\(\int (a+b \tan (c+d x))^{4/3} \, dx\) [689]

Optimal result

Integrand size = 14, antiderivative size = 327 \[ \int (a+b \tan (c+d x))^{4/3} \, dx=-\frac {1}{4} (a-i b)^{4/3} x-\frac {1}{4} (a+i b)^{4/3} x-\frac {i \sqrt {3} (a-i b)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}+\frac {i \sqrt {3} (a+i b)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac {3 i (a-i b)^{4/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 i (a+i b)^{4/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3563, 3620, 3618, 59, 631, 210, 31} \[ \int (a+b \tan (c+d x))^{4/3} \, dx=-\frac {i \sqrt {3} (a-i b)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}+\frac {i \sqrt {3} (a+i b)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac {3 i (a-i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac {3 i (a+i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}+\frac {i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {1}{4} x (a-i b)^{4/3}-\frac {1}{4} x (a+i b)^{4/3} \]

[In]

Int[(a + b*Tan[c + d*x])^(4/3),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[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 210

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3563

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\int \frac {a^2-b^2+2 a b \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx \\ & = \frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac {1}{2} (a-i b)^2 \int \frac {1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx+\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx \\ & = \frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac {\left (i (a-i b)^2\right ) \text {Subst}\left (\int \frac {1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {\left (i (a+i b)^2\right ) \text {Subst}\left (\int \frac {1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = -\frac {1}{4} (a-i b)^{4/3} x-\frac {1}{4} (a+i b)^{4/3} x+\frac {i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}-\frac {\left (3 i (a-i b)^{4/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {\left (3 i (a-i b)^{5/3}\right ) \text {Subst}\left (\int \frac {1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {\left (3 i (a+i b)^{4/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {\left (3 i (a+i b)^{5/3}\right ) \text {Subst}\left (\int \frac {1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d} \\ & = -\frac {1}{4} (a-i b)^{4/3} x-\frac {1}{4} (a+i b)^{4/3} x+\frac {i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac {3 i (a-i b)^{4/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 i (a+i b)^{4/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac {\left (3 i (a-i b)^{4/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 d}-\frac {\left (3 i (a+i b)^{4/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 d} \\ & = -\frac {1}{4} (a-i b)^{4/3} x-\frac {1}{4} (a+i b)^{4/3} x-\frac {i \sqrt {3} (a-i b)^{4/3} \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)^{4/3} \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)^{4/3} \log (\cos (c+d x))}{4 d}-\frac {i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac {3 i (a-i b)^{4/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 i (a+i b)^{4/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 b \sqrt [3]{a+b \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.12 \[ \int (a+b \tan (c+d x))^{4/3} \, dx=\frac {(i a+b) \left (2 \sqrt [3]{a-i b} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )-\sqrt [3]{a-i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )+6 \sqrt [3]{a+b \tan (c+d x)}\right )-(i a-b) \left (2 \sqrt [3]{a+i b} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )-\sqrt [3]{a+i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )+6 \sqrt [3]{a+b \tan (c+d x)}\right )}{4 d} \]

[In]

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

[Out]

((I*a + b)*(2*(a - I*b)^(1/3)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] - (a - I*b)^(1/3)*(2*Sqrt[3]*A
rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a
+ b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]) + 6*(a + b*Tan[c + d*x])^(1/3)) - (I*a - b)*(2*(a + I*
b)^(1/3)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] - (a + I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*
Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] + Log[(a + I*b)^(2/3) + (a + I*b)^(1/3)*(a + b*Tan[c + d*x])^(1
/3) + (a + b*Tan[c + d*x])^(2/3)]) + 6*(a + b*Tan[c + d*x])^(1/3)))/(4*d)

Maple [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.28

method	result	size
derivativedivides	\(\frac {b \left (3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (2 \textit {\_R}^{3} a -a^{2}-b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}\right )}{d}\)	\(90\)
default	\(\frac {b \left (3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (2 \textit {\_R}^{3} a -a^{2}-b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}\right )}{d}\)	\(90\)

[In]

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

[Out]

1/d*b*(3*(a+b*tan(d*x+c))^(1/3)+1/2*sum((2*_R^3*a-a^2-b^2)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=Root
Of(_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. 1567 vs. \(2 (235) = 470\).

