\(\int \tan ^3(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\) [682]

Optimal result

Integrand size = 23, antiderivative size = 373 \[ \int \tan ^3(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {1}{4} i \sqrt [3]{c-i d} x+\frac {1}{4} i \sqrt [3]{c+i d} x+\frac {\sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{2 f}+\frac {\sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )}{2 f}-\frac {\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}-\frac {\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}-\frac {3 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f}-\frac {3 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f}-\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}-\frac {9 c (c+d \tan (e+f x))^{4/3}}{28 d^2 f}+\frac {3 \tan (e+f x) (c+d \tan (e+f x))^{4/3}}{7 d f} \] Output:

-1/4*I*(c-I*d)^(1/3)*x+1/4*I*(c+I*d)^(1/3)*x+1/2*3^(1/2)*(c-I*d)^(1/3)*arc 
tan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c-I*d)^(1/3))*3^(1/2))/f+1/2*3^(1/2)* 
(c+I*d)^(1/3)*arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c+I*d)^(1/3))*3^(1/2 
))/f-1/4*(c-I*d)^(1/3)*ln(cos(f*x+e))/f-1/4*(c+I*d)^(1/3)*ln(cos(f*x+e))/f 
-3/4*(c-I*d)^(1/3)*ln((c-I*d)^(1/3)-(c+d*tan(f*x+e))^(1/3))/f-3/4*(c+I*d)^ 
(1/3)*ln((c+I*d)^(1/3)-(c+d*tan(f*x+e))^(1/3))/f-3*(c+d*tan(f*x+e))^(1/3)/ 
f-9/28*c*(c+d*tan(f*x+e))^(4/3)/d^2/f+3/7*tan(f*x+e)*(c+d*tan(f*x+e))^(4/3 
)/d/f
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.18 \[ \int \tan ^3(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {14 \sqrt {3} \sqrt [3]{c-i d} d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )+14 \sqrt {3} \sqrt [3]{c+i d} d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-14 \sqrt [3]{c-i d} d^2 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )-14 \sqrt [3]{c+i d} d^2 \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+7 \sqrt [3]{c-i d} d^2 \log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )+7 \sqrt [3]{c+i d} d^2 \log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )-9 c^2 \sqrt [3]{c+d \tan (e+f x)}-84 d^2 \sqrt [3]{c+d \tan (e+f x)}+3 c d \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)}+12 d^2 \tan ^2(e+f x) \sqrt [3]{c+d \tan (e+f x)}}{28 d^2 f} \] Input:

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

Output:

(14*Sqrt[3]*(c - I*d)^(1/3)*d^2*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3)) 
/(c - I*d)^(1/3))/Sqrt[3]] + 14*Sqrt[3]*(c + I*d)^(1/3)*d^2*ArcTan[(1 + (2 
*(c + d*Tan[e + f*x])^(1/3))/(c + I*d)^(1/3))/Sqrt[3]] - 14*(c - I*d)^(1/3 
)*d^2*Log[(c - I*d)^(1/3) - (c + d*Tan[e + f*x])^(1/3)] - 14*(c + I*d)^(1/ 
3)*d^2*Log[(c + I*d)^(1/3) - (c + d*Tan[e + f*x])^(1/3)] + 7*(c - I*d)^(1/ 
3)*d^2*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)] + 7*(c + I*d)^(1/3)*d^2*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)] - 
 9*c^2*(c + d*Tan[e + f*x])^(1/3) - 84*d^2*(c + d*Tan[e + f*x])^(1/3) + 3* 
c*d*Tan[e + f*x]*(c + d*Tan[e + f*x])^(1/3) + 12*d^2*Tan[e + f*x]^2*(c + d 
*Tan[e + f*x])^(1/3))/(28*d^2*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.99 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.86, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 4049, 27, 3042, 4113, 27, 3042, 4011, 3042, 4022, 3042, 4020, 25, 69, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(3*Tan[e + f*x]*(c + d*Tan[e + f*x])^(4/3))/(7*d*f) - ((9*c*(c + d*Tan[e + 
 f*x])^(4/3))/(4*d*f) + 7*d*(((-1/2*I)*(I*c + d)*(((-I)*Sqrt[3]*ArcTanh[Ta 
n[e + f*x]/Sqrt[3]])/(c - I*d)^(2/3) - Log[1 - I*Tan[e + f*x]]/(2*(c - I*d 
)^(2/3)) + (3*Log[(c - I*d)^(1/3) - I*Tan[e + f*x]])/(2*(c - I*d)^(2/3)))) 
/f - ((I/2)*(I*c - d)*((I*Sqrt[3]*ArcTanh[Tan[e + f*x]/Sqrt[3]])/(c + I*d) 
^(2/3) - Log[1 + I*Tan[e + f*x]]/(2*(c + I*d)^(2/3)) + (3*Log[(c + I*d)^(1 
/3) + I*Tan[e + f*x]])/(2*(c + I*d)^(2/3))))/f + (3*(c + d*Tan[e + f*x])^( 
1/3))/f))/(7*d)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

