^2)^(1/ 
2)/a^(1/2)))/f
 

Reduce [F]

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

int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e),x)
 

Output:

(sqrt(sec(e + f*x)**2*b + a) + int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f* 
x))/(sec(e + f*x)**2*b + a),x)*a*f)/f