\(\int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx\) [22]

Optimal result

Integrand size = 12, antiderivative size = 245 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\arctan \left (\sqrt {3}+\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3555, 3557, 335, 301, 648, 632, 210, 642, 209} \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\arctan \left (\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt {3}\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}} \]

[In]

Int[(b*Tan[c + d*x])^(-4/3),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 301

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(
2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 +
2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))*Int[1/(r^2 + s^2*x^2), x] +
Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 632

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

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 3555

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

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = -\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {\int (b \tan (c+d x))^{2/3} \, dx}{b^2} \\ & = -\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {x^{2/3}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = -\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {3 \text {Subst}\left (\int \frac {x^4}{b^2+x^6} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b d} \\ & = -\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {-\frac {\sqrt [3]{b}}{2}+\frac {\sqrt {3} x}{2}}{b^{2/3}-\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b^{4/3} d}-\frac {\text {Subst}\left (\int \frac {-\frac {\sqrt [3]{b}}{2}-\frac {\sqrt {3} x}{2}}{b^{2/3}+\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b^{4/3} d}-\frac {\text {Subst}\left (\int \frac {1}{b^{2/3}+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b d} \\ & = -\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {\sqrt {3} \text {Subst}\left (\int \frac {-\sqrt {3} \sqrt [3]{b}+2 x}{b^{2/3}-\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \text {Subst}\left (\int \frac {\sqrt {3} \sqrt [3]{b}+2 x}{b^{2/3}+\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b^{4/3} d}-\frac {\text {Subst}\left (\int \frac {1}{b^{2/3}-\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b d}-\frac {\text {Subst}\left (\int \frac {1}{b^{2/3}+\sqrt {3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b d} \\ & = -\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 \sqrt {3} b^{4/3} d}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 \sqrt {3} b^{4/3} d} \\ & = -\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\arctan \left (\frac {1}{3} \left (3 \sqrt {3}-\frac {6 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )\right )}{2 b^{4/3} d}-\frac {\arctan \left (\frac {1}{3} \left (3 \sqrt {3}+\frac {6 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=\frac {-6-i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-\sqrt [6]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+\sqrt [6]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-(-1)^{5/6} \log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+(-1)^{5/6} \log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{2 b d \sqrt [3]{b \tan (c+d x)}} \]

[In]

Integrate[(b*Tan[c + d*x])^(-4/3),x]

[Out]

(-6 - I*Log[1 - I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + I*Log[1 + I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c
+ d*x]^2)^(1/6) - (-1)^(1/6)*Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + (-1)^(1/6)*Lo
g[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) - (-1)^(5/6)*Log[1 - (-1)^(5/6)*(Tan[c + d*x]^
2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + (-1)^(5/6)*Log[1 + (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6
))/(2*b*d*(b*Tan[c + d*x])^(1/3))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {3 b \left (-\frac {1}{b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 \left (b^{2}\right )^{\frac {1}{6}}}}{b^{2}}\right )}{d}\)	\(215\)
default	\(\frac {3 b \left (-\frac {1}{b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 \left (b^{2}\right )^{\frac {1}{6}}}}{b^{2}}\right )}{d}\)	\(215\)

[In]

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

[Out]

3/d*b*(-1/b^2/(b*tan(d*x+c))^(1/3)-(1/12/b^2*3^(1/2)*(b^2)^(5/6)*ln(-(b*tan(d*x+c))^(2/3)+3^(1/2)*(b^2)^(1/6)*
(b*tan(d*x+c))^(1/3)-(b^2)^(1/3))+1/6/(b^2)^(1/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)-3^(1/2))-1/12/b^2*
3^(1/2)*(b^2)^(5/6)*ln((b*tan(d*x+c))^(2/3)+3^(1/2)*(b^2)^(1/6)*(b*tan(d*x+c))^(1/3)+(b^2)^(1/3))+1/6/(b^2)^(1
/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)+3^(1/2))+1/3/(b^2)^(1/6)*arctan((b*tan(d*x+c))^(1/3)/(b^2)^(1/6)
))/b^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (187) = 374\).

Time = 0.25 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.76 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=-\frac {2 \, b^{2} d \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (b^{7} d^{5} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) - 2 \, b^{2} d \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (-b^{7} d^{5} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) - {\left (\sqrt {-3} b^{2} d - b^{2} d\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} b^{7} d^{5} + b^{7} d^{5}\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) + {\left (\sqrt {-3} b^{2} d - b^{2} d\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} b^{7} d^{5} + b^{7} d^{5}\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) - {\left (\sqrt {-3} b^{2} d + b^{2} d\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} b^{7} d^{5} - b^{7} d^{5}\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) + {\left (\sqrt {-3} b^{2} d + b^{2} d\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} b^{7} d^{5} - b^{7} d^{5}\right )} \left (-\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \tan \left (d x + c\right ) + 12 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}}}{4 \, b^{2} d \tan \left (d x + c\right )} \]

[In]

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

[Out]

