Integrand size = 23, antiderivative size = 166 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} f}-\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} b^{2/3} f}+\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+a^{2/3} \cos ^2(e+f x)\right )}{6 \sqrt [3]{a} b^{2/3} f}+\frac {\log \left (b+a \cos ^3(e+f x)\right )}{3 a f} \] Output:
1/3*arctan(1/3*(b^(1/3)-2*a^(1/3)*cos(f*x+e))*3^(1/2)/b^(1/3))*3^(1/2)/a^( 1/3)/b^(2/3)/f-1/3*ln(b^(1/3)+a^(1/3)*cos(f*x+e))/a^(1/3)/b^(2/3)/f+1/6*ln (b^(2/3)-a^(1/3)*b^(1/3)*cos(f*x+e)+a^(2/3)*cos(f*x+e)^2)/a^(1/3)/b^(2/3)/ f+1/3*ln(b+a*cos(f*x+e)^3)/a/f
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.41 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {-3 \log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )+\text {RootSum}\left [-a-b+3 a \text {$\#$1}-3 b \text {$\#$1}-3 a \text {$\#$1}^2-3 b \text {$\#$1}^2+a \text {$\#$1}^3-b \text {$\#$1}^3\&,\frac {-a \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-b \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 a \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}-2 b \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}+a \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2-b \log \left (-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2}{a-b-2 a \text {$\#$1}-2 b \text {$\#$1}+a \text {$\#$1}^2-b \text {$\#$1}^2}\&\right ]}{3 a f} \] Input:
Integrate[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^3),x]
Output:
(-3*Log[Sec[(e + f*x)/2]^2] + RootSum[-a - b + 3*a*#1 - 3*b*#1 - 3*a*#1^2 - 3*b*#1^2 + a*#1^3 - b*#1^3 & , (-(a*Log[-#1 + Tan[(e + f*x)/2]^2]) - b*L og[-#1 + Tan[(e + f*x)/2]^2] - 4*a*Log[-#1 + Tan[(e + f*x)/2]^2]*#1 - 2*b* Log[-#1 + Tan[(e + f*x)/2]^2]*#1 + a*Log[-#1 + Tan[(e + f*x)/2]^2]*#1^2 - b*Log[-#1 + Tan[(e + f*x)/2]^2]*#1^2)/(a - b - 2*a*#1 - 2*b*#1 + a*#1^2 - b*#1^2) & ])/(3*a*f)
Time = 0.42 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4626, 2410, 750, 16, 792, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^3}{a+b \sec (e+f x)^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {1-\cos ^2(e+f x)}{a \cos ^3(e+f x)+b}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 2410 |
| \(\displaystyle -\frac {\int \frac {1}{a \cos ^3(e+f x)+b}d\cos (e+f x)-\int \frac {\cos ^2(e+f x)}{a \cos ^3(e+f x)+b}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle -\frac {\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{3 b^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}}d\cos (e+f x)}{3 b^{2/3}}-\int \frac {\cos ^2(e+f x)}{a \cos ^3(e+f x)+b}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{3 b^{2/3}}-\int \frac {\cos ^2(e+f x)}{a \cos ^3(e+f x)+b}d\cos (e+f x)+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}}{f}\) |
| \(\Big \downarrow \) 792 |
| \(\displaystyle -\frac {\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle -\frac {\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)-\frac {\int -\frac {\sqrt [3]{a} \left (\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)\right )}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{2 \sqrt [3]{a}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)+\frac {\int \frac {\sqrt [3]{a} \left (\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)\right )}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{2 \sqrt [3]{a}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)+\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {\frac {\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}\right )}{\sqrt [3]{a}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\frac {\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}}d\cos (e+f x)-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {\frac {-\frac {\log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{2 \sqrt [3]{a}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}}{f}\) |
Input:
Int[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^3),x]
Output:
-((Log[b^(1/3) + a^(1/3)*Cos[e + f*x]]/(3*a^(1/3)*b^(2/3)) + (-((Sqrt[3]*A rcTan[(1 - (2*a^(1/3)*Cos[e + f*x])/b^(1/3))/Sqrt[3]])/a^(1/3)) - Log[b^(2 /3) - a^(1/3)*b^(1/3)*Cos[e + f*x] + a^(2/3)*Cos[e + f*x]^2]/(2*a^(1/3)))/ (3*b^(2/3)) - Log[b + a*Cos[e + f*x]^3]/(3*a))/f)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveConten t[a + b*x^n, x]]/(b*n), x] /; FreeQ[{a, b, m, n}, x] && EqQ[m, n - 1]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x, 2]}, Int[(A + B*x)/(a + b*x^3), x] + Si mp[C Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] || !RationalQ[ a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.70 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68
| method | result | size |
| risch | \(-\frac {i x}{a}-\frac {2 i e}{a f}+i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 b^{2} a^{3} f^{3} \textit {\_Z}^{3}+27 i b^{2} a^{2} f^{2} \textit {\_Z}^{2}-9 \textit {\_Z} a \,b^{2} f +i a^{2}-i b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\left (-6 i b f \textit {\_R} +\frac {2 b}{a}\right ) {\mathrm e}^{i \left (f x +e \right )}+1\right )\right )\) | \(113\) |
| derivativedivides | \(\frac {-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}}{f}\) | \(133\) |
| default | \(\frac {-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}}{f}\) | \(133\) |
Input:
int(tan(f*x+e)^3/(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)
Output:
-I*x/a-2*I/a/f*e+I*sum(_R*ln(exp(2*I*(f*x+e))+(-6*I*b*f*_R+2*b/a)*exp(I*(f *x+e))+1),_R=RootOf(27*b^2*a^3*f^3*_Z^3+27*I*b^2*a^2*f^2*_Z^2-9*_Z*a*b^2*f +I*a^2-I*b^2))
Result contains complex when optimal does not.
