\(\int (a+b \tan (c+d x))^{5/3} \, dx\) [688]

Optimal result

Integrand size = 14, antiderivative size = 329 \[ \int (a+b \tan (c+d x))^{5/3} \, dx=-\frac {1}{4} (a-i b)^{5/3} x-\frac {1}{4} (a+i b)^{5/3} x+\frac {i \sqrt {3} (a-i b)^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}-\frac {i \sqrt {3} (a+i b)^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {i (a-i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}+\frac {3 i (a-i b)^{5/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 i (a+i b)^{5/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 b (a+b \tan (c+d x))^{2/3}}{2 d} \] Output:

-1/4*(a-I*b)^(5/3)*x-1/4*(a+I*b)^(5/3)*x+1/2*I*3^(1/2)*(a-I*b)^(5/3)*arcta 
n(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))/d-1/2*I*3^(1/2)* 
(a+I*b)^(5/3)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(1/3))*3^(1/2 
))/d+1/4*I*(a-I*b)^(5/3)*ln(cos(d*x+c))/d-1/4*I*(a+I*b)^(5/3)*ln(cos(d*x+c 
))/d+3/4*I*(a-I*b)^(5/3)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d-3/4*I* 
(a+I*b)^(5/3)*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 = 0.64 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.91 \[ \int (a+b \tan (c+d x))^{5/3} \, dx=\frac {(i a+b) \left (2 \sqrt {3} (a-i b)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-(a-i b)^{2/3} \log (i+\tan (c+d x))+3 \left ((a-i b)^{2/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+(a+b \tan (c+d x))^{2/3}\right )\right )+(-i a+b) \left (2 \sqrt {3} (a+i b)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-(a+i b)^{2/3} \log (i-\tan (c+d x))+3 \left ((a+i b)^{2/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+(a+b \tan (c+d x))^{2/3}\right )\right )}{4 d} \] Input:

Integrate[(a + b*Tan[c + d*x])^(5/3),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.79, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 3963, 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.

\(\displaystyle \int (a+b \tan (c+d x))^{5/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{5/3}dx\)

\(\Big \downarrow \) 3963

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(a + b*Tan[c + d*x])^(5/3),x]
 

Output:

((I/2)*(a - I*b)^2*((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)^2*(((-I)*Sq 
rt[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 3963

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d 
*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 
+ b^2, 0] && GtQ[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 = 0.70 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.27

method	result	size
derivativedivides	\(\frac {b \left (\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-2 a \,\textit {\_R}^{4}+\left (a^{2}+b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}\right )}{d}\)	\(89\)
default	\(\frac {b \left (\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-2 a \,\textit {\_R}^{4}+\left (a^{2}+b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}\right )}{d}\)	\(89\)

Input:

int((a+b*tan(d*x+c))^(5/3),x,method=_RETURNVERBOSE)
 

Output:

1/d*b*(3/2*(a+b*tan(d*x+c))^(2/3)-1/2*sum((-2*a*_R^4+(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. 2071 vs. \(2 (235) = 470\).

Time = 0.21 (sec) , antiderivative size = 2071, normalized size of antiderivative = 6.29 \[ \int (a+b \tan (c+d x))^{5/3} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(5/3),x, algorithm="fricas")
 

Output:

