Integrand size = 35, antiderivative size = 249 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 \sqrt {a} e}-\frac {b \text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{4 a^{3/2} e}+\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \cot ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}+\frac {\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)} \tan ^2(d+e x)}{2 a e} \]
-1/4*b*arctanh(1/2*(2*a+b*cot(e*x+d)^2)/a^(1/2)/(a+b*cot(e*x+d)^2+c*cot(e* x+d)^4)^(1/2))/a^(3/2)/e-1/2*arctanh(1/2*(2*a+b*cot(e*x+d)^2)/a^(1/2)/(a+b *cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2))/e/a^(1/2)+1/2*arctanh(1/2*(2*a-b+(b-2 *c)*cot(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2))/e /(a-b+c)^(1/2)+1/2*(a+b*cot(e*x+d)^2+c*cot(e*x+d)^4)^(1/2)*tan(e*x+d)^2/a/ e
Time = 1.38 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\frac {\cot ^2(d+e x) \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)} \left (-\left ((2 a+b) \sqrt {a-b+c} \text {arctanh}\left (\frac {b+2 a \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}}\right )\right )+2 \sqrt {a} \left (a \text {arctanh}\left (\frac {b-2 c+(2 a-b) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}}\right )+\sqrt {a-b+c} \sqrt {c+b \tan ^2(d+e x)+a \tan ^4(d+e x)}\right )\right )}{4 a^{3/2} \sqrt {a-b+c} e \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \]
(Cot[d + e*x]^2*Sqrt[c + b*Tan[d + e*x]^2 + a*Tan[d + e*x]^4]*(-((2*a + b) *Sqrt[a - b + c]*ArcTanh[(b + 2*a*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[c + b*Ta n[d + e*x]^2 + a*Tan[d + e*x]^4])]) + 2*Sqrt[a]*(a*ArcTanh[(b - 2*c + (2*a - b)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[c + b*Tan[d + e*x]^2 + a*Tan [d + e*x]^4])] + Sqrt[a - b + c]*Sqrt[c + b*Tan[d + e*x]^2 + a*Tan[d + e*x ]^4])))/(4*a^(3/2)*Sqrt[a - b + c]*e*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d + e*x]^4])
Time = 0.51 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4184, 1578, 1289, 2009}
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.
| \(\displaystyle \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cot (d+e x)^3 \sqrt {a+b \cot (d+e x)^2+c \cot (d+e x)^4}}dx\) |
| \(\Big \downarrow \) 4184 |
| \(\displaystyle -\frac {\int \frac {\tan ^3(d+e x)}{\left (\cot ^2(d+e x)+1\right ) \sqrt {c \cot ^4(d+e x)+b \cot ^2(d+e x)+a}}d\cot (d+e x)}{e}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {\int \frac {\tan ^2(d+e x)}{\left (\cot ^2(d+e x)+1\right ) \sqrt {c \cot ^4(d+e x)+b \cot ^2(d+e x)+a}}d\cot ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 1289 |
| \(\displaystyle -\frac {\int \left (\frac {\tan ^2(d+e x)}{\sqrt {c \cot ^4(d+e x)+b \cot ^2(d+e x)+a}}-\frac {\tan (d+e x)}{\sqrt {c \cot ^4(d+e x)+b \cot ^2(d+e x)+a}}+\frac {1}{\left (\cot ^2(d+e x)+1\right ) \sqrt {c \cot ^4(d+e x)+b \cot ^2(d+e x)+a}}\right )d\cot ^2(d+e x)}{2 e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {b \text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{2 a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b \cot ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{\sqrt {a}}-\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \cot ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}\right )}{\sqrt {a-b+c}}-\frac {\tan (d+e x) \sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}}{a}}{2 e}\) |
