Integrand size = 35, antiderivative size = 182 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 c^{3/2} e}+\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 c e} \]
-1/4*(b+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^(3/2)/e-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)+1 /2*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)/c/e
Time = 3.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\frac {2 \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}+\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{c^{3/2}}-\frac {2 \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}}{4 e} \]
-1/4*((2*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*S qrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/Sqrt[a - b + c] + ((b + 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])])/c^(3/2) - (2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e *x]^4])/c)/e
Time = 0.47 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4183, 1578, 1267, 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.
\(\displaystyle \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (d+e x)^5}{\sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle \frac {\int \frac {\tan ^5(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {\int \frac {\tan ^4(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\) |
\(\Big \downarrow \) 1267 |
\(\displaystyle \frac {\frac {\int -\frac {(b+2 c) \tan ^2(d+e x)+b}{2 \left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{c}+\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}}{2 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {\int \frac {(b+2 c) \tan ^2(d+e x)+b}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 c}}{2 e}\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {(b+2 c) \int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)-2 c \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 c}}{2 e}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {2 (b+2 c) \int \frac {1}{4 c-\tan ^4(d+e x)}d\frac {2 c \tan ^2(d+e x)+b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}-2 c \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 c}}{2 e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-2 c \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 c}}{2 e}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {4 c \int \frac {1}{4 (a-b+c)-\tan ^4(d+e x)}d\frac {(b-2 c) \tan ^2(d+e x)+2 a-b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{2 c}}{2 e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}-\frac {\frac {2 c \text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}+\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{2 c}}{2 e}\) |
(-1/2*((2*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])])/Sqrt[a - b + c] + ((b + 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 + Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]/c)/(2*e)
3.1.36.3.1 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_))^(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 = 0.11 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.27
method | result | size |
derivativedivides | \(\frac {\frac {\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{4 c^{\frac {3}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}}{e}\) | \(232\) |
default | \(\frac {\frac {\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{4 c^{\frac {3}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}}{e}\) | \(232\) |
1/e*(1/2/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)-1/4*b/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*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 = 1.32 (sec) , antiderivative size = 1226, normalized size of antiderivative = 6.74 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Too large to display} \]
[1/8*(2*sqrt(a - b + c)*c^2*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 - b^2 + (2*a - b)*c + 2*c^2)*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) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e), 1/4*(s qrt(a - b + c)*c^2*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 - b^2 + (2*a - b)*c + 2*c^2)*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*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e), -1/8*(4*sqrt(-a + b - c)*c^2*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 - b^2 + (2*a - b)*c + 2*c^2)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*...
\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\tan ^{5}{\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]
Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]