Integrand size = 35, antiderivative size = 141 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}+\frac {\text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {c} e} \]
1/2*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))/e/c^(1/2)+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))/e/(a-b+c)^(1/2)
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\frac {\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}+\frac {\text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{2 e} \]
(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])]/Sqrt[a - b + 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])/(2*e)
Time = 0.40 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4183, 1578, 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.
| \(\displaystyle \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (d+e x)^3}{\sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}}dx\) |
| \(\Big \downarrow \) 4183 |
| \(\displaystyle \frac {\int \frac {\tan ^3(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{e}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {\int \frac {\tan ^2(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)-\int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {2 \int \frac {1}{4 c-\tan ^4(d+e x)}d\frac {2 c \tan ^2(d+e x)+b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}-\int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-\int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {2 \int \frac {1}{4 (a-b+c)-\tan ^4(d+e x)}d\frac {(b-2 c) \tan ^2(d+e x)+2 a-b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac {\text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{2 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}+\frac {\text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{2 e}\) |
(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])]/Sqrt[a - b + 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])/(2*e)
3.1.37.3.1 Defintions of rubi rules used
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[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 = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}+\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}}{e}\) | \(153\) |
| default | \(\frac {\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}+\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}}{e}\) | \(153\) |
1/e*(1/2*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))/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 = 1.09 (sec) , antiderivative size = 993, normalized size of antiderivative = 7.04 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Too large to display} \]
[1/4*((a - b + c)*sqrt(c)*log(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) + sqrt(a - b + c)*c*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*s qrt(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)))/(((a - b)*c + c^2)*e), -1/4*(2*(a - b + 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)) - sqrt( a - b + c)*c*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)))/(((a - b )*c + c^2)*e), 1/4*(2*sqrt(-a + b - c)*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)) + (a - b + c)*sqrt(c)*log(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))/(((a - b)*c + c^2)*e), 1/2*(sqrt( -a + b - c)*c*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)...
\[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\tan ^{3}{\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]
\[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int { \frac {\tan \left (e x + d\right )^{3}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^3}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]