\(\int \frac {\tan ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [421]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f} \]

[Out]

-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f-1/2*(3*a+5*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b
+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+1/2*(3*a+2*b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/a/b^2/f-(a+b)*tan(f*x+e)
^3/a/b/f/(a+b+b*tan(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4226, 2000, 481, 596, 537, 223, 212, 385, 209} \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b^{5/2} f}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a b^2 f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b \tan ^2(e+f x)+b}} \]

[In]

Int[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f)) - ((3*a + 5*b)*ArcTanh[(Sqrt[b]*T
an[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*b^(5/2)*f) - ((a + b)*Tan[e + f*x]^3)/(a*b*f*Sqrt[a + b + b*T
an[e + f*x]^2]) + ((3*a + 2*b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*b^2*f)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

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

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
 && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
 c, d, e, f, n}, x]

Rule 596

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
 x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q +
 1) + 1))), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 2000

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4226

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(3 a+2 b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a b f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {(a+b) (3 a+2 b)+a (3 a+5 b) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 a b^2 f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f}-\frac {(3 a+5 b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b^2 f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a f}-\frac {(3 a+5 b) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^2 f} \\ & = -\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.40 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (\frac {2 b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}+\frac {a (3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{4 \sqrt {2} a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 a^2+6 a b+2 b^2+\left (3 a^2+4 a b+2 b^2\right ) \cos (2 (e+f x))\right ) \sec ^4(e+f x) \tan (e+f x)}{8 a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

Integrate[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-1/4*(((2*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[a] + (a*(3*a + 5*b)*ArcTanh[
(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e +
f*x]^3)/(Sqrt[2]*a*b^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2*(e + f*x)])*(3*a^2 + 6*a*b + 2*b^
2 + (3*a^2 + 4*a*b + 2*b^2)*Cos[2*(e + f*x)])*Sec[e + f*x]^4*Tan[e + f*x])/(8*a*b^2*f*(a + b*Sec[e + f*x]^2)^(
3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2791\) vs. \(2(152)=304\).

Time = 13.01 (sec) , antiderivative size = 2792, normalized size of antiderivative = 16.23

method result size
default \(\text {Expression too large to display}\) \(2792\)

[In]

int(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f/b^(7/2)/(-a)^(1/2)/a/(a+b*sec(f*x+e)^2)^(3/2)*(3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(
1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*(-a)^(1/
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2*sec(f*x+e)^2+5*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e
)-1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3*sec(f*x+e)^2+3*ln(4*(((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)
/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2*sec(f*x+e)^2+3*ln(-4*(((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin
(f*x+e)*a+a+b)/(sin(f*x+e)-1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2*sec(f*x+e)^3+5*l
n(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))
^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3*sec(f
*x+e)^2+5*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
*a*b^3*sec(f*x+e)^3+5*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2)*a*b^3*sec(f*x+e)^3+3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)*a^2*b^2*sec(f*x+e)^3+3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f
*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*(-a)^(1/2)*((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b*sec(f*x+e)-4*(-a)^(1/2)*b^(7/2)*a*tan(f*x+e)-8*(-a)^(1/2)*b^(5/2)*a
^2*tan(f*x+e)-4*(-a)^(1/2)*b^(9/2)*tan(f*x+e)*sec(f*x+e)^2+5*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a^2*b^2*sec(f*x+e)+5*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2*b^2*sec(f*x+e)+3*ln(4*(((b+a*cos(f*x+e)^2)/
(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-
b)/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b*sec(f*x+e)-6*(-a)^(1/2)*b^(3/2
)*a^3*tan(f*x+e)+4*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*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(9/2)*sec(f
*x+e)^2+4*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*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(9/2)*sec(f*x+e)^3+4
*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+co
s(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)*a+3*ln(-4*(((b+a*cos(f*
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x
+e)*a+a+b)/(sin(f*x+e)-1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b+3*ln(4*(((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e
)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b+5*(-a)^(1/2)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a^2*b^2+5*(-a)^(1/2)*((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2*b^2-8*(-a)^(1/2)*b^(7/2)*a*
tan(f*x+e)*sec(f*x+e)^2-2*(-a)^(1/2)*b^(7/2)*a*tan(f*x+e)*sec(f*x+e)^4+4*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*sin(f*x+e)*a)*((b
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)*a*sec(f*x+e)-8*(-a)^(1/2)*b^(5/2)*a^2*tan(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. 398 vs. \(2 (152) = 304\).

Time = 1.76 (sec) , antiderivative size = 1895, normalized size of antiderivative = 11.02 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*((a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f
*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 -
32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x +
e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - ((3*a^4 + 5*a^3*b)*cos(f*x + e)^3 + (3*a^3*b + 5*a
^2*b^2)*cos(f*x + e))*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 - 4*((a -
 b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^
2)/cos(f*x + e)^4) - 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/
cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)), -1/8*(2*((3*a^4 + 5*a^3*b)*
cos(f*x + e)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*
x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + (a*b
^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 +
 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7
*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5
*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos
(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)),
 1/8*(2*(a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos
(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*co
s(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + ((3*a^4 + 5*a^3*b)*cos(f*x + e
)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(
f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)
*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*
cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)), 1/4*((a
*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^
3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)
^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - ((3*a^4 + 5*a^3*b)*cos(f*x + e)^3 + (3*a
^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt
((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + 2*(a^2*b^2 + (3*a^3*b + 4
*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f
*x + e)^3 + a^2*b^4*f*cos(f*x + e))]

Sympy [F]

\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(tan(f*x+e)**6/(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral(tan(e + f*x)**6/(a + b*sec(e + f*x)**2)**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]

[In]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2),x)

[Out]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2), x)