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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4224, 457, 81, 52, 65, 214} \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b f}-\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f} \]

[In]

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

[Out]

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

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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 214

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4224

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]}, Dist[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] || IntegersQ[2*n, p])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}}{x} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(-1+x) (a+b x)^{3/2}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b f}-\frac {\text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 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}-\frac {a \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 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}-\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 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}-\frac {a^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sec ^2(e+f x)}\right )}{b f} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {15 a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )-15 a b \sqrt {a+b \sec ^2(e+f x)}-5 b \left (a+b \sec ^2(e+f x)\right )^{3/2}+3 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{15 b f} \]

[In]

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

[Out]

(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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(517\) vs. \(2(88)=176\).

Time = 14.22 (sec) , antiderivative size = 518, normalized size of antiderivative = 4.98

method	result	size
default	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (15 \cos \left (f x +e \right )^{3} a^{\frac {3}{2}} \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 \cos \left (f x +e \right ) a +4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) b +3 \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}-20 \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 b +3 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}-20 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +6 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -5 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \right )+6 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -5 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+3 b^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )+3 b^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )^{2}\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 (b +a \cos \left (f x +e \right )^{2}\right ) \left (1+\cos \left (f x +e \right )\right )}\)	\(518\)

[In]

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

[Out]

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)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)*(15*cos(f*x+e)^3*a^(3/2)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+
4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*b+3*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*a^2-20*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+3*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)*a^2-20*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+6*cos(f*x+e)*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)+6*((b+a
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+3*b^2*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)+3*b^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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 = 1.97 (sec) , antiderivative size = 443, normalized size of antiderivative = 4.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b \cos \left (f x + e\right )^{4} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, b f \cos \left (f x + e\right )^{4}}, -\frac {15 \, \sqrt {-a} a b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{4} - 4 \, {\left ({\left (3 \, a^{2} - 20 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (6 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, b f \cos \left (f x + e\right )^{4}}\right ] \]

[In]

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

[Out]

[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)/c
os(f*x + e)^2))/(b*f*cos(f*x + e)^4)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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 = 1.67 (sec) , antiderivative size = 1431, normalized size of antiderivative = 13.76 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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*
(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))^8*(7*a^2 - 8*a*b - 4*b^2)*sqrt(a + b)*sgn(cos(f*x + e)) +
20*(15*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(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*sqrt(a + b)*sgn(cos(f*x + e)) - 20*(21*a^5 - 21*a^4*b -
69*a^3*b^2 + 225*a^2*b^3 + 4*a*b^4 + 8*b^5)*(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)) + 20*(15*a^5 - 39*a^4*b + 57*a^3*b^2 + 87*a^2*b^3 - 244*a*b^4 - 28*b^5)*(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)) - 5*(21*a^6 - 60*a^5*b + 222*a^4*b^2 - 172*a^3*b^3 - 491
*a^2*b^4 + 848*a*b^5 + 96*b^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))*sgn(cos(f*x + e)) + (15*a^
6 - 60*a^5*b + 390*a^4*b^2 - 980*a^3*b^3 + 1543*a^2*b^4 - 984*a*b^5 - 132*b^6)*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 + 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)^5)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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