Time = 0.34 (sec) , antiderivative size = 1567, normalized size of antiderivative = 4.79 \[ \int (a+b \tan (c+d x))^{4/3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(2*d*(-(4*a^3*b - 4*a*b^3 + d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)*
log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b^6)*(b*tan(d*x + c) + a)^(1/3) + (a*d^4*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b
^4 - 12*a^2*b^6 + b^8)/d^6) - (a^4*b - 6*a^2*b^3 + b^5)*d)*(-(4*a^3*b - 4*a*b^3 + d^3*sqrt(-(a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)) - (sqrt(-3)*d + d)*(-(4*a^3*b - 4*a*b^3 + d^3*sqrt(-(a^8 -
12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)*log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b^6)*(b*tan(d*
x + c) + a)^(1/3) + 1/2*(sqrt(-3)*(a^4*b - 6*a^2*b^3 + b^5)*d + (a^4*b - 6*a^2*b^3 + b^5)*d - (sqrt(-3)*a*d^4
+ a*d^4)*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))*(-(4*a^3*b - 4*a*b^3 + d^3*sqrt(-(a^8
- 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)) + (sqrt(-3)*d - d)*(-(4*a^3*b - 4*a*b^3 + d^3*
sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)*log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b
^6)*(b*tan(d*x + c) + a)^(1/3) - 1/2*(sqrt(-3)*(a^4*b - 6*a^2*b^3 + b^5)*d - (a^4*b - 6*a^2*b^3 + b^5)*d - (sq
rt(-3)*a*d^4 - a*d^4)*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))*(-(4*a^3*b - 4*a*b^3 + d^
3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)) + 2*d*(-(4*a^3*b - 4*a*b^3 - d^3*
sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)*log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b
^6)*(b*tan(d*x + c) + a)^(1/3) - (a*d^4*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6) + (a^4*b
- 6*a^2*b^3 + b^5)*d)*(-(4*a^3*b - 4*a*b^3 - d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6
))/d^3)^(1/3)) - (sqrt(-3)*d + d)*(-(4*a^3*b - 4*a*b^3 - d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6
+ b^8)/d^6))/d^3)^(1/3)*log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b^6)*(b*tan(d*x + c) + a)^(1/3) + 1/2*(sqrt(-3)*(a
^4*b - 6*a^2*b^3 + b^5)*d + (a^4*b - 6*a^2*b^3 + b^5)*d + (sqrt(-3)*a*d^4 + a*d^4)*sqrt(-(a^8 - 12*a^6*b^2 + 3
8*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))*(-(4*a^3*b - 4*a*b^3 - d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b
^6 + b^8)/d^6))/d^3)^(1/3)) + (sqrt(-3)*d - d)*(-(4*a^3*b - 4*a*b^3 - d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4
- 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)*log((a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b^6)*(b*tan(d*x + c) + a)^(1/3) - 1/2
*(sqrt(-3)*(a^4*b - 6*a^2*b^3 + b^5)*d - (a^4*b - 6*a^2*b^3 + b^5)*d + (sqrt(-3)*a*d^4 - a*d^4)*sqrt(-(a^8 - 1
2*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^6))*(-(4*a^3*b - 4*a*b^3 - d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b
^4 - 12*a^2*b^6 + b^8)/d^6))/d^3)^(1/3)) + 12*(b*tan(d*x + c) + a)^(1/3)*b)/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

undef

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 924, normalized size of antiderivative = 2.83 \[ \int (a+b \tan (c+d x))^{4/3} \, dx=\text {Too large to display} \]

[In]

int((a + b*tan(c + d*x))^(4/3),x)

[Out]

log(a*(a + b*tan(c + d*x))^(1/3) - b*(a + b*tan(c + d*x))^(1/3)*1i + d*(-((a - b*1i)^4*1i)/d^3)^(1/3)*1i)*(-(4
*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)/(8*d^3))^(1/3) + log(a*(a + b*tan(c + d*x))^(1/3)*1i - b*(a +
b*tan(c + d*x))^(1/3) + d*(((a*1i - b)^4*1i)/d^3)^(1/3))*((4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)/
(8*d^3))^(1/3) + log((b^4*((3^(1/2)*1i)/2 - 1/2)*(a - b*1i)^2*(((a*1i - b)^4*1i)/d^3)^(1/3)*(a*b^4 + a^4*b*1i
+ a^5 + b^5*1i - a^2*b^3*6i - 6*a^3*b^2)*486i)/d^3 - (486*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3)*(a^4 +
b^4 - 6*a^2*b^2))/d^4)*((3^(1/2)*1i)/2 - 1/2)*((4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)/(8*d^3))^(1/
3) - log((486*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3)*(a^4 + b^4 - 6*a^2*b^2))/d^4 + (b^4*((3^(1/2)*1i)/2
+ 1/2)*(a - b*1i)^2*(((a*1i - b)^4*1i)/d^3)^(1/3)*(a*b^4 + a^4*b*1i + a^5 + b^5*1i - a^2*b^3*6i - 6*a^3*b^2)*
486i)/d^3)*((3^(1/2)*1i)/2 + 1/2)*((4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)/(8*d^3))^(1/3) + (3*b*(a
+ b*tan(c + d*x))^(1/3))/d + log(- (486*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3)*(a^4 + b^4 - 6*a^2*b^2))
/d^4 - (486*b^4*((3^(1/2)*1i)/2 - 1/2)*(a + b*1i)^2*(-((a - b*1i)^4*1i)/d^3)^(1/3)*(a*b^4*1i + a^4*b + a^5*1i
+ b^5 - 6*a^2*b^3 - a^3*b^2*6i))/d^3)*((3^(1/2)*1i)/2 - 1/2)*(-(4*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*
6i)/(8*d^3))^(1/3) - log((486*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3)*(a^4 + b^4 - 6*a^2*b^2))/d^4 - (486
*b^4*((3^(1/2)*1i)/2 + 1/2)*(a + b*1i)^2*(-((a - b*1i)^4*1i)/d^3)^(1/3)*(a*b^4*1i + a^4*b + a^5*1i + b^5 - 6*a
^2*b^3 - a^3*b^2*6i))/d^3)*((3^(1/2)*1i)/2 + 1/2)*(-(4*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)/(8*d^3)
)^(1/3)