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

Optimal. Leaf size=245 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt {3}\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}} \]

[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)

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Rubi [A]  time = 0.44, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3474, 3476, 329, 295, 634, 618, 204, 628, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt {3}\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[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 203

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

Rule 204

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

Rule 295

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)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
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 329

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 618

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 628

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 634

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 3474

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 3476

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*} \int \frac {1}{(b \tan (c+d x))^{4/3}} \, dx &=-\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 {\operatorname {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 \operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {Subst}\left (\int \frac {1}{b^{2/3}+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b d}\\ &=-\frac {\tan ^{-1}\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} \operatorname {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} \operatorname {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 {\operatorname {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 {\operatorname {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 {\tan ^{-1}\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 {\operatorname {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 {\operatorname {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 {\tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\tan ^{-1}\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 {\tan ^{-1}\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*}

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Mathematica [C]  time = 0.06, size = 38, normalized size = 0.16 \[ -\frac {3 \, _2F_1\left (-\frac {1}{6},1;\frac {5}{6};-\tan ^2(c+d x)\right )}{b d \sqrt [3]{b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(-4/3),x]

[Out]

(-3*Hypergeometric2F1[-1/6, 1, 5/6, -Tan[c + d*x]^2])/(b*d*(b*Tan[c + d*x])^(1/3))

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fricas [B]  time = 0.62, size = 701, normalized size = 2.86 \[ \frac {12 \, \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, {\left (b^{2} d \cos \left (d x + c\right )^{2} - b^{2} d\right )} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \arctan \left (2 \, \sqrt {\sqrt {3} b^{7} d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + b^{6} d^{4} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {2}{3}} + \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}}} b d \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} - 2 \, b d \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} - \sqrt {3}\right ) + 4 \, {\left (b^{2} d \cos \left (d x + c\right )^{2} - b^{2} d\right )} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \arctan \left (2 \, \sqrt {-\sqrt {3} b^{7} d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + b^{6} d^{4} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {2}{3}} + \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}}} b d \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} - 2 \, b d \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} + \sqrt {3}\right ) + 8 \, {\left (b^{2} d \cos \left (d x + c\right )^{2} - b^{2} d\right )} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \arctan \left (\sqrt {b^{6} d^{4} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {2}{3}} + \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}}} b d \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} - b d \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}}\right ) + {\left (\sqrt {3} b^{2} d \cos \left (d x + c\right )^{2} - \sqrt {3} b^{2} d\right )} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (\sqrt {3} b^{7} d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + b^{6} d^{4} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {2}{3}} + \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}}\right ) - {\left (\sqrt {3} b^{2} d \cos \left (d x + c\right )^{2} - \sqrt {3} b^{2} d\right )} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {1}{6}} \log \left (-\sqrt {3} b^{7} d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {5}{6}} + b^{6} d^{4} \left (\frac {1}{b^{8} d^{6}}\right )^{\frac {2}{3}} + \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}}\right )}{4 \, {\left (b^{2} d \cos \left (d x + c\right )^{2} - b^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*(b*sin(d*x + c)/cos(d*x + c))^(2/3)*cos(d*x + c)*sin(d*x + c) + 4*(b^2*d*cos(d*x + c)^2 - b^2*d)*(1/(b
^8*d^6))^(1/6)*arctan(2*sqrt(sqrt(3)*b^7*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^8*d^6))^(5/6) + b^6*d^4
*(1/(b^8*d^6))^(2/3) + (b*sin(d*x + c)/cos(d*x + c))^(2/3))*b*d*(1/(b^8*d^6))^(1/6) - 2*b*d*(b*sin(d*x + c)/co
s(d*x + c))^(1/3)*(1/(b^8*d^6))^(1/6) - sqrt(3)) + 4*(b^2*d*cos(d*x + c)^2 - b^2*d)*(1/(b^8*d^6))^(1/6)*arctan
(2*sqrt(-sqrt(3)*b^7*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^8*d^6))^(5/6) + b^6*d^4*(1/(b^8*d^6))^(2/3)
 + (b*sin(d*x + c)/cos(d*x + c))^(2/3))*b*d*(1/(b^8*d^6))^(1/6) - 2*b*d*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1
/(b^8*d^6))^(1/6) + sqrt(3)) + 8*(b^2*d*cos(d*x + c)^2 - b^2*d)*(1/(b^8*d^6))^(1/6)*arctan(sqrt(b^6*d^4*(1/(b^
8*d^6))^(2/3) + (b*sin(d*x + c)/cos(d*x + c))^(2/3))*b*d*(1/(b^8*d^6))^(1/6) - b*d*(b*sin(d*x + c)/cos(d*x + c
))^(1/3)*(1/(b^8*d^6))^(1/6)) + (sqrt(3)*b^2*d*cos(d*x + c)^2 - sqrt(3)*b^2*d)*(1/(b^8*d^6))^(1/6)*log(sqrt(3)
*b^7*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^8*d^6))^(5/6) + b^6*d^4*(1/(b^8*d^6))^(2/3) + (b*sin(d*x +
c)/cos(d*x + c))^(2/3)) - (sqrt(3)*b^2*d*cos(d*x + c)^2 - sqrt(3)*b^2*d)*(1/(b^8*d^6))^(1/6)*log(-sqrt(3)*b^7*
d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^8*d^6))^(5/6) + b^6*d^4*(1/(b^8*d^6))^(2/3) + (b*sin(d*x + c)/co
s(d*x + c))^(2/3)))/(b^2*d*cos(d*x + c)^2 - b^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(d*x + c))^(-4/3), x)

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maple [A]  time = 0.12, size = 227, normalized size = 0.93 \[ -\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}-\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{4 d \,b^{3}}-\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{2 d b \left (b^{2}\right )^{\frac {1}{6}}}-\frac {\arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{d b \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{4 d \,b^{3}}-\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{2 d b \left (b^{2}\right )^{\frac {1}{6}}}-\frac {3}{b d \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/b^3*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
/2/d/b/(b^2)^(1/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)-3^(1/2))-1/d/b/(b^2)^(1/6)*arctan((b*tan(d*x+c))^
(1/3)/(b^2)^(1/6))+1/4/d/b^3*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))-1/2/d/b/(b^2)^(1/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)+3^(1/2))-3/b/d/(b*tan(d*x+c))^(
1/3)

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maxima [A]  time = 0.73, size = 182, normalized size = 0.74 \[ \frac {\frac {\sqrt {3} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {\sqrt {3} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (\frac {\sqrt {3} b^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (-\frac {\sqrt {3} b^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {4 \, \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {12}{\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}}{4 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.55, size = 278, normalized size = 1.13 \[ -\frac {3}{b\,d\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}-\frac {{\left (-1\right )}^{1/6}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{b^{1/3}}\right )\,1{}\mathrm {i}}{b^{4/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6-972\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{4/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6-972\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{4/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6+1944\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{4/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left (972\,b^{12}\,d^6+1944\,{\left (-1\right )}^{1/6}\,b^{35/3}\,d^6\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{4/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(c + d*x))**(-4/3), x)

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