\(\int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\) [681]

Optimal result

Integrand size = 23, antiderivative size = 525 \[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {1}{4} \sqrt [3]{c-\sqrt {-d^2}} x-\frac {1}{4} \sqrt [3]{c+\sqrt {-d^2}} x-\frac {\sqrt {3} \sqrt {-d^2} \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 d f}+\frac {\sqrt {3} \sqrt {-d^2} \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 d f}+\frac {\sqrt {-d^2} \sqrt [3]{c-\sqrt {-d^2}} \log (\cos (e+f x))}{4 d f}-\frac {\sqrt {-d^2} \sqrt [3]{c+\sqrt {-d^2}} \log (\cos (e+f x))}{4 d f}+\frac {3 \sqrt {-d^2} \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d f}-\frac {3 \sqrt {-d^2} \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d f}+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{140 d^3 f}-\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{35 d^2 f}+\frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f} \] Output:

-1/4*(c-(-d^2)^(1/2))^(1/3)*x-1/4*(c+(-d^2)^(1/2))^(1/3)*x-1/2*3^(1/2)*(-d 
^2)^(1/2)*(c-(-d^2)^(1/2))^(1/3)*arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c 
-(-d^2)^(1/2))^(1/3))*3^(1/2))/d/f+1/2*3^(1/2)*(-d^2)^(1/2)*(c+(-d^2)^(1/2 
))^(1/3)*arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c+(-d^2)^(1/2))^(1/3))*3^ 
(1/2))/d/f+1/4*(-d^2)^(1/2)*(c-(-d^2)^(1/2))^(1/3)*ln(cos(f*x+e))/d/f-1/4* 
(-d^2)^(1/2)*(c+(-d^2)^(1/2))^(1/3)*ln(cos(f*x+e))/d/f+3/4*(-d^2)^(1/2)*(c 
-(-d^2)^(1/2))^(1/3)*ln((c-(-d^2)^(1/2))^(1/3)-(c+d*tan(f*x+e))^(1/3))/d/f 
-3/4*(-d^2)^(1/2)*(c+(-d^2)^(1/2))^(1/3)*ln((c+(-d^2)^(1/2))^(1/3)-(c+d*ta 
n(f*x+e))^(1/3))/d/f+3/140*(9*c^2-35*d^2)*(c+d*tan(f*x+e))^(4/3)/d^3/f-9/3 
5*c*tan(f*x+e)*(c+d*tan(f*x+e))^(4/3)/d^2/f+3/10*tan(f*x+e)^2*(c+d*tan(f*x 
+e))^(4/3)/d/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 14.73 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.71 \[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {-i \sqrt [3]{c-i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )+i \sqrt [3]{c+i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )+\frac {3 \sqrt [3]{c+d \tan (e+f x)} \left (9 c^3-37 c d^2-d \left (3 c^2+49 d^2\right ) \tan (e+f x)+2 d^2 \sec ^2(e+f x) (c+7 d \tan (e+f x))\right )}{35 d^3}}{4 f} \] Input:

Integrate[Tan[e + f*x]^4*(c + d*Tan[e + f*x])^(1/3),x]
 

Output:

((-I)*(c - I*d)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3) 
)/(c - I*d)^(1/3))/Sqrt[3]] - 2*Log[(c - I*d)^(1/3) - (c + d*Tan[e + f*x]) 
^(1/3)] + Log[(c - I*d)^(2/3) + (c - I*d)^(1/3)*(c + d*Tan[e + f*x])^(1/3) 
 + (c + d*Tan[e + f*x])^(2/3)]) + I*(c + I*d)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + 
 (2*(c + d*Tan[e + f*x])^(1/3))/(c + I*d)^(1/3))/Sqrt[3]] - 2*Log[(c + I*d 
)^(1/3) - (c + d*Tan[e + f*x])^(1/3)] + Log[(c + I*d)^(2/3) + (c + I*d)^(1 
/3)*(c + d*Tan[e + f*x])^(1/3) + (c + d*Tan[e + f*x])^(2/3)]) + (3*(c + d* 
Tan[e + f*x])^(1/3)*(9*c^3 - 37*c*d^2 - d*(3*c^2 + 49*d^2)*Tan[e + f*x] + 
2*d^2*Sec[e + f*x]^2*(c + 7*d*Tan[e + f*x])))/(35*d^3))/(4*f)
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 497, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4114, 3042, 3966, 485, 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.

