\(\int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) [473]

Optimal result

Integrand size = 25, antiderivative size = 379 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {1}{4} (a-i b)^{2/3} (A-i B) x-\frac {1}{4} (a+i b)^{2/3} (A+i B) x+\frac {\sqrt {3} (a-i b)^{2/3} (i A+B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}-\frac {\sqrt {3} (a+i b)^{2/3} (i A-B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}-\frac {(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 (a+i b)^{2/3} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.69 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {i \left ((A-i B) \left ((a-i b)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )+3 (a+b \tan (c+d x))^{2/3}\right )-(A+i B) \left ((a+i b)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-\log (i-\tan (c+d x))+3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )+3 (a+b \tan (c+d x))^{2/3}\right )\right )}{4 d} \] Input:

Integrate[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]
 

Output:

((I/4)*((A - I*B)*((a - I*b)^(2/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c 
+ d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] - Log[I + Tan[c + d*x]] + 3*Log[( 
a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]) + 3*(a + b*Tan[c + d*x])^(2/ 
3)) - (A + I*B)*((a + I*b)^(2/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + 
d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - Log[I - Tan[c + d*x]] + 3*Log[(a 
+ I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]) + 3*(a + b*Tan[c + d*x])^(2/3) 
)))/d
 

Rubi [A] (warning: unable to verify)

Time = 0.72 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.71, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4011, 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.

\(\displaystyle \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]
 

Output:

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

Maple [C] (verified)

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

Time = 0.78 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.26

method	result	size
derivativedivides	\(\frac {\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}} B}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{4}+B \left (-a^{2}-b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\)	\(99\)
default	\(\frac {\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}} B}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{4}+B \left (-a^{2}-b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\)	\(99\)
parts	\(\frac {A b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{3}-a}\right )}{2 d}+\frac {B \left (\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3} a +a^{2}+b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R} \left (-\textit {\_R}^{3}+a \right )}\right )}{2}\right )}{d}\)	\(144\)

Input:

int((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(3/2*(a+b*tan(d*x+c))^(2/3)*B+1/2*sum(((A*b+B*a)*_R^4+B*(-a^2-b^2)*_R) 
/(_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. 5405 vs. \(2 (281) = 562\).

Time = 1.72 (sec) , antiderivative size = 5405, normalized size of antiderivative = 14.26 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}\, dx \] Input:

integrate((a+b*tan(d*x+c))**(2/3)*(A+B*tan(d*x+c)),x)
 

Output:

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(2/3), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(2/3), x)
 

Giac [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 20.26 (sec) , antiderivative size = 3945, normalized size of antiderivative = 10.41 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

int((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(2/3),x)
 

Output:

log((((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2/3)* 
((((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(1/3)*(19 
44*a*b^4*((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2 
/3)*(a^2 + b^2) - (1944*B^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/ 
d^2))/2 + (972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3))/4 + (243*B^5*b^4 
*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(((-4*B^6*a^2* 
b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/(8*d^6))^(1/3) + log(((-(2*(-B 
^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(2/3)*(((-(2*(-B^6 
*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(1/3)*(1944*a*b^4*(- 
(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(2/3)*(a^2 + 
 b^2) - (1944*B^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2))/2 + 
(972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3))/4 + (243*B^5*b^4*(a^2 - b^ 
2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(-((-4*B^6*a^2*b^2*d^6)^ 
(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/(8*d^6))^(1/3) + log(((((-A^6*d^6*(a^2 
- b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3)*(((1944*a*b^4*(((-A^6*d^6*(a^2 
 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 + b^2) + (1944*A^2*b^4*( 
a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2)*(((-A^6*d^6*(a^2 - b^2)^2)^( 
1/2) - 2*A^3*a*b*d^3)/d^6)^(1/3))/2 + (972*A^3*b^5*(3*a^4 - b^4 + 2*a^2*b^ 
2))/d^3))/4 + (486*A^5*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 
)*((2*A^6*a^2*b^2*d^6 - A^6*b^4*d^6 - A^6*a^4*d^6)^(1/2)/(8*d^6) - (A^3...
 

Reduce [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\left (\int \left (a +\tan \left (d x +c \right ) b \right )^{\frac {2}{3}}d x \right ) a +\left (\int \left (a +\tan \left (d x +c \right ) b \right )^{\frac {2}{3}} \tan \left (d x +c \right )d x \right ) b \] Input:

int((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x)
 

Output:

int((tan(c + d*x)*b + a)**(2/3),x)*a + int((tan(c + d*x)*b + a)**(2/3)*tan 
(c + d*x),x)*b