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])
 

rule 1082

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]
 

rule 3042

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

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

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

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 4113

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

Maple [C] (verified)

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

Time = 0.67 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.32

method	result	size
derivativedivides	\(\frac {\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{3}}}{7 d^{2}}-\frac {3 c \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4 d^{2}}-3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3} c +c^{2}+d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f}\)	\(120\)
default	\(\frac {\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{3}}}{7 d^{2}}-\frac {3 c \left (c +d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}{4 d^{2}}-3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3} c +c^{2}+d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f}\)	\(120\)

Input:

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

Output:

1/f*(3/7/d^2*(c+d*tan(f*x+e))^(7/3)-3/4/d^2*c*(c+d*tan(f*x+e))^(4/3)-3*(c+ 
d*tan(f*x+e))^(1/3)+1/2*sum((-_R^3*c+c^2+d^2)/(_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 = 0.13 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.46 \[ \int \tan ^3(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/28*(14*d^2*f*(-(f^3*sqrt(-d^2/f^6) + c)/f^3)^(1/3)*log(f*(-(f^3*sqrt(-d^ 
2/f^6) + c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) + 14*d^2*f*((f^3*sqrt 
(-d^2/f^6) - c)/f^3)^(1/3)*log(f*((f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3) + (d 
*tan(f*x + e) + c)^(1/3)) - 7*(sqrt(-3)*d^2*f + d^2*f)*(-(f^3*sqrt(-d^2/f^ 
6) + c)/f^3)^(1/3)*log(-1/2*(sqrt(-3)*f + f)*(-(f^3*sqrt(-d^2/f^6) + c)/f^ 
3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) + 7*(sqrt(-3)*d^2*f - d^2*f)*(-(f^3 
*sqrt(-d^2/f^6) + c)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f - f)*(-(f^3*sqrt(-d^2/ 
f^6) + c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) - 7*(sqrt(-3)*d^2*f + d 
^2*f)*((f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3)*log(-1/2*(sqrt(-3)*f + f)*((f^3 
*sqrt(-d^2/f^6) - c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) + 7*(sqrt(-3 
)*d^2*f - d^2*f)*((f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f 
- f)*((f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) + 
3*(4*d^2*tan(f*x + e)^2 + c*d*tan(f*x + e) - 3*c^2 - 28*d^2)*(d*tan(f*x + 
e) + c)^(1/3))/(d^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(tan(f*x+e)^3*(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)^3, x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 890, normalized size of antiderivative = 2.39 \[ \int \tan ^3(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

log((c + d*tan(e + f*x))^(1/3) + f*(-(c - d*1i)/f^3)^(1/3))*(-(c - d*1i)/( 
8*f^3))^(1/3) + log((c + d*tan(e + f*x))^(1/3) + f*(-(c + d*1i)/f^3)^(1/3) 
)*(-(c + d*1i)/(8*f^3))^(1/3) + ((3*c^2)/(d^2*f) - (3*(c^2 + d^2))/(d^2*f) 
)*(c + d*tan(e + f*x))^(1/3) + (3*(c + d*tan(e + f*x))^(7/3))/(7*d^2*f) + 
log((((3^(1/2)*1i)/2 - 1/2)*((972*(d^8 - c^4*d^4))/f^3 + (((3^(1/2)*1i)/2 
+ 1/2)*((3888*c*d^4*(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 - d*1i)/f^3)^(1/3)*(c^2 + d^2))*(-(c - d*1i)/ 
f^3)^(2/3))/4)*(-(c - d*1i)/f^3)^(1/3))/2 + (486*(d^8 - c^4*d^4)*(c + d*ta 
n(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2 - 1/2)*(-(c - d*1i)/(8*f^3))^(1/3) 
 + log((((3^(1/2)*1i)/2 - 1/2)*((972*(d^8 - c^4*d^4))/f^3 + (((3^(1/2)*1i) 
/2 + 1/2)*((3888*c*d^4*(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 + d*1i)/f^3)^(1/3)*(c^2 + d^2))*(-(c + d*1 
i)/f^3)^(2/3))/4)*(-(c + d*1i)/f^3)^(1/3))/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 + d*1i)/(8*f^3))^(1 
/3) - log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4 - (((3^(1/2 
)*1i)/2 + 1/2)*((972*(d^8 - c^4*d^4))/f^3 - (((3^(1/2)*1i)/2 - 1/2)*((3888 
*c*d^4*(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 - d*1i)/f^3)^(1/3)*(c^2 + d^2))*(-(c - d*1i)/f^3)^(2/3))/4 
)*(-(c - d*1i)/f^3)^(1/3))/2)*((3^(1/2)*1i)/2 + 1/2)*(-(c - d*1i)/(8*f^3)) 
^(1/3) - log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4 - (((...
 

Reduce [F]

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

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

Output:

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