\(\int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx\) [694]

Optimal result

Integrand size = 14, antiderivative size = 336 \[ \int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx=-\frac {x}{4 (a-i b)^{4/3}}-\frac {x}{4 (a+i b)^{4/3}}+\frac {i \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 (a-i b)^{4/3} d}-\frac {i \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 (a+i b)^{4/3} d}+\frac {i \log (\cos (c+d x))}{4 (a-i b)^{4/3} d}-\frac {i \log (\cos (c+d x))}{4 (a+i b)^{4/3} d}+\frac {3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{4/3} d}-\frac {3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{4/3} d}-\frac {3 b}{\left (a^2+b^2\right ) d \sqrt [3]{a+b \tan (c+d x)}} \] Output:

-1/4*x/(a-I*b)^(4/3)-1/4*x/(a+I*b)^(4/3)+1/2*I*3^(1/2)*arctan(1/3*(1+2*(a+ 
b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))/(a-I*b)^(4/3)/d-1/2*I*3^(1/2)* 
arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(1/3))*3^(1/2))/(a+I*b)^(4/ 
3)/d+1/4*I*ln(cos(d*x+c))/(a-I*b)^(4/3)/d-1/4*I*ln(cos(d*x+c))/(a+I*b)^(4/ 
3)/d+3/4*I*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a-I*b)^(4/3)/d-3/4*I* 
ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a+I*b)^(4/3)/d-3*b/(a^2+b^2)/d/( 
a+b*tan(d*x+c))^(1/3)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.32 \[ \int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx=\frac {3 i \left ((a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{2 \left (a^2+b^2\right ) d \sqrt [3]{a+b \tan (c+d x)}} \] Input:

Integrate[(a + b*Tan[c + d*x])^(-4/3),x]
 

Output:

(((3*I)/2)*((a + I*b)*Hypergeometric2F1[-1/3, 1, 2/3, (a + b*Tan[c + d*x]) 
/(a - I*b)] - (a - I*b)*Hypergeometric2F1[-1/3, 1, 2/3, (a + b*Tan[c + d*x 
])/(a + I*b)]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.81, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 3964, 3042, 4022, 3042, 4020, 25, 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.

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{4/3}}dx\)

\(\Big \downarrow \) 3964

\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {3 b}{d \left (a^2+b^2\right ) \sqrt [3]{a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {3 b}{d \left (a^2+b^2\right ) \sqrt [3]{a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle -\frac {3 b}{d \left (a^2+b^2\right ) \sqrt [3]{a+b \tan (c+d x)}}+\frac {\frac {i (a+i b) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 \sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 \sqrt [3]{a-i b}}\right )}{2 d}-\frac {i (a-i b) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 \sqrt [3]{a+i b}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 \sqrt [3]{a+i b}}\right )}{2 d}}{a^2+b^2}\)

Input:

Int[(a + b*Tan[c + d*x])^(-4/3),x]
 

Output:

(((I/2)*(a + I*b)*((I*Sqrt[3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a - I*b)^(1/ 
3) - Log[1 - I*Tan[c + d*x]]/(2*(a - I*b)^(1/3)) + (3*Log[(a - I*b)^(1/3) 
- I*Tan[c + d*x]])/(2*(a - I*b)^(1/3))))/d - ((I/2)*(a - I*b)*(((-I)*Sqrt[ 
3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a + I*b)^(1/3) - Log[1 + I*Tan[c + d*x] 
]/(2*(a + I*b)^(1/3)) + (3*Log[(a + I*b)^(1/3) + I*Tan[c + d*x]])/(2*(a + 
I*b)^(1/3))))/d)/(a^2 + b^2) - (3*b)/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^( 
1/3))
 

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 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 3964

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) 
 Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, 
 b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
 

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]
 

Maple [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.30

method	result	size
derivativedivides	\(\frac {b \left (-\frac {3}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{4}+2 \textit {\_R} a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 a^{2}+2 b^{2}}\right )}{d}\)	\(101\)
default	\(\frac {b \left (-\frac {3}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{4}+2 \textit {\_R} a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 a^{2}+2 b^{2}}\right )}{d}\)	\(101\)

Input:

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

Output:

1/d*b*(-3/(a^2+b^2)/(a+b*tan(d*x+c))^(1/3)+1/2/(a^2+b^2)*sum((-_R^4+2*_R*a 
)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+ 
b^2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4277 vs. \(2 (244) = 488\).

