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[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

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Rubi [A]  time = 0.417682, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, 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 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]

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 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 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 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 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 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 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])

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*}

Mathematica [C]  time = 0.0269183, 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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Maple [A]  time = 0.086, size = 215, normalized size = 0.9 \begin{align*} 3\,{\frac{b\sqrt [3]{b\tan \left ( dx+c \right ) }}{d}}-{\frac{b\sqrt{3}}{4\,d}\sqrt [6]{{b}^{2}}\ln \left ( \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt{3}\sqrt [6]{{b}^{2}}\sqrt [3]{b\tan \left ( dx+c \right ) }+\sqrt [3]{{b}^{2}} \right ) }-{\frac{b}{2\,d}\sqrt [6]{{b}^{2}}\arctan \left ( 2\,{\frac{\sqrt [3]{b\tan \left ( dx+c \right ) }}{\sqrt [6]{{b}^{2}}}}+\sqrt{3} \right ) }+{\frac{b\sqrt{3}}{4\,d}\sqrt [6]{{b}^{2}}\ln \left ( \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-\sqrt{3}\sqrt [6]{{b}^{2}}\sqrt [3]{b\tan \left ( dx+c \right ) }+\sqrt [3]{{b}^{2}} \right ) }-{\frac{b}{2\,d}\sqrt [6]{{b}^{2}}\arctan \left ( 2\,{\frac{\sqrt [3]{b\tan \left ( dx+c \right ) }}{\sqrt [6]{{b}^{2}}}}-\sqrt{3} \right ) }-{\frac{b}{d}\sqrt [6]{{b}^{2}}\arctan \left ({\sqrt [3]{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [6]{{b}^{2}}}}} \right ) } \end{align*}

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))+1/4/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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.45459, size = 1481, normalized size = 6.09 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan{\left (c + d x \right )}\right )^{\frac{4}{3}}\, dx \end{align*}

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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Giac [A]  time = 1.41056, size = 282, normalized size = 1.16 \begin{align*} -\frac{1}{4} \, b{\left (\frac{\sqrt{3}{\left | b \right |}^{\frac{1}{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 )}{d} - \frac{\sqrt{3}{\left | b \right |}^{\frac{1}{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 )}{d} + \frac{2 \,{\left | b \right |}^{\frac{1}{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 )}{d} + \frac{2 \,{\left | b \right |}^{\frac{1}{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 )}{d} + \frac{4 \,{\left | b \right |}^{\frac{1}{3}} \arctan \left (\frac{\left (b \tan \left (d x + c\right )\right )^{\frac{1}{3}}}{{\left | b \right |}^{\frac{1}{3}}}\right )}{d} - \frac{12 \, \left (b \tan \left (d x + c\right )\right )^{\frac{1}{3}}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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