Time = 0.71 (sec) , antiderivative size = 1052, normalized size of antiderivative = 6.34 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="fricas")
Output:
-1/12*(2*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^ 2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f*log(-1/2*(3*(I*sqrt(3) + 1)*( -1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*b*f - a*cos(f*x + e) - b) - ((3*(I*sqrt(3) + 1)*(-1/54/(a^3* f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)) *a*f + 3*sqrt(1/3)*a*f*sqrt(-((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/( a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2 )/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a^2*f^2)) + 6)*log(1/2*(3*(I*s qrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^ 2*f^3))^(1/3) - 2/(a*f))*a*b*f + 3/2*sqrt(1/3)*a*b*f*sqrt(-((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3)) ^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54 /(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/( a^2*f^2)) - 2*a*cos(f*x + e) + b) - ((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f - 3*sqrt(1/3)*a*f*sqrt(-((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2* f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*( I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3 *b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a^2*f^2)) + 6)*log(-1/2*(3*(I*sqr...
\[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{a + b \sec ^{3}{\left (e + f x \right )}}\, dx \] Input:
integrate(tan(f*x+e)**3/(a+b*sec(f*x+e)**3),x)
Output:
Integral(tan(e + f*x)**3/(a + b*sec(e + f*x)**3), x)
Time = 0.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=-\frac {\frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {2 \, b}{a}\right )} + 2 \, b\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {b}{a}\right )^{\frac {1}{3}} - 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {3 \, {\left (2 \, \left (\frac {b}{a}\right )^{\frac {2}{3}} + 1\right )} \log \left (\cos \left (f x + e\right )^{2} - \left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{a \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (\left (\frac {b}{a}\right )^{\frac {2}{3}} - 1\right )} \log \left (\left (\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right )\right )}{a \left (\frac {b}{a}\right )^{\frac {2}{3}}}}{18 \, f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="maxima")
Output:
-1/18*(2*sqrt(3)*(a*(3*(b/a)^(1/3) - 2*b/a) + 2*b)*arctan(-1/3*sqrt(3)*((b /a)^(1/3) - 2*cos(f*x + e))/(b/a)^(1/3))/(a*b) - 3*(2*(b/a)^(2/3) + 1)*log (cos(f*x + e)^2 - (b/a)^(1/3)*cos(f*x + e) + (b/a)^(2/3))/(a*(b/a)^(2/3)) - 6*((b/a)^(2/3) - 1)*log((b/a)^(1/3) + cos(f*x + e))/(a*(b/a)^(2/3)))/f
Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {\left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right ) \right |}\right )}{3 \, b f} + \frac {\log \left ({\left | a \cos \left (f x + e\right )^{3} + b \right |}\right )}{3 \, a f} - \frac {\sqrt {3} \left (-a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {b}{a}\right )^{\frac {1}{3}} + 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{3 \, a b f} - \frac {\left (-a^{2} b\right )^{\frac {1}{3}} \log \left (\cos \left (f x + e\right )^{2} + \left (-\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 \, a b f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="giac")
Output:
1/3*(-b/a)^(1/3)*log(abs(-(-b/a)^(1/3) + cos(f*x + e)))/(b*f) + 1/3*log(ab s(a*cos(f*x + e)^3 + b))/(a*f) - 1/3*sqrt(3)*(-a^2*b)^(1/3)*arctan(1/3*sqr t(3)*((-b/a)^(1/3) + 2*cos(f*x + e))/(-b/a)^(1/3))/(a*b*f) - 1/6*(-a^2*b)^ (1/3)*log(cos(f*x + e)^2 + (-b/a)^(1/3)*cos(f*x + e) + (-b/a)^(2/3))/(a*b* f)
Time = 18.91 (sec) , antiderivative size = 1620, normalized size of antiderivative = 9.76 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
int(tan(e + f*x)^3/(a + b/cos(e + f*x)^3),x)
Output:
symsum(log(262144*(a - b)^2*(8*a - 8*b + 4*root(27*a^3*b^2*z^3 - 27*a^2*b^ 2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a^2 + 4*root(27*a^3*b^2*z^3 - 27*a^2* b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*b^2 - 3*root(27*a^3*b^2*z^3 - 27*a^ 2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^3 - 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a*b^2 - 36*root(27*a^3*b^2 *z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^3*b + 28*root(27* a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a*b + 36*root( 27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^2*b^2)* (16*a^2*tan(e/2 + (f*x)/2)^2 + 32*b^2*tan(e/2 + (f*x)/2)^2 - 4*root(27*a^3 *b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a^3 - 4*root(27*a ^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*b^3 - 8*a^2 + 8 *b^2 + 3*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^4 - 3*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^4*tan(e/2 + (f*x)/2)^2 + 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^ 2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a*b^3 + 3*root(27*a^3*b^2*z^3 - 27*a^2* b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^3*b + 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^4*b - 48*a*b*tan(e/2 + ( f*x)/2)^2 + 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^ 2, z, k)^2*a^2*b^2 - 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^2*b^3 + 14*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 +...
\[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\int \frac {\tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{3} b +a}d x \] Input:
int(tan(f*x+e)^3/(a+b*sec(f*x+e)^3),x)
Output:
int(tan(e + f*x)**3/(sec(e + f*x)**3*b + a),x)