Integrand size = 35, antiderivative size = 270 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=-\frac {\sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}+\frac {\left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e} \]
1/32*(b^3+2*b^2*c-4*b*(a-2*c)*c-8*c^2*(a+2*c))*arctanh(1/2*(b+2*c*tan(e*x+ d)^2)/c^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/c^(5/2)/e-1/2*arcta nh(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)^(1/2)/e-1/16*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1 /2)*((b-2*c)*(b+4*c)+2*c*(b+2*c)*tan(e*x+d)^2)/c^2/e+1/6*(a+b*tan(e*x+d)^2 +c*tan(e*x+d)^4)^(3/2)/c/e
Time = 6.06 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.07 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {-48 c^{5/2} \sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+3 \left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+\frac {1}{4} \sqrt {c} \left (-9 b^2+24 a c-16 b c+84 c^2-4 \left (3 b^2+6 b c-8 c (a+2 c)\right ) \cos (2 (d+e x))+\left (-3 b^2+8 a c-8 b c+44 c^2\right ) \cos (4 (d+e x))\right ) \sec ^4(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{96 c^{5/2} e} \]
(-48*c^(5/2)*Sqrt[a - b + c]*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])] + 3*(b^ 3 + 2*b^2*c - 4*b*(a - 2*c)*c - 8*c^2*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] + (Sqrt [c]*(-9*b^2 + 24*a*c - 16*b*c + 84*c^2 - 4*(3*b^2 + 6*b*c - 8*c*(a + 2*c)) *Cos[2*(d + e*x)] + (-3*b^2 + 8*a*c - 8*b*c + 44*c^2)*Cos[4*(d + e*x)])*Se c[d + e*x]^4*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/4)/(96*c^(5/2) *e)
Time = 0.62 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4183, 1578, 1267, 27, 1231, 27, 1269, 1092, 219, 1154, 219}
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.
| \(\displaystyle \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (d+e x)^5 \sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}dx\) |
| \(\Big \downarrow \) 4183 |
| \(\displaystyle \frac {\int \frac {\tan ^5(d+e x) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan (d+e x)}{e}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {\int \frac {\tan ^4(d+e x) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 1267 |
| \(\displaystyle \frac {\frac {\int -\frac {3 \left ((b+2 c) \tan ^2(d+e x)+b\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{2 \left (\tan ^2(d+e x)+1\right )}d\tan ^2(d+e x)}{3 c}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}}{2 e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\int \frac {\left ((b+2 c) \tan ^2(d+e x)+b\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan ^2(d+e x)}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\int \frac {\left (b^3+2 c b^2-4 (a-2 c) c b-8 c^2 (a+2 c)\right ) \tan ^2(d+e x)+(b-2 c) \left (b^2+4 c b-4 a c\right )}{2 \left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{4 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\int \frac {\left (b^3+2 c b^2-4 (a-2 c) c b-8 c^2 (a+2 c)\right ) \tan ^2(d+e x)+(b-2 c) \left (b^2+4 c b-4 a c\right )}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)+16 c^2 (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {2 \left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \int \frac {1}{4 c-\tan ^4(d+e x)}d\frac {2 c \tan ^2(d+e x)+b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+16 c^2 (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {16 c^2 (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)+\frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{8 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-32 c^2 (a-b+c) \int \frac {1}{4 (a-b+c)-\tan ^4(d+e x)}d\frac {(b-2 c) \tan ^2(d+e x)+2 a-b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}}{8 c}}{2 c}}{2 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-16 c^2 \sqrt {a-b+c} \text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{8 c}}{2 c}}{2 e}\) |
((a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2)/(3*c) - (-1/8*(-16*c^2*Sq rt[a - b + c]*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])] + ((b^3 + 2*b^2*c - 4* b*(a - 2*c)*c - 8*c^2*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[ c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/Sqrt[c])/c + (((b - 2* c)*(b + 4*c) + 2*c*(b + 2*c)*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^2 + c *Tan[d + e*x]^4])/(4*c))/(2*c))/(2*e)
3.1.27.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b *x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d + e*x)^(n - 2)*(b*d*e*(p + 1) + a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) - e*(2*c*d - b*e)*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g , m, p}, x] && IGtQ[n, 1] && IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[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] && Int egerQ[(m - 1)/2]
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Simp[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[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Time = 0.29 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.69
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}\right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}-\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}+\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}+\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}-\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\) | \(455\) |
| default | \(\frac {\frac {\left (a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}\right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}-\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}+\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}+\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}-\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\) | \(455\) |
1/e*(1/6*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2)/c-1/4*b/c*(1/4*(b+2*c*tan (e*x+d)^2)/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)+1/8*(4*a*c-b^2)/c^(3/ 2)*ln((1/2*b+c*tan(e*x+d)^2)/c^(1/2)+(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/ 2)))-1/8*(b+2*c*tan(e*x+d)^2)/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)-1/ 16*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*tan(e*x+d)^2)/c^(1/2)+(a+b*tan(e*x+d)^2 +c*tan(e*x+d)^4)^(1/2))+1/2*(c*(1+tan(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2) +a-b+c)^(1/2)+1/4*(b-2*c)*ln((1/2*b-c+(1+tan(e*x+d)^2)*c)/c^(1/2)+(c*(1+ta n(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))/c^(1/2)-1/2*(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)))
Time = 2.89 (sec) , antiderivative size = 1405, normalized size of antiderivative = 5.20 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Too large to display} \]
[1/192*(48*sqrt(a - b + c)*c^3*log(((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*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) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d) ^2 + 1)) - 3*(b^3 - 8*(a - b)*c^2 - 16*c^3 - 2*(2*a*b - b^2)*c)*sqrt(c)*lo g(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^2 + b^2 - 4*sqrt(c*tan(e*x + d )^4 + b*tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c) + 4* (8*c^3*tan(e*x + d)^4 - 3*b^2*c + 2*(4*a - 3*b)*c^2 + 24*c^3 + 2*(b*c^2 - 6*c^3)*tan(e*x + d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a))/(c^3 *e), 1/96*(24*sqrt(a - b + c)*c^3*log(((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*tan( e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sq rt(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)) - 3*(b^3 - 8*(a - b)*c^2 - 16*c^3 - 2*(2*a*b - b^2)*c)*sqrt(-c )*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 + b*c*tan(e*x + d)^2 + a*c)) + 2*(8 *c^3*tan(e*x + d)^4 - 3*b^2*c + 2*(4*a - 3*b)*c^2 + 24*c^3 + 2*(b*c^2 - 6* c^3)*tan(e*x + d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a))/(c^3*e ), -1/192*(96*sqrt(-a + b - c)*c^3*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*t an(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c...
\[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{5}{\left (d + e x \right )}\, dx \]
\[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{5} \,d x } \]
Timed out. \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int {\mathrm {tan}\left (d+e\,x\right )}^5\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \]