\(\int \frac {\tan ^5(d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\) [46]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}+\frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

[Out]

-1/2*arctanh(1/2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/(a-b+c)^(
3/2)/e+(a*(2*a-b)+((a-b)*b+2*a*c)*tan(e*x+d)^2)/(a-b+c)/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3781, 1265, 1660, 12, 738, 212} \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\frac {(b (a-b)+2 a c) \tan ^2(d+e x)+a (2 a-b)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}} \]

[In]

Int[Tan[d + e*x]^5/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]

[Out]

-1/2*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x
]^4])]/((a - b + c)^(3/2)*e) + (a*(2*a - b) + ((a - b)*b + 2*a*c)*Tan[d + e*x]^2)/((a - b + c)*(b^2 - 4*a*c)*e
*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 1660

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*
c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 3781

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}} \, dx,x,\tan (d+e x)\right )}{e} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1+x) \left (a+b x+c x^2\right )^{3/2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e} \\ & = \frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\text {Subst}\left (\int -\frac {b^2-4 a c}{2 (a-b+c) (1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{\left (b^2-4 a c\right ) e} \\ & = \frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 (a-b+c) e} \\ & = \frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\text {Subst}\left (\int \frac {1}{4 a-4 b+4 c-x^2} \, dx,x,\frac {2 a-b-(-b+2 c) \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c) e} \\ & = -\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}+\frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.90 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}+\frac {2 \sqrt {2} \left (2 a^2-b^2+2 a c+\left (2 a^2+b^2-2 a (b+c)\right ) \cos (2 (d+e x))\right ) \sec ^2(d+e x)}{(a-b+c) \left (-b^2+4 a c\right ) \sqrt {(3 a+b+3 c+4 (a-c) \cos (2 (d+e x))+(a-b+c) \cos (4 (d+e x))) \sec ^4(d+e x)}}}{2 e} \]

[In]

Integrate[Tan[d + e*x]^5/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]

[Out]

-1/2*(ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*
x]^4])]/(a - b + c)^(3/2) + (2*Sqrt[2]*(2*a^2 - b^2 + 2*a*c + (2*a^2 + b^2 - 2*a*(b + c))*Cos[2*(d + e*x)])*Se
c[d + e*x]^2)/((a - b + c)*(-b^2 + 4*a*c)*Sqrt[(3*a + b + 3*c + 4*(a - c)*Cos[2*(d + e*x)] + (a - b + c)*Cos[4
*(d + e*x)])*Sec[d + e*x]^4]))/e

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(508\) vs. \(2(147)=294\).

Time = 0.10 (sec) , antiderivative size = 509, normalized size of antiderivative = 3.20

method	result	size
derivativedivides	\(\frac {-\frac {2 a +b \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}-\frac {b +2 c \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}+\frac {2 c \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \sqrt {a -b +c}}-\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}+\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}+\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}-\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{e}\)	\(509\)
default	\(\frac {-\frac {2 a +b \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}-\frac {b +2 c \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}+\frac {2 c \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \sqrt {a -b +c}}-\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}+\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}+\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}-\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{e}\)	\(509\)

[In]