-1/4*(2*b^2*d*(-1/(b^8*d^6))^(1/6)*log(b^7*d^5*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x + c))^(1/3))*tan(d*x + c) - 2
*b^2*d*(-1/(b^8*d^6))^(1/6)*log(-b^7*d^5*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x + c))^(1/3))*tan(d*x + c) - (sqrt(-
3)*b^2*d - b^2*d)*(-1/(b^8*d^6))^(1/6)*log(1/2*(sqrt(-3)*b^7*d^5 + b^7*d^5)*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x
+ c))^(1/3))*tan(d*x + c) + (sqrt(-3)*b^2*d - b^2*d)*(-1/(b^8*d^6))^(1/6)*log(-1/2*(sqrt(-3)*b^7*d^5 + b^7*d^5
)*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x + c))^(1/3))*tan(d*x + c) - (sqrt(-3)*b^2*d + b^2*d)*(-1/(b^8*d^6))^(1/6)*
log(1/2*(sqrt(-3)*b^7*d^5 - b^7*d^5)*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x + c))^(1/3))*tan(d*x + c) + (sqrt(-3)*b
^2*d + b^2*d)*(-1/(b^8*d^6))^(1/6)*log(-1/2*(sqrt(-3)*b^7*d^5 - b^7*d^5)*(-1/(b^8*d^6))^(5/6) + (b*tan(d*x + c
))^(1/3))*tan(d*x + c) + 12*(b*tan(d*x + c))^(2/3))/(b^2*d*tan(d*x + c))

Sympy [F]

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

[In]

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

[Out]

Integral((b*tan(c + d*x))**(-4/3), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=\frac {\frac {\sqrt {3} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {\sqrt {3} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (\frac {\sqrt {3} b^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (-\frac {\sqrt {3} b^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {4 \, \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {12}{\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}}{4 \, b d} \]

[In]

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

[Out]

1/4*(sqrt(3)*log(sqrt(3)*(b*tan(d*x + c))^(1/3)*b^(1/3) + (b*tan(d*x + c))^(2/3) + b^(2/3))/b^(1/3) - sqrt(3)*
log(-sqrt(3)*(b*tan(d*x + c))^(1/3)*b^(1/3) + (b*tan(d*x + c))^(2/3) + b^(2/3))/b^(1/3) - 2*arctan((sqrt(3)*b^
(1/3) + 2*(b*tan(d*x + c))^(1/3))/b^(1/3))/b^(1/3) - 2*arctan(-(sqrt(3)*b^(1/3) - 2*(b*tan(d*x + c))^(1/3))/b^
(1/3))/b^(1/3) - 4*arctan((b*tan(d*x + c))^(1/3)/b^(1/3))/b^(1/3) - 12/(b*tan(d*x + c))^(1/3))/(b*d)

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=\frac {1}{4} \, b {\left (\frac {\sqrt {3} {\left | b \right |}^{\frac {5}{3}} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{b^{4} d} - \frac {\sqrt {3} {\left | b \right |}^{\frac {5}{3}} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{b^{4} d} - \frac {2 \, {\left | b \right |}^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b^{4} d} - \frac {2 \, {\left | b \right |}^{\frac {5}{3}} \arctan \left (-\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b^{4} d} - \frac {4 \, {\left | b \right |}^{\frac {5}{3}} \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b^{4} d} - \frac {12}{\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{2} d}\right )} \]

[In]

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

[Out]

1/4*b*(sqrt(3)*abs(b)^(5/3)*log(sqrt(3)*(b*tan(d*x + c))^(1/3)*abs(b)^(1/3) + (b*tan(d*x + c))^(2/3) + abs(b)^
(2/3))/(b^4*d) - sqrt(3)*abs(b)^(5/3)*log(-sqrt(3)*(b*tan(d*x + c))^(1/3)*abs(b)^(1/3) + (b*tan(d*x + c))^(2/3
) + abs(b)^(2/3))/(b^4*d) - 2*abs(b)^(5/3)*arctan((sqrt(3)*abs(b)^(1/3) + 2*(b*tan(d*x + c))^(1/3))/abs(b)^(1/
3))/(b^4*d) - 2*abs(b)^(5/3)*arctan(-(sqrt(3)*abs(b)^(1/3) - 2*(b*tan(d*x + c))^(1/3))/abs(b)^(1/3))/(b^4*d) -
4*abs(b)^(5/3)*arctan((b*tan(d*x + c))^(1/3)/abs(b)^(1/3))/(b^4*d) - 12/((b*tan(d*x + c))^(1/3)*b^2*d))

Mupad [B] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx=-\frac {3}{b\,d\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}-\frac {{\left (-1\right )}^{1/6}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{b^{1/3}}\right )\,1{}\mathrm {i}}{b^{4/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6-972\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{4/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6-972\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{4/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6+1944\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{4/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6+1944\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{4/3}\,d} \]

[In]

int(1/(b*tan(c + d*x))^(4/3),x)

[Out]

((-1)^(1/6)*log(972*b^12*d^6 + 1944*(-1)^(1/6)*b^(35/3)*d^6*((3^(1/2)*1i)/4 - 1/4)*(b*tan(c + d*x))^(1/3))*((3
^(1/2)*1i)/4 - 1/4))/(b^(4/3)*d) - ((-1)^(1/6)*atan(((-1)^(2/3)*(b*tan(c + d*x))^(1/3))/b^(1/3))*1i)/(b^(4/3)*
d) - ((-1)^(1/6)*log(972*b^12*d^6 - 972*(-1)^(1/6)*b^(35/3)*d^6*((3^(1/2)*1i)/2 - 1/2)*(b*tan(c + d*x))^(1/3))
*((3^(1/2)*1i)/2 - 1/2))/(2*b^(4/3)*d) - ((-1)^(1/6)*log(972*b^12*d^6 - 972*(-1)^(1/6)*b^(35/3)*d^6*((3^(1/2)*
1i)/2 + 1/2)*(b*tan(c + d*x))^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(2*b^(4/3)*d) - 3/(b*d*(b*tan(c + d*x))^(1/3)) +
((-1)^(1/6)*log(972*b^12*d^6 + 1944*(-1)^(1/6)*b^(35/3)*d^6*((3^(1/2)*1i)/4 + 1/4)*(b*tan(c + d*x))^(1/3))*((3
^(1/2)*1i)/4 + 1/4))/(b^(4/3)*d)