Time = 0.21 (sec) , antiderivative size = 4277, normalized size of antiderivative = 12.73 \[ \int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

integrate(1/(a+b*tan(d*x+c))**(4/3),x)
 

Output:

Integral((a + b*tan(c + d*x))**(-4/3), x)
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

integrate((b*tan(d*x + c) + a)^(-4/3), x)
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 3.86 (sec) , antiderivative size = 3130, normalized size of antiderivative = 9.32 \[ \int \frac {1}{(a+b \tan (c+d x))^{4/3}} \, dx=\text {Too large to display} \] Input:

int(1/(a + b*tan(c + d*x))^(4/3),x)
 

Output:

(log((((a + b*tan(c + d*x))^(1/3)*(38880*a^4*b^18*d^7 - 1944*a^2*b^20*d^7 
- 1944*b^22*d^7 + 163296*a^6*b^16*d^7 + 299376*a^8*b^14*d^7 + 299376*a^10* 
b^12*d^7 + 163296*a^12*b^10*d^7 + 38880*a^14*b^8*d^7 - 1944*a^16*b^6*d^7 - 
 1944*a^18*b^4*d^7) - ((-1/(a^4*d^3*1i + b^4*d^3*1i + 4*a*b^3*d^3 - 4*a^3* 
b*d^3 - a^2*b^2*d^3*6i))^(2/3)*(7776*a*b^24*d^9 + 77760*a^3*b^22*d^9 + 349 
920*a^5*b^20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^ 
11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 349920*a^17*b 
^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4*d^9))/4)*(-1/(a^4*d^3*1i + b^4 
*d^3*1i + 4*a*b^3*d^3 - 4*a^3*b*d^3 - a^2*b^2*d^3*6i))^(1/3))/2 - 972*b^21 
*d^6 - 3888*a^2*b^19*d^6 + 27216*a^6*b^15*d^6 + 68040*a^8*b^13*d^6 + 81648 
*a^10*b^11*d^6 + 54432*a^12*b^9*d^6 + 19440*a^14*b^7*d^6 + 2916*a^16*b^5*d 
^6)*(-1/(a^4*d^3*1i + b^4*d^3*1i + 4*a*b^3*d^3 - 4*a^3*b*d^3 - a^2*b^2*d^3 
*6i))^(1/3))/2 + log(((a + b*tan(c + d*x))^(1/3)*(38880*a^4*b^18*d^7 - 194 
4*a^2*b^20*d^7 - 1944*b^22*d^7 + 163296*a^6*b^16*d^7 + 299376*a^8*b^14*d^7 
 + 299376*a^10*b^12*d^7 + 163296*a^12*b^10*d^7 + 38880*a^14*b^8*d^7 - 1944 
*a^16*b^6*d^7 - 1944*a^18*b^4*d^7) - (-1i/(8*(a^4*d^3 + b^4*d^3 + a*b^3*d^ 
3*4i - a^3*b*d^3*4i - 6*a^2*b^2*d^3)))^(2/3)*(7776*a*b^24*d^9 + 77760*a^3* 
b^22*d^9 + 349920*a^5*b^20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^ 
9 + 1959552*a^11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 
 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4*d^9))*(-1i/(8...
 

Reduce [F]

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

int(1/(a+b*tan(d*x+c))^(4/3),x)
 

Output:

int(1/((tan(c + d*x)*b + a)**(1/3)*tan(c + d*x)*b + (tan(c + d*x)*b + a)** 
(1/3)*a),x)