\(\displaystyle \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^4 \sqrt [3]{c+d \tan (e+f x)}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {3 \int -\frac {2}{3} \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \left (3 c \tan ^2(e+f x)+5 d \tan (e+f x)+3 c\right )dx}{10 d}+\frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \left (3 c \tan ^2(e+f x)+5 d \tan (e+f x)+3 c\right )dx}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \left (3 c \tan (e+f x)^2+5 d \tan (e+f x)+3 c\right )dx}{5 d}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {3 \int -\frac {1}{3} \sqrt [3]{c+d \tan (e+f x)} \left (9 c^2+\left (9 c^2-35 d^2\right ) \tan ^2(e+f x)\right )dx}{7 d}+\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {\int \sqrt [3]{c+d \tan (e+f x)} \left (9 c^2+\left (9 c^2-35 d^2\right ) \tan ^2(e+f x)\right )dx}{7 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {\int \sqrt [3]{c+d \tan (e+f x)} \left (9 c^2+\left (9 c^2-35 d^2\right ) \tan (e+f x)^2\right )dx}{7 d}}{5 d}\)

\(\Big \downarrow \) 4114

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {35 d^2 \int \sqrt [3]{c+d \tan (e+f x)}dx+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{4 d f}}{7 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {35 d^2 \int \sqrt [3]{c+d \tan (e+f x)}dx+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{4 d f}}{7 d}}{5 d}\)

\(\Big \downarrow \) 3966

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {\frac {35 d^3 \int \frac {\sqrt [3]{c+d \tan (e+f x)}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{4 d f}}{7 d}}{5 d}\)

\(\Big \downarrow \) 485

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {\frac {35 d^3 \int \left (\frac {\sqrt [3]{c+d \tan (e+f x)} \sqrt {-d^2}}{2 d^2 \left (\sqrt {-d^2}-d \tan (e+f x)\right )}+\frac {\sqrt [3]{c+d \tan (e+f x)} \sqrt {-d^2}}{2 d^2 \left (d \tan (e+f x)+\sqrt {-d^2}\right )}\right )d(d \tan (e+f x))}{f}+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{4 d f}}{7 d}}{5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \tan ^2(e+f x) (c+d \tan (e+f x))^{4/3}}{10 d f}-\frac {\frac {9 c \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f}-\frac {\frac {35 d^3 \left (\frac {\sqrt {3} \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt {3} \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt {-d^2}-d \tan (e+f x)\right )}{4 \sqrt {-d^2}}+\frac {\sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt {-d^2}+d \tan (e+f x)\right )}{4 \sqrt {-d^2}}-\frac {3 \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}+\frac {3 \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}\right )}{f}+\frac {3 \left (9 c^2-35 d^2\right ) (c+d \tan (e+f x))^{4/3}}{4 d f}}{7 d}}{5 d}\)

Input:

Int[Tan[e + f*x]^4*(c + d*Tan[e + f*x])^(1/3),x]
 

Output:

