\(\int (a+b \sec ^2(e+f x))^{3/2} \tan ^3(e+f x) \, dx\) [390]

Optimal result

Integrand size = 25, antiderivative size = 104 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}-\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}-\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b f} \] 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+1/5*(a+b*sec(f*x+e)^2)^(5/2)/b/f
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {15 a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )-15 a b \sqrt {a+b \sec ^2(e+f x)}-5 b \left (a+b \sec ^2(e+f x)\right )^{3/2}+3 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{15 b f} \] Input:

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

Output:

(15*a^(3/2)*b*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] - 15*a*b*Sqrt[a 
+ b*Sec[e + f*x]^2] - 5*b*(a + b*Sec[e + f*x]^2)^(3/2) + 3*(a + b*Sec[e + 
f*x]^2)^(5/2))/(15*b*f)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4627, 25, 354, 90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

\(\displaystyle -\frac {\int \cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec ^2(e+f x)-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle -\frac {a \int \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec ^2(e+f x)-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle -\frac {a \left (a \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+2 \sqrt {a+b \sec ^2(e+f x)}\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {a \left (\frac {2 a \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sec ^2(e+f x)}\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {a \left (2 \sqrt {a+b \sec ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )\right )-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\)

Input:

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

Output:

-1/2*((2*(a + b*Sec[e + f*x]^2)^(3/2))/3 - (2*(a + b*Sec[e + f*x]^2)^(5/2) 
)/(5*b) + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*Sq 
rt[a + b*Sec[e + f*x]^2]))/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. \(336\) vs. \(2(88)=176\).

Time = 32.08 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.24

method	result	size
default	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (15 a^{\frac {3}{2}} \ln \left (4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) b \cos \left (f x +e \right )^{3}+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} \left (3 \cos \left (f x +e \right )^{3}+3 \cos \left (f x +e \right )^{2}\right )+\left (-20 \cos \left (f x +e \right )^{3}-20 \cos \left (f x +e \right )^{2}+6 \cos \left (f x +e \right )+6\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \left (-5 \cos \left (f x +e \right )-5+3 \sec \left (f x +e \right )+3 \sec \left (f x +e \right )^{2}\right )\right )}{15 f b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (a \cos \left (f x +e \right )^{3}+a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right ) b +b \right )}\)	\(337\)

Input:

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

Output:

1/15/f/b*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(15*a^(3/2)*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*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)*b*cos(f*x+e)^3+((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*(3*cos(f*x+e)^3+3*cos(f*x+e)^2)+ 
(-20*cos(f*x+e)^3-20*cos(f*x+e)^2+6*cos(f*x+e)+6)*((b+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*a*b+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*(- 
5*cos(f*x+e)-5+3*sec(f*x+e)+3*sec(f*x+e)^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (88) = 176\).

Time = 1.70 (sec) , antiderivative size = 443, normalized size of antiderivative = 4.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b \cos \left (f x + e\right )^{4} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, b f \cos \left (f x + e\right )^{4}}, -\frac {15 \, \sqrt {-a} a b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{4} - 4 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, b f \cos \left (f x + e\right )^{4}}\right ] \] Input:

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

Output:

[1/120*(15*a^(3/2)*b*cos(f*x + e)^4*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*((3*a^2 - 20*a*b)*cos(f*x + e)^4 + (6*a*b - 5*b^2)*cos(f*x + 
e)^2 + 3*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + 
e)^4), -1/60*(15*sqrt(-a)*a*b*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)^4 - 4* 
((3*a^2 - 20*a*b)*cos(f*x + e)^4 + (6*a*b - 5*b^2)*cos(f*x + e)^2 + 3*b^2) 
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1431 vs. \(2 (88) = 176\).

Time = 1.37 (sec) , antiderivative size = 1431, normalized size of antiderivative = 13.76 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

-2/15*(15*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*(15*(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))^9*a^2*sgn(cos(f*x + e)) - 15*(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))^8*(7*a^2 - 8*a*b - 4*b^2)*sqrt(a + b)*sgn(cos(f*x + e)) + 20*(1 
5*a^3 - 21*a^2*b - 12*a*b^2 + 8*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))^7*sgn(cos(f*x + e)) - 
20*(21*a^3 - 63*a^2*b + 60*a*b^2 - 4*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))^6*sqrt(a + b)*sgn 
(cos(f*x + e)) + 2*(105*a^4 - 630*a^3*b + 1065*a^2*b^2 + 360*a*b^3 - 16*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))^5*sgn(cos(f*x + e)) + 10*(21*a^4 + 42*a^3*b - 303*a^2*b^ 
2 + 336*a*b^3 + 4*b^4)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {6 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2} b -4 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} b +15 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2} a -34 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, a -9 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a b f -30 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a^{2} f}{30 f} \] Input:

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

Output:

(6*sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**2*b - 4*sqrt( 
sec(e + f*x)**2*b + a)*sec(e + f*x)**2*b + 15*sqrt(sec(e + f*x)**2*b + a)* 
tan(e + f*x)**2*a - 34*sqrt(sec(e + f*x)**2*b + a)*a - 9*int((sqrt(sec(e + 
 f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**3)/(sec(e + f*x)**2*b + a),x 
)*a*b*f - 30*int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x))/(sec(e + f*x)* 
*2*b + a),x)*a**2*f)/(30*f)