\(\int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx\) [477]

Optimal result

Integrand size = 27, antiderivative size = 148 \[ \int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx=-\frac {i x}{2 \sqrt [3]{c-i d}}-\frac {\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{\sqrt [3]{c-i d} f}-\frac {\log (\cos (e+f x))}{2 \sqrt [3]{c-i d} f}-\frac {3 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{2 \sqrt [3]{c-i d} f} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.70 \[ \int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )+\log (i+\tan (e+f x))-3 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{2 \sqrt [3]{c-i d} f} \] Input:

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

Output:

(-2*Sqrt[3]*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3))/(c - I*d)^(1/3))/Sq 
rt[3]] + Log[I + Tan[e + f*x]] - 3*Log[(c - I*d)^(1/3) - (c + d*Tan[e + f* 
x])^(1/3)])/(2*(c - I*d)^(1/3)*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.71, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4020, 67, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i \int \frac {1}{(1-i \tan (e+f x)) \sqrt [3]{c+d \tan (e+f x)}}d(-\tan (e+f x))}{f}\)

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle -\frac {i \left (\frac {i \sqrt {3} \arctan \left (\frac {\tan (e+f x)}{\sqrt {3}}\right )}{\sqrt [3]{c-i d}}+\frac {i \log (\tan (e+f x)+i)}{2 \sqrt [3]{c-i d}}-\frac {3 i \log \left (\tan (e+f x)+\sqrt [3]{c-i d}\right )}{2 \sqrt [3]{c-i d}}\right )}{f}\)

Input:

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

Output:

((-I)*((I*Sqrt[3]*ArcTan[Tan[e + f*x]/Sqrt[3]])/(c - I*d)^(1/3) + ((I/2)*L 
og[I + Tan[e + f*x]])/(c - I*d)^(1/3) - (((3*I)/2)*Log[(c - I*d)^(1/3) + T 
an[e + f*x]])/(c - I*d)^(1/3)))/f
 

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 67

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

Maple [C] (verified)

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

Time = 0.64 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28

method	result	size
derivativedivides	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+i d -c \right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}}}{f}\)	\(42\)
default	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+i d -c \right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}}}{f}\)	\(42\)
parts	\(\frac {i d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R} \left (\textit {\_R}^{3}-c \right )}\right )}{2 f}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}}}{2 f}\)	\(105\)

Input:

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

Output:

-1/f*sum(1/_R*ln((c+d*tan(f*x+e))^(1/3)-_R),_R=RootOf(_Z^3+I*d-c))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (109) = 218\).

Time = 0.08 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.88 \[ \int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx=\frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {3} {\left (i \, c + d\right )} f^{2} + {\left (c - i \, d\right )} f^{2}\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {3} {\left (-i \, c - d\right )} f^{2} + {\left (c - i \, d\right )} f^{2}\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) + \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (-{\left (c - i \, d\right )} f^{2} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) \] Input:

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

Output:

1/2*(I*sqrt(3) - 1)*(-I/((I*c + d)*f^3))^(1/3)*log(1/2*(sqrt(3)*(I*c + d)* 
f^2 + (c - I*d)*f^2)*(-I/((I*c + d)*f^3))^(2/3) + (((c - I*d)*e^(2*I*f*x + 
 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)) + 1/2*(-I*sqrt(3) - 1 
)*(-I/((I*c + d)*f^3))^(1/3)*log(1/2*(sqrt(3)*(-I*c - d)*f^2 + (c - I*d)*f 
^2)*(-I/((I*c + d)*f^3))^(2/3) + (((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d 
)/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)) + (-I/((I*c + d)*f^3))^(1/3)*log(-(c - 
 I*d)*f^2*(-I/((I*c + d)*f^3))^(2/3) + (((c - I*d)*e^(2*I*f*x + 2*I*e) + c 
 + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^(1/3))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (109) = 218\).

