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

Optimal. Leaf size=243 \[ -\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{d}+\frac {b^{4/3} \tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 d}-\frac {b^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt {3}\right )}{2 d}+\frac {\sqrt {3} b^{4/3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 d}-\frac {\sqrt {3} b^{4/3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 d}+\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 243, 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 = {3473, 3476, 329, 209, 634, 618, 204, 628, 203} \[ -\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{d}+\frac {b^{4/3} \tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 d}-\frac {b^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt {3}\right )}{2 d}+\frac {\sqrt {3} b^{4/3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 d}-\frac {\sqrt {3} b^{4/3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 d}+\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(4/3),x]

[Out]

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

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 209

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x] +
Int[(r + s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && 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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 (b \tan (c+d x))^{4/3} \, dx &=\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}-b^2 \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx\\ &=\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x^{2/3} \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^6} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{d}\\ &=\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}-\frac {b^{4/3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{b}-\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 )}{d}-\frac {b^{4/3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{b}+\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 )}{d}-\frac {b^{5/3} \operatorname {Subst}\left (\int \frac {1}{b^{2/3}+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{d}\\ &=-\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{d}+\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}+\frac {\left (\sqrt {3} b^{4/3}\right ) \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 d}-\frac {\left (\sqrt {3} b^{4/3}\right ) \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 d}-\frac {b^{5/3} \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 d}-\frac {b^{5/3} \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 d}\\ &=-\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{d}+\frac {\sqrt {3} b^{4/3} \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 d}-\frac {\sqrt {3} b^{4/3} \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 d}+\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}-\frac {b^{4/3} \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} d}+\frac {b^{4/3} \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} d}\\ &=-\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{d}+\frac {b^{4/3} \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 d}-\frac {b^{4/3} \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 d}+\frac {\sqrt {3} b^{4/3} \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 d}-\frac {\sqrt {3} b^{4/3} \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 d}+\frac {3 b \sqrt [3]{b \tan (c+d x)}}{d}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 38, normalized size = 0.16 \[ -\frac {3 b \sqrt [3]{b \tan (c+d x)} \left (\, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-\tan ^2(c+d x)\right )-1\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(4/3),x]

[Out]

(-3*b*(-1 + Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2])*(b*Tan[c + d*x])^(1/3))/d

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fricas [B]  time = 1.20, size = 588, normalized size = 2.42 \[ -\frac {\sqrt {3} \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{6}} d \log \left (\sqrt {3} \left (\frac {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}} + b^{2} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} + \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{3}} d^{2}\right ) - \sqrt {3} \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{6}} d \log \left (-\sqrt {3} \left (\frac {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}} + b^{2} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} + \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{3}} d^{2}\right ) - 4 \, \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{6}} d \arctan \left (-\frac {\sqrt {3} b^{8} + 2 \, \left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} b d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} - 2 \, \sqrt {\sqrt {3} \left (\frac {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}} + b^{2} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} + \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{3}} d^{2}} \left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} d^{5}}{b^{8}}\right ) - 4 \, \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} b^{8} - 2 \, \left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} b d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} + 2 \, \sqrt {-\sqrt {3} \left (\frac {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}} + b^{2} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} + \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{3}} d^{2}} \left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} d^{5}}{b^{8}}\right ) - 8 \, \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{6}} d \arctan \left (-\frac {\left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} b d^{5} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}} - \sqrt {b^{2} \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {2}{3}} + \left (\frac {b^{8}}{d^{6}}\right )^{\frac {1}{3}} d^{2}} \left (\frac {b^{8}}{d^{6}}\right )^{\frac {5}{6}} d^{5}}{b^{8}}\right ) - 12 \, b \left (\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{\frac {1}{3}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.14, size = 215, normalized size = 0.88 \[ \frac {3 b \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {b \sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}-\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{4 d}-\frac {b \left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{2 d}-\frac {b \left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{d}-\frac {b \sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{4 d}-\frac {b \left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 185, normalized size = 0.76 \[ -\frac {\sqrt {3} b^{\frac {7}{3}} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right ) - \sqrt {3} b^{\frac {7}{3}} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right ) + 2 \, b^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} b^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right ) + 2 \, b^{\frac {7}{3}} \arctan \left (-\frac {\sqrt {3} b^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right ) + 4 \, b^{\frac {7}{3}} \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right ) - 12 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{2}}{4 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(3)*b^(7/3)*log(sqrt(3)*(b*tan(d*x + c))^(1/3)*b^(1/3) + (b*tan(d*x + c))^(2/3) + b^(2/3)) - sqrt(3)
*b^(7/3)*log(-sqrt(3)*(b*tan(d*x + c))^(1/3)*b^(1/3) + (b*tan(d*x + c))^(2/3) + b^(2/3)) + 2*b^(7/3)*arctan((s
qrt(3)*b^(1/3) + 2*(b*tan(d*x + c))^(1/3))/b^(1/3)) + 2*b^(7/3)*arctan(-(sqrt(3)*b^(1/3) - 2*(b*tan(d*x + c))^
(1/3))/b^(1/3)) + 4*b^(7/3)*arctan((b*tan(d*x + c))^(1/3)/b^(1/3)) - 12*(b*tan(d*x + c))^(1/3)*b^2)/(b*d)

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mupad [B]  time = 3.08, size = 247, normalized size = 1.02 \[ \frac {3\,b\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d}-\frac {{\left (-1\right )}^{1/6}\,b^{4/3}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{5/6}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\,1{}\mathrm {i}}{b^{1/3}}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/6}\,b^{4/3}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}+2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}-\frac {{\left (-1\right )}^{1/6}\,b^{4/3}\,\ln \left (2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}-{\left (-1\right )}^{1/6}\,b^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {{\left (-1\right )}^{1/6}\,b^{4/3}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}-2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d}+\frac {{\left (-1\right )}^{1/6}\,b^{4/3}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}+2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}-{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x))^(4/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(c + d*x))**(4/3), x)

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