(3*Tan[e + f*x]^2*(c + d*Tan[e + f*x])^(4/3))/(10*d*f) - ((9*c*Tan[e + f*x 
]*(c + d*Tan[e + f*x])^(4/3))/(7*d*f) - ((35*d^3*((Sqrt[3]*(c - Sqrt[-d^2] 
)^(1/3)*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3))/(c - Sqrt[-d^2])^(1/3)) 
/Sqrt[3]])/(2*Sqrt[-d^2]) - (Sqrt[3]*(c + Sqrt[-d^2])^(1/3)*ArcTan[(1 + (2 
*(c + d*Tan[e + f*x])^(1/3))/(c + Sqrt[-d^2])^(1/3))/Sqrt[3]])/(2*Sqrt[-d^ 
2]) - ((c + Sqrt[-d^2])^(1/3)*Log[Sqrt[-d^2] - d*Tan[e + f*x]])/(4*Sqrt[-d 
^2]) + ((c - Sqrt[-d^2])^(1/3)*Log[Sqrt[-d^2] + d*Tan[e + f*x]])/(4*Sqrt[- 
d^2]) - (3*(c - Sqrt[-d^2])^(1/3)*Log[(c - Sqrt[-d^2])^(1/3) - (c + d*Tan[ 
e + f*x])^(1/3)])/(4*Sqrt[-d^2]) + (3*(c + Sqrt[-d^2])^(1/3)*Log[(c + Sqrt 
[-d^2])^(1/3) - (c + d*Tan[e + f*x])^(1/3)])/(4*Sqrt[-d^2])))/f + (3*(9*c^ 
2 - 35*d^2)*(c + d*Tan[e + f*x])^(4/3))/(4*d*f))/(7*d))/(5*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 485

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand 
Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & 
&  !IntegerQ[2*n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3966

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Su 
bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && NeQ[a^2 + b^2, 0]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4114

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m 
 + 1))), x] + Simp[(A - C)   Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a 
, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.25

method	result	size
derivativedivides	\(\frac {\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {10}{3}}}{10}-\frac {6 c \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{3}}}{7}+\frac {3 c^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4}-\frac {3 d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4}+\frac {d^{4} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f \,d^{3}}\)	\(130\)
default	\(\frac {\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {10}{3}}}{10}-\frac {6 c \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{3}}}{7}+\frac {3 c^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4}-\frac {3 d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4}+\frac {d^{4} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f \,d^{3}}\)	\(130\)

Input:

int(tan(f*x+e)^4*(c+d*tan(f*x+e))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

1/f/d^3*(3/10*(c+d*tan(f*x+e))^(10/3)-6/7*c*(c+d*tan(f*x+e))^(7/3)+3/4*c^2 
*(c+d*tan(f*x+e))^(4/3)-3/4*d^2*(c+d*tan(f*x+e))^(4/3)+1/2*d^4*sum(_R^3/(_ 
R^5-_R^2*c)*ln((c+d*tan(f*x+e))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*c+c^2+d^2) 
))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 663, normalized size of antiderivative = 1.26 \[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^4*(c+d*tan(f*x+e))^(1/3),x, algorithm="fricas")
 

Output:

1/140*(70*d^3*f*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*log(f^4*(-(f^3*sqrt( 
-c^2/f^6) + d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) + 
 70*d^3*f*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*log(-f^4*((f^3*sqrt(-c^2/f^ 
6) - d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) - 35*(sq 
rt(-3)*d^3*f + d^3*f)*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*log(-1/2*(sqrt 
(-3)*f^4 + f^4)*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d* 
tan(f*x + e) + c)^(1/3)*c) + 35*(sqrt(-3)*d^3*f - d^3*f)*(-(f^3*sqrt(-c^2/ 
f^6) + d)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f^4 - f^4)*(-(f^3*sqrt(-c^2/f^6) + 
d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) - 35*(sqrt(-3 
)*d^3*f + d^3*f)*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f^ 
4 + f^4)*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x 
+ e) + c)^(1/3)*c) + 35*(sqrt(-3)*d^3*f - d^3*f)*((f^3*sqrt(-c^2/f^6) - d) 
/f^3)^(1/3)*log(-1/2*(sqrt(-3)*f^4 - f^4)*((f^3*sqrt(-c^2/f^6) - d)/f^3)^( 
1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) + 3*(14*d^3*tan(f*x + 
e)^3 + 2*c*d^2*tan(f*x + e)^2 + 9*c^3 - 35*c*d^2 - (3*c^2*d + 35*d^3)*tan( 
f*x + e))*(d*tan(f*x + e) + c)^(1/3))/(d^3*f)
 