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/e*(-1/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*(2*a+b*tan(e*x+d)^2)/(4*a*c-b^2)-1/(a+b*tan(e*x+d)^2+c*tan(e*x
+d)^4)^(1/2)*(b+2*c*tan(e*x+d)^2)/(4*a*c-b^2)+2*c/((-4*a*c+b^2)^(1/2)-b+2*c)/((-4*a*c+b^2)^(1/2)+b-2*c)/(a-b+c
)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+c)^(1/2)*(c*(1+tan(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2
)+a-b+c)^(1/2))/(1+tan(e*x+d)^2))-2*c/((-4*a*c+b^2)^(1/2)-b+2*c)/(-4*a*c+b^2)/(tan(e*x+d)^2-1/2*(-b+(-4*a*c+b^
2)^(1/2))/c)*(c*(tan(e*x+d)^2-1/2*(-b+(-4*a*c+b^2)^(1/2))/c)^2+(-4*a*c+b^2)^(1/2)*(tan(e*x+d)^2-1/2*(-b+(-4*a*
c+b^2)^(1/2))/c))^(1/2)+2*c/((-4*a*c+b^2)^(1/2)+b-2*c)/(-4*a*c+b^2)/(tan(e*x+d)^2+1/2*(b+(-4*a*c+b^2)^(1/2))/c
)*(c*(tan(e*x+d)^2+1/2*(b+(-4*a*c+b^2)^(1/2))/c)^2-(-4*a*c+b^2)^(1/2)*(tan(e*x+d)^2+1/2*(b+(-4*a*c+b^2)^(1/2))
/c))^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (147) = 294\).

Time = 0.65 (sec) , antiderivative size = 1095, normalized size of antiderivative = 6.89 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(((b^2*c - 4*a*c^2)*tan(e*x + d)^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*tan(e*x + d)^2)*sqrt(a - b + c)*l
og(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 - 4*sqrt(c*t
an(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b
^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*a^3
- 3*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)*tan(e*x + d)^2 + (2*a^2 - a*b)*c
))/((4*a*c^4 + (8*a^2 - 8*a*b - b^2)*c^3 + 2*(2*a^3 - 4*a^2*b + a*b^2 + b^3)*c^2 - (a^2*b^2 - 2*a*b^3 + b^4)*c
)*e*tan(e*x + d)^4 - (a^2*b^3 - 2*a*b^4 + b^5 - 4*a*b*c^3 - (8*a^2*b - 8*a*b^2 - b^3)*c^2 - 2*(2*a^3*b - 4*a^2
*b^2 + a*b^3 + b^4)*c)*e*tan(e*x + d)^2 - (a^3*b^2 - 2*a^2*b^3 + a*b^4 - 4*a^2*c^3 - (8*a^3 - 8*a^2*b - a*b^2)
*c^2 - 2*(2*a^4 - 4*a^3*b + a^2*b^2 + a*b^3)*c)*e), 1/2*(((b^2*c - 4*a*c^2)*tan(e*x + d)^4 + a*b^2 - 4*a^2*c +
(b^3 - 4*a*b*c)*tan(e*x + d)^2)*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((
b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(
e*x + d)^2 + a^2 - a*b + a*c)) - 2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*a^3 - 3*a^2*b + a*b^2 + (a
^2*b - 2*a*b^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)*tan(e*x + d)^2 + (2*a^2 - a*b)*c))/((4*a*c^4 + (8*a^2
- 8*a*b - b^2)*c^3 + 2*(2*a^3 - 4*a^2*b + a*b^2 + b^3)*c^2 - (a^2*b^2 - 2*a*b^3 + b^4)*c)*e*tan(e*x + d)^4 - (
a^2*b^3 - 2*a*b^4 + b^5 - 4*a*b*c^3 - (8*a^2*b - 8*a*b^2 - b^3)*c^2 - 2*(2*a^3*b - 4*a^2*b^2 + a*b^3 + b^4)*c)
*e*tan(e*x + d)^2 - (a^3*b^2 - 2*a^2*b^3 + a*b^4 - 4*a^2*c^3 - (8*a^3 - 8*a^2*b - a*b^2)*c^2 - 2*(2*a^4 - 4*a^
3*b + a^2*b^2 + a*b^3)*c)*e)]

Sympy [F]

\[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (d + e x \right )}}{\left (a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(tan(e*x+d)**5/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(3/2),x)

[Out]

Integral(tan(d + e*x)**5/(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)**(3/2), x)

Maxima [F]

\[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (e x + d\right )^{5}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(e*x + d)^5/(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)^(3/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{{\left (c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2),x)

[Out]

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2), x)