\(\int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx\) [377]

Optimal result

Integrand size = 25, antiderivative size = 80 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}-\frac {\sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b f} \] 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+1/3*(a+b*sec(f*x+e)^2)^(3/2)/b/f
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a+b \sec ^2(e+f x)} \left (a-3 b+b \sec ^2(e+f x)\right )}{3 b f} \] Input:

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

Output:

(3*Sqrt[a]*b*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + Sqrt[a + b*Sec[ 
e + f*x]^2]*(a - 3*b + b*Sec[e + f*x]^2))/(3*b*f)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4627, 25, 354, 90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/2*(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + 
b*Sec[e + f*x]^2] - (2*(a + b*Sec[e + f*x]^2)^(3/2))/(3*b))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(68)=136\).

Time = 10.57 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.92

method	result	size
default	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (3 \sqrt {a}\, \ln \left (4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) b \cos \left (f x +e \right )+\left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a +b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (-3 \cos \left (f x +e \right )-3+\sec \left (f x +e \right )+\sec \left (f x +e \right )^{2}\right )\right )}{3 f b \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(234\)

Input:

int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x,method=_RETURNVERBOSE)
 

Output:

1/3/f/b*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*(3*a^(1/2)*ln(4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)+4*cos(f*x+e)*a)*b*cos(f*x+e)+(1+cos(f*x+e))*((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*a+b*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-3*cos(f 
*x+e)-3+sec(f*x+e)+sec(f*x+e)^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (68) = 136\).

Time = 0.41 (sec) , antiderivative size = 386, normalized size of antiderivative = 4.82 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx=\left [\frac {3 \, \sqrt {a} b \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, b f \cos \left (f x + e\right )^{2}}, -\frac {3 \, \sqrt {-a} b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, b f \cos \left (f x + e\right )^{2}}\right ] \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="fricas")
 

Output:

[1/24*(3*sqrt(a)*b*cos(f*x + e)^2*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*c 
os(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*((a - 3*b)*cos(f*x + e)^2 + b)*sqrt((a*cos(f*x + e)^2 + b)/cos( 
f*x + e)^2))/(b*f*cos(f*x + e)^2), -1/12*(3*sqrt(-a)*b*arctan(1/4*(8*a^2*c 
os(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*((a - 3*b)*cos(f*x + e)^2 + b)*sqrt((a*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^2)]
 

Sympy [F]

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

integrate((a+b*sec(f*x+e)**2)**(1/2)*tan(f*x+e)**3,x)
 

Output:

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

Maxima [F]

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

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 800 vs. \(2 (68) = 136\).

Time = 0.45 (sec) , antiderivative size = 800, normalized size of antiderivative = 10.00 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="giac")
 

Output:

-2/3*(3*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*tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))/sqrt(-a))/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*a - 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))^4*(3*a - 4*b)*sqrt(a + b) + 2*(sq 
rt(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*(3*a^2 - 9*a*b + 8*b^2) + 6*(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*(a^2 + a*b - 4* 
b^2)*sqrt(a + b) - 3*(3*a^3 - 2*a^2*b - 5*a*b^2 + 16*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*a^3 - 6*a^2*b + 19*a*b^2 - 20*b^3)*sqrt(a + 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...
 

Mupad [F(-1)]

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

int(tan(e + f*x)^3*(a + b/cos(e + f*x)^2)^(1/2),x)
 

Output:

int(tan(e + f*x)^3*(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x)
 

Output:

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