-1/2*(ArcTanh[(2*a + b*Cot[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Cot[d + e*x]^ 2 + c*Cot[d + e*x]^4])]/Sqrt[a] + (b*ArcTanh[(2*a + b*Cot[d + e*x]^2)/(2*S qrt[a]*Sqrt[a + b*Cot[d + e*x]^2 + c*Cot[d + e*x]^4])])/(2*a^(3/2)) - ArcT anh[(2*a - b + (b - 2*c)*Cot[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Cot [d + e*x]^2 + c*Cot[d + e*x]^4])]/Sqrt[a - b + c] - (Sqrt[a + b*Cot[d + e* x]^2 + c*Cot[d + e*x]^4]*Tan[d + e*x])/a)/e
3.1.21.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 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[cot[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*(cot[(d_.) + (e_.)*(x_)]*( f_.))^(n_.) + (c_.)*(cot[(d_.) + (e_.)*(x_)]*(f_.))^(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*Cot[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]
\[\int \frac {\tan \left (e x +d \right )^{3}}{\sqrt {a +b \cot \left (e x +d \right )^{2}+c \cot \left (e x +d \right )^{4}}}d x\]
Time = 1.38 (sec) , antiderivative size = 1444, normalized size of antiderivative = 5.80 \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\text {Too large to display} \]
[1/8*(2*sqrt(a - b + c)*a^2*log(((8*a^2 - 8*a*b + b^2 + 4*a*c)*tan(e*x + d )^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + b^2 + 4*(a - 2*b)*c + 8*c^2 + 4*((2*a - b)*tan(e*x + d)^4 + (b - 2*c)*tan(e*x + d)^2)*sqrt(a - b + c)*sqrt((a*tan(e*x + d)^4 + b*tan(e*x + d)^2 + c)/tan(e*x + d)^4))/( tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) + 4*(a^2 - a*b + a*c)*sqrt((a*tan( e*x + d)^4 + b*tan(e*x + d)^2 + c)/tan(e*x + d)^4)*tan(e*x + d)^2 + (2*a^2 - a*b - b^2 + (2*a + b)*c)*sqrt(a)*log(8*a^2*tan(e*x + d)^4 + 8*a*b*tan(e *x + d)^2 + b^2 + 4*a*c - 4*(2*a*tan(e*x + d)^4 + b*tan(e*x + d)^2)*sqrt(a )*sqrt((a*tan(e*x + d)^4 + b*tan(e*x + d)^2 + c)/tan(e*x + d)^4)))/((a^3 - a^2*b + a^2*c)*e), 1/4*(sqrt(a - b + c)*a^2*log(((8*a^2 - 8*a*b + b^2 + 4 *a*c)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + b^ 2 + 4*(a - 2*b)*c + 8*c^2 + 4*((2*a - b)*tan(e*x + d)^4 + (b - 2*c)*tan(e* x + d)^2)*sqrt(a - b + c)*sqrt((a*tan(e*x + d)^4 + b*tan(e*x + d)^2 + c)/t an(e*x + d)^4))/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) + 2*(a^2 - a*b + a*c)*sqrt((a*tan(e*x + d)^4 + b*tan(e*x + d)^2 + c)/tan(e*x + d)^4)*tan(e* x + d)^2 + (2*a^2 - a*b - b^2 + (2*a + b)*c)*sqrt(-a)*arctan(1/2*(2*a*tan( e*x + d)^4 + b*tan(e*x + d)^2)*sqrt(-a)*sqrt((a*tan(e*x + d)^4 + b*tan(e*x + d)^2 + c)/tan(e*x + d)^4)/(a^2*tan(e*x + d)^4 + a*b*tan(e*x + d)^2 + a* c)))/((a^3 - a^2*b + a^2*c)*e), 1/8*(4*a^2*sqrt(-a + b - c)*arctan(-1/2*(( 2*a - b)*tan(e*x + d)^4 + (b - 2*c)*tan(e*x + d)^2)*sqrt(-a + b - c)*sq...
\[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\int \frac {\tan ^{3}{\left (d + e x \right )}}{\sqrt {a + b \cot ^{2}{\left (d + e x \right )} + c \cot ^{4}{\left (d + e x \right )}}}\, dx \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\sqrt {a+b \cot ^2(d+e x)+c \cot ^4(d+e x)}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^3}{\sqrt {c\,{\mathrm {cot}\left (d+e\,x\right )}^4+b\,{\mathrm {cot}\left (d+e\,x\right )}^2+a}} \,d x \]