1/4*(2*d*(-(5*a^4*b - 10*a^2*b^3 + b^5 + d^3*sqrt(-(a^10 - 20*a^8*b^2 + 11 
0*a^6*b^4 - 100*a^4*b^6 + 25*a^2*b^8)/d^6))/d^3)^(1/3)*log(((3*a^2*b - b^3 
)*d^5*sqrt(-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 100*a^4*b^6 + 25*a^2*b^8)/d 
^6) + (a^8 - 13*a^6*b^2 + 35*a^4*b^4 - 15*a^2*b^6)*d^2)*(-(5*a^4*b - 10*a^ 
2*b^3 + b^5 + d^3*sqrt(-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 100*a^4*b^6 + 2 
5*a^2*b^8)/d^6))/d^3)^(2/3) + (a^11 - 7*a^9*b^2 - 22*a^7*b^4 - 14*a^5*b^6 
+ 5*a^3*b^8 + 5*a*b^10)*(b*tan(d*x + c) + a)^(1/3)) + (sqrt(-3)*d - d)*(-( 
5*a^4*b - 10*a^2*b^3 + b^5 + d^3*sqrt(-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 
100*a^4*b^6 + 25*a^2*b^8)/d^6))/d^3)^(1/3)*log(-1/2*(sqrt(-3)*(a^8 - 13*a^ 
6*b^2 + 35*a^4*b^4 - 15*a^2*b^6)*d^2 + (a^8 - 13*a^6*b^2 + 35*a^4*b^4 - 15 
*a^2*b^6)*d^2 + (sqrt(-3)*(3*a^2*b - b^3)*d^5 + (3*a^2*b - b^3)*d^5)*sqrt( 
-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 100*a^4*b^6 + 25*a^2*b^8)/d^6))*(-(5*a 
^4*b - 10*a^2*b^3 + b^5 + d^3*sqrt(-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 100 
*a^4*b^6 + 25*a^2*b^8)/d^6))/d^3)^(2/3) + (a^11 - 7*a^9*b^2 - 22*a^7*b^4 - 
 14*a^5*b^6 + 5*a^3*b^8 + 5*a*b^10)*(b*tan(d*x + c) + a)^(1/3)) - (sqrt(-3 
)*d + d)*(-(5*a^4*b - 10*a^2*b^3 + b^5 + d^3*sqrt(-(a^10 - 20*a^8*b^2 + 11 
0*a^6*b^4 - 100*a^4*b^6 + 25*a^2*b^8)/d^6))/d^3)^(1/3)*log(1/2*(sqrt(-3)*( 
a^8 - 13*a^6*b^2 + 35*a^4*b^4 - 15*a^2*b^6)*d^2 - (a^8 - 13*a^6*b^2 + 35*a 
^4*b^4 - 15*a^2*b^6)*d^2 + (sqrt(-3)*(3*a^2*b - b^3)*d^5 - (3*a^2*b - b^3) 
*d^5)*sqrt(-(a^10 - 20*a^8*b^2 + 110*a^6*b^4 - 100*a^4*b^6 + 25*a^2*b^8...
 

Sympy [F]

\[ \int (a+b \tan (c+d x))^{5/3} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{3}}\, dx \] Input:

integrate((a+b*tan(d*x+c))**(5/3),x)
 

Output:

Integral((a + b*tan(c + d*x))**(5/3), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int (a+b \tan (c+d x))^{5/3} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(5/3),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(5/3), x)
 

Giac [F]

\[ \int (a+b \tan (c+d x))^{5/3} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(5/3),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 7.52 (sec) , antiderivative size = 1540, normalized size of antiderivative = 4.68 \[ \int (a+b \tan (c+d x))^{5/3} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(c + d*x))^(5/3),x)
 

Output:

log(((-(a*1i + b)^5/d^3)^(2/3)*(((-(a*1i + b)^5/d^3)^(1/3)*(1944*a*b^4*(-( 
a*1i + b)^5/d^3)^(2/3)*(a^2 + b^2) + (1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*( 
a + b*tan(c + d*x))^(1/3))/d^2))/2 + (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^ 
4 - 7*a^4*b^2))/d^3))/4 + (243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan( 
c + d*x))^(1/3))/d^5)*(-(a*b^4*5i + 5*a^4*b + a^5*1i + b^5 - 10*a^2*b^3 - 
a^3*b^2*10i)/(8*d^3))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*(((a + b*1i)^ 
5*1i)/d^3)^(2/3) + (1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x) 
)^(1/3))/d^2)*(((a + b*1i)^5*1i)/d^3)^(1/3))/2 + (1944*a*b^5*(3*a^6 + 3*b^ 
6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3)*(((a + b*1i)^5*1i)/d^3)^(2/3))/4 + (243*b 
^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5)*((a*b^4*5i 
 - 5*a^4*b + a^5*1i - b^5 + 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) - log 
((243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5 - (( 
(3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 
 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*((3^(1/2)*1i)/2 - 1 
/2)*(a^2 + b^2)*(((a + b*1i)^5*1i)/d^3)^(2/3))*(((a + b*1i)^5*1i)/d^3)^(1/ 
3))/2 - (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3)*(((a + b 
*1i)^5*1i)/d^3)^(2/3))/4)*((3^(1/2)*1i)/2 + 1/2)*((a*b^4*5i - 5*a^4*b + a^ 
5*1i - b^5 + 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) + log((243*b^5*(3*a^ 
2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 
+ 1/2)*((((3^(1/2)*1i)/2 - 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a...
 

Reduce [F]

\[ \int (a+b \tan (c+d x))^{5/3} \, dx=\left (\int \left (a +\tan \left (d x +c \right ) b \right )^{\frac {2}{3}}d x \right ) a +\left (\int \left (a +\tan \left (d x +c \right ) b \right )^{\frac {2}{3}} \tan \left (d x +c \right )d x \right ) b \] Input:

int((a+b*tan(d*x+c))^(5/3),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