Integrand size = 33, antiderivative size = 548 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\frac {\sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {a-c-\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}-\frac {\sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {a-c+\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}+\frac {b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2} e}-\frac {b \left (5 b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{7/2} e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}+\frac {\left (15 b^2-16 a c-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e} \]
1/2*b*arctanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^ 2)^(1/2))/c^(3/2)/e-1/16*b*(-12*a*c+5*b^2)*arctanh(1/2*(b+2*c*tan(e*x+d))/ c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/c^(7/2)/e+1/2*arctanh(1/2*( a-c-(a^2-2*a*c+b^2+c^2)^(1/2)+b*tan(e*x+d))*2^(1/2)/(a-c-(a^2-2*a*c+b^2+c^ 2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(a-c-(a^2-2*a*c+b^2 +c^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-1/2*arctanh(1/2*(a- c+(a^2-2*a*c+b^2+c^2)^(1/2)+b*tan(e*x+d))*2^(1/2)/(a-c+(a^2-2*a*c+b^2+c^2) ^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(a-c+(a^2-2*a*c+b^2+c ^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-(a+b*tan(e*x+d)+c*tan (e*x+d)^2)^(1/2)/c/e+1/3*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)*tan(e*x+d)^ 2/c/e+1/24*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)*(15*b^2-16*a*c-10*b*c*tan (e*x+d))/c^3/e
Result contains complex when optimal does not.
Time = 6.14 (sec) , antiderivative size = 456, normalized size of antiderivative = 0.83 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\frac {-\frac {2 \sqrt {a+i b-c} \text {arctanh}\left (\frac {2 a+i b-(-b-2 i c) \tan (d+e x)}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 a+4 i b-4 c}-\frac {2 \sqrt {a-i b-c} \text {arctanh}\left (\frac {2 a-i b-(-b+2 i c) \tan (d+e x)}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 a-4 i b-4 c}+\frac {b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2}}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c}+\frac {\frac {\left (-\frac {15 b^3}{4}+9 a b c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 c^{5/2}}+\frac {\left (\frac {15 b^2}{4}-4 a c-\frac {5}{2} b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c^2}}{3 c}}{e} \]
((-2*Sqrt[a + I*b - c]*ArcTanh[(2*a + I*b - (-b - (2*I)*c)*Tan[d + e*x])/( 2*Sqrt[a + I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(4*a + (4*I)*b - 4*c) - (2*Sqrt[a - I*b - c]*ArcTanh[(2*a - I*b - (-b + (2*I)*c)* Tan[d + e*x])/(2*Sqrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x ]^2])])/(4*a - (4*I)*b - 4*c) + (b*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[ c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(2*c^(3/2)) - Sqrt[a + b *Tan[d + e*x] + c*Tan[d + e*x]^2]/c + (Tan[d + e*x]^2*Sqrt[a + b*Tan[d + e *x] + c*Tan[d + e*x]^2])/(3*c) + ((((-15*b^3)/4 + 9*a*b*c)*ArcTanh[(b + 2* c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/ (4*c^(5/2)) + (((15*b^2)/4 - 4*a*c - (5*b*c*Tan[d + e*x])/2)*Sqrt[a + b*Ta n[d + e*x] + c*Tan[d + e*x]^2])/(2*c^2))/(3*c))/e
Time = 1.14 (sec) , antiderivative size = 531, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 4183, 7276, 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.
\(\displaystyle \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (d+e x)^5}{\sqrt {a+b \tan (d+e x)+c \tan (d+e x)^2}}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle \frac {\int \frac {\tan ^5(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\frac {\tan ^3(d+e x)}{\sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}+\frac {\tan (d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}-\frac {\tan (d+e x)}{\sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}\right )d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \text {arctanh}\left (\frac {-\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2-2 a c+b^2+c^2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \text {arctanh}\left (\frac {\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2-2 a c+b^2+c^2}}-\frac {b \left (5 b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{7/2}}+\frac {b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2}}+\frac {\left (-16 a c+15 b^2-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c}}{e}\) |
((Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2] + b*Tan[d + e*x])/(Sqrt[2]*Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*Sq rt[a^2 + b^2 - 2*a*c + c^2]) - (Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2] ]*ArcTanh[(a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2] + b*Tan[d + e*x])/(Sqrt[2 ]*Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c* Tan[d + e*x]^2])])/(Sqrt[2]*Sqrt[a^2 + b^2 - 2*a*c + c^2]) + (b*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2 ])])/(2*c^(3/2)) - (b*(5*b^2 - 12*a*c)*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*S qrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(16*c^(7/2)) - Sqrt[ a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]/c + (Tan[d + e*x]^2*Sqrt[a + b*Tan[ d + e*x] + c*Tan[d + e*x]^2])/(3*c) + ((15*b^2 - 16*a*c - 10*b*c*Tan[d + e *x])*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(24*c^3))/e
3.1.10.3.1 Defintions of rubi rules used
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]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.65 (sec) , antiderivative size = 9581953, normalized size of antiderivative = 17485.32
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 5228 vs. \(2 (485) = 970\).
Time = 1.20 (sec) , antiderivative size = 10457, normalized size of antiderivative = 19.08 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\int \frac {\tan ^{5}{\left (d + e x \right )}}{\sqrt {a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}}}\, dx \]
\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\int { \frac {\tan \left (e x + d\right )^{5}}{\sqrt {c \tan \left (e x + d\right )^{2} + b \tan \left (e x + d\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^2+b\,\mathrm {tan}\left (d+e\,x\right )+a}} \,d x \]