Sympy [F]

\[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\int \sqrt [3]{c + d \tan {\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx \] Input:

integrate(tan(f*x+e)**4*(c+d*tan(f*x+e))**(1/3),x)
 

Output:

Integral((c + d*tan(e + f*x))**(1/3)*tan(e + f*x)**4, x)
 

Maxima [F]

\[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} \tan \left (f x + e\right )^{4} \,d x } \] Input:

integrate(tan(f*x+e)^4*(c+d*tan(f*x+e))^(1/3),x, algorithm="maxima")
 

Output:

integrate((d*tan(f*x + e) + c)^(1/3)*tan(f*x + e)^4, x)
 

Giac [F(-1)]

Timed out. \[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)^4*(c+d*tan(f*x+e))^(1/3),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 17.16 (sec) , antiderivative size = 1015, normalized size of antiderivative = 1.93 \[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

int(tan(e + f*x)^4*(c + d*tan(e + f*x))^(1/3),x)
 

Output:

log((c + d*tan(e + f*x))^(1/3) + f*(-(c*1i + d)/f^3)^(1/3)*1i)*(-(c*1i + d 
)/(8*f^3))^(1/3) + (c + d*tan(e + f*x))^(1/3)*(2*c*((6*c^2)/(d^3*f) - (3*( 
c^2 + d^2))/(d^3*f)) - (12*c^3)/(d^3*f) + (6*c*(c^2 + d^2))/(d^3*f)) + ((3 
*c^2)/(2*d^3*f) - (3*(c^2 + d^2))/(4*d^3*f))*(c + d*tan(e + f*x))^(4/3) + 
log(d*(c + d*tan(e + f*x))^(1/3)*1i - c*(c + d*tan(e + f*x))^(1/3) + f^4*( 
(c*1i - d)/f^3)^(4/3) + 2*d*f*((c*1i - d)/f^3)^(1/3))*((c*1i - d)/(8*f^3)) 
^(1/3) - log((((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*((((c*1i - d)/ 
f^3)^(2/3)*((3^(1/2)*1i)/2 - 1/2)*((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f* 
x))^(1/3))/f - 3888*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*(c 
^2 + d^2)))/4 + (1944*c*d^5*(c^2 + d^2))/f^3))/2 - (486*(d^8 - c^4*d^4)*(c 
 + d*tan(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2 + 1/2)*((c*1i - d)/(8*f^3)) 
^(1/3) + log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4 - (((c*1 
i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*((((c*1i - d)/f^3)^(2/3)*((3^(1/2 
)*1i)/2 + 1/2)*((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f + 3888 
*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(c^2 + d^2)))/4 - (19 
44*c*d^5*(c^2 + d^2))/f^3))/2)*((3^(1/2)*1i)/2 - 1/2)*((c*1i - d)/(8*f^3)) 
^(1/3) + (3*(c + d*tan(e + f*x))^(10/3))/(10*d^3*f) - log((((3^(1/2)*1i)/2 
 + 1/2)*((((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f - 3888*c*d^ 
4*((3^(1/2)*1i)/2 + 1/2)*(-(c*1i + d)/f^3)^(1/3)*(c^2 + d^2))*((3^(1/2)*1i 
)/2 - 1/2)*(-(c*1i + d)/f^3)^(2/3))/4 + (1944*c*d^5*(c^2 + d^2))/f^3)*(...
 

Reduce [F]

\[ \int \tan ^4(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\int \left (d \tan \left (f x +e \right )+c \right )^{\frac {1}{3}} \tan \left (f x +e \right )^{4}d x \] Input:

int(tan(f*x+e)^4*(c+d*tan(f*x+e))^(1/3),x)
 

Output:

int((tan(e + f*x)*d + c)**(1/3)*tan(e + f*x)**4,x)