\(\int \frac {\tan ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [408]

Optimal result

Integrand size = 25, antiderivative size = 173 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f} \]

[Out]

1/8*(3*a^2+10*a*b+15*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-arctan(a^(1/2)*tan(
f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/a^(1/2)-1/8*(3*a+7*b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/4*(a+
b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/b/f

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 173, 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, 490, 596, 537, 223, 212, 385, 209} \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 b^{5/2} f}-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a} f}-\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b f} \]

[In]

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

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[a]*f)) + ((3*a^2 + 10*a*b + 15*b^2)*ArcT
anh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*b^(5/2)*f) - ((3*a + 7*b)*Tan[e + f*x]*Sqrt[a +
b + b*Tan[e + f*x]^2])/(8*b^2*f) + (Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*b*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 490

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(2*n -
1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && 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 ) \sqrt {a+b \left (1+x^2\right )}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f}-\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(3 a+7 b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{4 b f} \\ & = -\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f}+\frac {\text {Subst}\left (\int \frac {(a+b) (3 a+7 b)+\left (3 a^2+10 a b+15 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 b^2 f} \\ & = -\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b 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 )}{f}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 b^2 f} \\ & = -\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b 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 )}{f}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \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 )}{8 b^2 f} \\ & = -\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {(3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.64 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (\frac {8 b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}-\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{8 \sqrt {2} b^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {(a+2 b+a \cos (2 (e+f x))) (3 a+5 b+3 (a+3 b) \cos (2 (e+f x))) \sec ^4(e+f x) \tan (e+f x)}{32 b^2 f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-1/8*(((8*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[a] - ((3*a^2 + 10*a*b + 15*b
^2)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]
]*Sec[e + f*x])/(Sqrt[2]*b^2*f*Sqrt[a + b*Sec[e + f*x]^2]) - ((a + 2*b + a*Cos[2*(e + f*x)])*(3*a + 5*b + 3*(a
+ 3*b)*Cos[2*(e + f*x)])*Sec[e + f*x]^4*Tan[e + f*x])/(32*b^2*f*Sqrt[a + b*Sec[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1929\) vs. \(2(151)=302\).

Time = 17.68 (sec) , antiderivative size = 1930, normalized size of antiderivative = 11.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1930\)

[In]

int(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/f/b^(9/2)/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)*(-16*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*a)*b^(9/2)-16*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*
a)*b^(9/2)*sec(f*x+e)-18*(-a)^(1/2)*b^(7/2)*a*tan(f*x+e)-6*(-a)^(1/2)*b^(5/2)*a^2*tan(f*x+e)-18*(-a)^(1/2)*b^(
9/2)*tan(f*x+e)*sec(f*x+e)^2+3*(-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+10*(-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*b^3+15*(-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))*b^4+3*(-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+10*(-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*b^3+15*(-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))*b^4-2*(-a)^(1/2)*b^(7/2)*a*tan(f*x+e)*sec(f*x+e)^2+3*(-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)+10*(-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*b^3*sec(f*x+e)+15*(-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))*b^4*sec(f*x+e)+3*(-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)*co
s(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)+10*(-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*b
^3*sec(f*x+e)+15*(-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))*b^4*sec(f*x+e)+4*(-a)^(1/2)*b^(9/2)*tan(f*x+e)*sec(f*x+e)^4)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (151) = 302\).

Time = 1.39 (sec) , antiderivative size = 1673, normalized size of antiderivative = 9.67 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*(4*sqrt(-a)*b^3*cos(f*x + e)^3*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^3 + 10*a^2*b + 15*a*b^2)*sqrt(b)*cos(f*x + e)^3*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*(2*a*b^2 - 3*(a^2*b + 3*
a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^3), -1/
16*(2*sqrt(-a)*b^3*cos(f*x + e)^3*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^3 + 10*a^2*b + 15*a*b^2)*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)))*cos(f*x + e)^3 - 2*(2*a*b^2 - 3*(a^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^3), 1/32*(8*sqrt(a)*b^3*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)))*cos(f*x + e)^3 + (3*a^3
+ 10*a^2*b + 15*a*b^2)*sqrt(b)*cos(f*x + e)^3*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)*si
n(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*(2*a*b^2 - 3*(a^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^3), 1/16*(4*sqrt(a)*b^3*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)))*cos(f*x +
e)^3 + (3*a^3 + 10*a^2*b + 15*a*b^2)*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)))*cos(f*x + e)^3 + 2*(2
*a*b^2 - 3*(a^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*b^3*
f*cos(f*x + e)^3)]

Sympy [F]

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

[In]

integrate(tan(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(1/2), x)