\(\int (a+b \sec ^2(e+f x))^{3/2} \tan (e+f x) \, dx\) [391]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f} \] Output:

-a^(3/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f+a*(a+b*sec(f*x+e)^2)^ 
(1/2)/f+1/3*(a+b*sec(f*x+e)^2)^(3/2)/f
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {2 b \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {a \cos ^2(e+f x)}{b}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f \sqrt {1+\frac {a \cos ^2(e+f x)}{b}} (a+2 b+a \cos (2 (e+f x)))} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x],x]
 

Output:

(2*b*Hypergeometric2F1[-3/2, -3/2, -1/2, -((a*Cos[e + f*x]^2)/b)]*(a + b*S 
ec[e + f*x]^2)^(3/2))/(3*f*Sqrt[1 + (a*Cos[e + f*x]^2)/b]*(a + 2*b + a*Cos 
[2*(e + f*x)]))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4627, 243, 60, 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.

\(\displaystyle \int \tan (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x) \left (a+b \sec (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int \cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\int \cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {a \int \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec ^2(e+f x)+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {a \left (a \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+2 \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {a \left (\frac {2 a \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (2 \sqrt {a+b \sec ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

Input:

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x],x]
 

Output:

((2*(a + b*Sec[e + f*x]^2)^(3/2))/3 + a*(-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 = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 f}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}+\frac {a \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f}\)	\(81\)
default	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 f}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}+\frac {a \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f}\)	\(81\)

Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e),x,method=_RETURNVERBOSE)
 

Output:

1/3*(a+b*sec(f*x+e)^2)^(3/2)/f-1/f*a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sec(f*x+ 
e)^2)^(1/2))/sec(f*x+e))+a*(a+b*sec(f*x+e)^2)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (66) = 132\).

Time = 0.41 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.78 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\left [\frac {3 \, a^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left (4 \, a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, f \cos \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {-a} a \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} + 4 \, {\left (4 \, a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )^{2}}\right ] \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="fricas")
 

Output:

[1/24*(3*a^(3/2)*cos(f*x + e)^2*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*(4*a*cos(f*x + e)^2 + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2))/(f*cos(f*x + e)^2), 1/12*(3*sqrt(-a)*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))*cos(f 
*x + e)^2 + 4*(4*a*cos(f*x + e)^2 + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
 + e)^2))/(f*cos(f*x + e)^2)]
 

Sympy [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx \] Input:

integrate((a+b*sec(f*x+e)**2)**(3/2)*tan(f*x+e),x)
 

Output:

Integral((a + b*sec(e + f*x)**2)**(3/2)*tan(e + f*x), x)
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (66) = 132\).

Time = 0.84 (sec) , antiderivative size = 867, normalized size of antiderivative = 11.12 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="giac")
 

Output:

2/3*(3*a^2*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*tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))/sqrt(-a))*sgn(cos(f*x 
+ e))/sqrt(-a) + 2*(3*(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))^5*(2*a*b + b^2)*sgn(cos(f*x + e)) - 3 
*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*t 
an(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))^4*(6*a*b - b^2)*sqrt(a + b)*sgn(cos(f*x + e)) + 2*(6*a^2*b 
 - 15*a*b^2 - b^3)*(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))^3*sgn(cos(f*x + e)) + 6*(2*a^2*b + 7*a*b 
^2 + b^3)*(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*sqrt(a + b)*sgn(cos(f*x + e)) - 3*(6*a^3*b - a^ 
2*b^2 - 28*a*b^3 - 5*b^4)*(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))*sgn(cos(f*x + e)) + (6*a^3*b - 21 
*a^2*b^2 + 28*a*b^3 + 7*b^4)*sqrt(a + b)*sgn(cos(f*x + e)))/((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 ...
 

Mupad [B] (verification not implemented)

Time = 18.61 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{3\,f}-\frac {a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sqrt {a}}\right )}{f}+\frac {a\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{f} \] Input:

int(tan(e + f*x)*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

(a + b/cos(e + f*x)^2)^(3/2)/(3*f) - (a^(3/2)*atanh((a + b/cos(e + f*x)^2) 
^(1/2)/a^(1/2)))/f + (a*(a + b/cos(e + f*x)^2)^(1/2))/f
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} b +4 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, a +3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a^{2} f}{3 f} \] Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e),x)
 

Output:

(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*b + 4*sqrt(sec(e + f*x)**2*b 
+ a)*a + 3*int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x))/(sec(e + f*x)**2 
*b + a),x)*a**2*f)/(3*f)