Time = 0.42 (sec) , antiderivative size = 795, normalized size of antiderivative = 5.37 \[ \int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-(c - I*d)^(2/3)*log((d*tan(f*x + e) + c)^(1/3) - (c - I*d)^(1/3))/(c*f - 
I*d*f) - (sqrt(3)*c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan( 
d/c))^2 - sqrt(3)*c*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan( 
d/c))^2 + 2*c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))* 
sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)))*arctan(1/3*sq 
rt(3)*(2*(d*tan(f*x + e) + c)^(1/3) + (c - I*d)^(1/3))/(c - I*d)^(1/3))/(( 
c^2 + d^2)^(2/3)*f) - I*(sqrt(3)*d*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d 
) - 1/3*arctan(d/c))^2 - sqrt(3)*d*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d 
) - 1/3*arctan(d/c))^2 + 2*d*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/ 
3*arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)) 
)*arctan(1/3*sqrt(3)*(2*(d*tan(f*x + e) + c)^(1/3) + (c - I*d)^(1/3))/(c - 
 I*d)^(1/3))/((c^2 + d^2)^(2/3)*f) - 1/2*(2*sqrt(3)*c*cos(1/6*pi*sgn(c)*sg 
n(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi* 
sgn(d) - 1/3*arctan(d/c)) - c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1 
/3*arctan(d/c))^2 + c*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arcta 
n(d/c))^2)*log((c^2 + d^2)^(1/3)*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) 
- 1/3*arctan(d/c))^2 + (c^2 + d^2)^(1/3)*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi 
*sgn(d) - 1/3*arctan(d/c))^2 + (d*tan(f*x + e) + c)^(2/3) + (d*tan(f*x + e 
) + c)^(1/3)*(c - I*d)^(1/3))/((c^2 + d^2)^(2/3)*f) - 1/2*I*(2*sqrt(3)*d*c 
os(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))*sin(1/6*pi*s...
 

Mupad [B] (verification not implemented)

Time = 17.68 (sec) , antiderivative size = 2982, normalized size of antiderivative = 20.15 \[ \int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(d^5*f*(c + d*tan(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c 
+ d*tan(e + f*x))^(1/3) - 3*2^(1/3)*c*d^4*f^6*(c^2 + d^2)*(-(11664*f^3*((c 
^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(d^6*f^3* 
(c^2 + d^2)^2))^(2/3))*(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d 
^9*11664i + c^2*d^7*11664i))/(93312*d^6*f^3*(c^2 + d^2)^2))*(-(f^3*((4*(72 
9*d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46656*c^4*d^6))/f^6 - ( 
11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(93 
312*f^3*(d^10 + 2*c^2*d^8 + c^4*d^6)))^(1/3) + log(d^5*f*(c + d*tan(e + f* 
x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3) - 3 
*2^(1/3)*c*d^4*f^6*(c^2 + d^2)*(-(d^9*11664i - 11664*f^3*((c^2*d^12*(c^2 + 
 d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i)/(d^6*f^3*(c^2 + d^2)^2))^(2/3))*(d^9 
*11664i - 11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i) 
)/(93312*d^6*f^3*(c^2 + d^2)^2))*(-(d^9*11664i - f^3*((4*(729*d^8 + 729*c^ 
2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46656*c^4*d^6))/f^6 - (11664*d^9 + 11 
664*c^2*d^7)^2/f^6)^(1/2) + c^2*d^7*11664i)/(93312*f^3*(d^10 + 2*c^2*d^8 + 
 c^4*d^6)))^(1/3) + (log(- ((-1/(f^3*(c - d*1i)))^(2/3)*(((-1/(f^3*(c - d* 
1i)))^(1/3)*((1944*d^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3))/f^2 - 1944* 
c*d^4*(-1/(f^3*(c - d*1i)))^(2/3)*(c^2 + d^2)))/2 - (972*d^4*(c^2 + d^2))/ 
f^3))/4 - (243*c*d^4*(c + d*tan(e + f*x))^(1/3))/f^5)*(-1/(c*f^3 - d*f^3*1 
i))^(1/3))/2 + log(((7776*c*d^4*(c^2 + d^2)*(-1i/(8*f^3*(c*1i - d)))^(2...
 

Reduce [F]

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

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

Output:

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