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

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

[Out]

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

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Rubi [A]  time = 0.381613, antiderivative size = 415, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3485, 712, 50, 55, 617, 204, 31} \[ -\frac{\sqrt{3} b \left (a-\sqrt{-b^2}\right )^{2/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d}+\frac{\sqrt{3} b \left (a+\sqrt{-b^2}\right )^{2/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d}-\frac{3 b \left (a-\sqrt{-b^2}\right )^{2/3} \log \left (\sqrt [3]{a-\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d}+\frac{3 b \left (a+\sqrt{-b^2}\right )^{2/3} \log \left (\sqrt [3]{a+\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d}-\frac{b \left (a-\sqrt{-b^2}\right )^{2/3} \log (\cos (c+d x))}{4 \sqrt{-b^2} d}+\frac{b \left (a+\sqrt{-b^2}\right )^{2/3} \log (\cos (c+d x))}{4 \sqrt{-b^2} d}-\frac{1}{4} x \left (a-\sqrt{-b^2}\right )^{2/3}-\frac{1}{4} x \left (a+\sqrt{-b^2}\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(2/3),x]

[Out]

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

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]

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 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 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 31

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

Rubi steps

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

Mathematica [C]  time = 0.298816, size = 224, normalized size = 0.54 \[ \frac{\frac{(b+i a) \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )-\log (\tan (c+d x)+i)\right )}{\sqrt [3]{a-i b}}+\frac{(b-i a) \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )-\log (-\tan (c+d x)+i)\right )}{\sqrt [3]{a+i b}}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.014, size = 60, normalized size = 0.1 \begin{align*}{\frac{b}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{{{\it \_R}}^{4}}{{{\it \_R}}^{5}-{{\it \_R}}^{2}a}\ln \left ( \sqrt [3]{a+b\tan \left ( dx+c \right ) }-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*b*sum(_R^4/(_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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{2}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 4.55609, size = 370, normalized size = 0.89 \begin{align*} -\frac{1}{24} \,{\left ({\left (i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, a^{2} - 432 \, a b - 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (d\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, a^{2} - 432 \, a b - 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (d\right ) +{\left (i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, a^{2} - 432 \, a b + 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (d\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, a^{2} - 432 \, a b + 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (d\right ) - 2 \, \left (\frac{-216 i \, a^{2} - 432 \, a b + 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (i \, a d + b d -{\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right ) - 2 \, \left (\frac{216 i \, a^{2} - 432 \, a b - 216 i \, b^{2}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (-i \, a d + b d -{\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((I*sqrt(3) + 1)*((216*I*a^2 - 432*a*b - 216*I*b^2)/(b^3*d^3))^(1/3)*log(d) + (-I*sqrt(3) + 1)*((216*I*a
^2 - 432*a*b - 216*I*b^2)/(b^3*d^3))^(1/3)*log(d) + (I*sqrt(3) + 1)*((-216*I*a^2 - 432*a*b + 216*I*b^2)/(b^3*d
^3))^(1/3)*log(d) + (-I*sqrt(3) + 1)*((-216*I*a^2 - 432*a*b + 216*I*b^2)/(b^3*d^3))^(1/3)*log(d) - 2*((-216*I*
a^2 - 432*a*b + 216*I*b^2)/(b^3*d^3))^(1/3)*log(I*a*d + b*d - (-I*a^2 - 2*a*b + I*b^2)^(1/3)*(b*tan(d*x + c) +
a)^(1/3)*d) - 2*((216*I*a^2 - 432*a*b - 216*I*b^2)/(b^3*d^3))^(1/3)*log(-I*a*d + b*d - (I*a^2 - 2*a*b - I*b^2
)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d))*b