3.693 \(\int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=415 \[ \frac {\sqrt {3} b \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {3 b \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}+\frac {3 b \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}+\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {x}{4 \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {x}{4 \left (a+\sqrt {-b^2}\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3485, 712, 57, 617, 204, 31} \[ \frac {\sqrt {3} b \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {3 b \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}+\frac {3 b \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \left (a-\sqrt {-b^2}\right )^{2/3}}+\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {x}{4 \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {x}{4 \left (a+\sqrt {-b^2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-2/3),x]

[Out]

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

Rule 31

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

Rule 57

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 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 617

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 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3485

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.57, size = 313, normalized size = 0.75 \[ \frac {i \left (-\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )+\log \left (\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3}\right )}{(a-i b)^{2/3}}+\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )+\log \left (\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3}\right )}{(a+i b)^{2/3}}+\frac {2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{(a-i b)^{2/3}}-\frac {2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{(a+i b)^{2/3}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(-2/3),x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^(-2/3), x)

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maple [C]  time = 0.19, size = 57, normalized size = 0.14 \[ \frac {b \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c))^(2/3),x)

[Out]

1/2/d*b*sum(1/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(-2/3), x)

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mupad [B]  time = 8.48, size = 1048, normalized size = 2.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x))^(2/3),x)

[Out]

(log(((-1i/(d^3*(a*1i - b)^2))^(4/3)*(a^2*b^5*d*486i - b^7*d*486i + 486*a^3*b^4*d - 486*a*b^6*d + (972*a*b^5*(
a + b*tan(c + d*x))^(1/3))/(-1i/(d^3*(a*1i - b)^2))^(1/3)))/d - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1/(
b^2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3))^(1/3))/2 + log((((7776*a*b^5*(a + b*tan(c + d*x))^(1/3))/d + 7776*a*b^4*
(a^2 + b^2)*(1i/(8*d^3*(a*1i + b)^2))^(1/3))*(1i/(8*d^3*(a*1i + b)^2))^(2/3) - (972*b^5)/d^3)*(1i/(8*d^3*(a*1i
+ b)^2))^(1/3) - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1i/(8*(b^2*d^3 - a^2*d^3 + a*b*d^3*2i)))^(1/3) +
(log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 + ((3^(1/2)*1i - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*
a*b^5*(a + b*tan(c + d*x))^(1/3))/d + 1944*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*(-1i/(d^3*(a*1i - b)^2))^(1/3))*
(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^3*(a*1i - b)^2))^(1/3))/4)*(3^(1/2)*1i - 1)*(1/(b^2*d^3*1i - a^2*d
^3*1i + 2*a*b*d^3))^(1/3))/4 - (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((3^(1/2)*1i + 1)*((972*b^5)/d^
3 - ((3^(1/2)*1i + 1)^2*((7776*a*b^5*(a + b*tan(c + d*x))^(1/3))/d - 1944*a*b^4*(3^(1/2)*1i + 1)*(a^2 + b^2)*(
-1i/(d^3*(a*1i - b)^2))^(1/3))*(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^3*(a*1i - b)^2))^(1/3))/4)*(3^(1/2)
*1i + 1)*(1/(b^2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3))^(1/3))/4 + (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 +
((3^(1/2)*1i - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*a*b^5*(a + b*tan(c + d*x))^(1/3))/d + 3888*a*b^4
*(3^(1/2)*1i - 1)*(a^2 + b^2)*(1i/(8*d^3*(a*1i + b)^2))^(1/3))*(1i/(8*d^3*(a*1i + b)^2))^(2/3))/4)*(1i/(8*d^3*
(a*1i + b)^2))^(1/3))/2)*(3^(1/2)*1i - 1)*(1i/(8*(b^2*d^3 - a^2*d^3 + a*b*d^3*2i)))^(1/3))/2 - (log((486*b^4*(
a + b*tan(c + d*x))^(1/3))/d^4 - ((3^(1/2)*1i + 1)*((972*b^5)/d^3 - ((3^(1/2)*1i + 1)^2*((7776*a*b^5*(a + b*ta
n(c + d*x))^(1/3))/d - 3888*a*b^4*(3^(1/2)*1i + 1)*(a^2 + b^2)*(1i/(8*d^3*(a*1i + b)^2))^(1/3))*(1i/(8*d^3*(a*
1i + b)^2))^(2/3))/4)*(1i/(8*d^3*(a*1i + b)^2))^(1/3))/2)*(3^(1/2)*1i + 1)*(1i/(8*(b^2*d^3 - a^2*d^3 + a*b*d^3
*2i)))^(1/3))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))**(2/3),x)

[Out]

Integral((a + b*tan(c + d*x))**(-2/3), x)

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