3.695 \(\int \frac{1}{(a+b \tan (c+d x))^{5/3}} \, dx\)

Optimal. Leaf size=338 \[ -\frac{3 b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d (a-i b)^{5/3}}+\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d (a+i b)^{5/3}}+\frac{3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{5/3}}-\frac{3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{5/3}}+\frac{i \log (\cos (c+d x))}{4 d (a-i b)^{5/3}}-\frac{i \log (\cos (c+d x))}{4 d (a+i b)^{5/3}}-\frac{x}{4 (a-i b)^{5/3}}-\frac{x}{4 (a+i b)^{5/3}} \]

[Out]

-x/(4*(a - I*b)^(5/3)) - x/(4*(a + I*b)^(5/3)) - ((I/2)*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a
- I*b)^(1/3))/Sqrt[3]])/((a - I*b)^(5/3)*d) + ((I/2)*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I
*b)^(1/3))/Sqrt[3]])/((a + I*b)^(5/3)*d) + ((I/4)*Log[Cos[c + d*x]])/((a - I*b)^(5/3)*d) - ((I/4)*Log[Cos[c +
d*x]])/((a + I*b)^(5/3)*d) + (((3*I)/4)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/((a - I*b)^(5/3)*d)
- (((3*I)/4)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/((a + I*b)^(5/3)*d) - (3*b)/(2*(a^2 + b^2)*d*
(a + b*Tan[c + d*x])^(2/3))

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Rubi [A]  time = 0.361153, antiderivative size = 338, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3483, 3539, 3537, 57, 617, 204, 31} \[ -\frac{3 b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d (a-i b)^{5/3}}+\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d (a+i b)^{5/3}}+\frac{3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{5/3}}-\frac{3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{5/3}}+\frac{i \log (\cos (c+d x))}{4 d (a-i b)^{5/3}}-\frac{i \log (\cos (c+d x))}{4 d (a+i b)^{5/3}}-\frac{x}{4 (a-i b)^{5/3}}-\frac{x}{4 (a+i b)^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-5/3),x]

[Out]

-x/(4*(a - I*b)^(5/3)) - x/(4*(a + I*b)^(5/3)) - ((I/2)*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a
- I*b)^(1/3))/Sqrt[3]])/((a - I*b)^(5/3)*d) + ((I/2)*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I
*b)^(1/3))/Sqrt[3]])/((a + I*b)^(5/3)*d) + ((I/4)*Log[Cos[c + d*x]])/((a - I*b)^(5/3)*d) - ((I/4)*Log[Cos[c +
d*x]])/((a + I*b)^(5/3)*d) + (((3*I)/4)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/((a - I*b)^(5/3)*d)
- (((3*I)/4)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/((a + I*b)^(5/3)*d) - (3*b)/(2*(a^2 + b^2)*d*
(a + b*Tan[c + d*x])^(2/3))

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

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 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 \frac{1}{(a+b \tan (c+d x))^{5/3}} \, dx &=-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac{\int \frac{a-b \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{a^2+b^2}\\ &=-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac{\int \frac{1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{2 (a-i b)}+\frac{\int \frac{1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{2 (a+i b)}\\ &=-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 (i a-b) d}-\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 (i a+b) d}\\ &=-\frac{x}{4 (a-i b)^{5/3}}-\frac{x}{4 (a+i b)^{5/3}}+\frac{i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac{i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{4/3} d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{4/3} d}\\ &=-\frac{x}{4 (a-i b)^{5/3}}-\frac{x}{4 (a+i b)^{5/3}}+\frac{i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac{i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}+\frac{3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac{3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac{(3 i) \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-i b}}\right )}{2 (a-i b)^{5/3} d}-\frac{(3 i) \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+i b}}\right )}{2 (a+i b)^{5/3} d}\\ &=-\frac{x}{4 (a-i b)^{5/3}}-\frac{x}{4 (a+i b)^{5/3}}-\frac{i \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 )}{2 (a-i b)^{5/3} d}+\frac{i \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 )}{2 (a+i b)^{5/3} d}+\frac{i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac{i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}+\frac{3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac{3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}-\frac{3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}\\ \end{align*}

Mathematica [C]  time = 0.147591, size = 106, normalized size = 0.31 \[ \frac{3 i \left ((a+i b) \, _2F_1\left (-\frac{2}{3},1;\frac{1}{3};\frac{a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \, _2F_1\left (-\frac{2}{3},1;\frac{1}{3};\frac{a+b \tan (c+d x)}{a+i b}\right )\right )}{4 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(-5/3),x]

[Out]

(((3*I)/4)*((a + I*b)*Hypergeometric2F1[-2/3, 1, 1/3, (a + b*Tan[c + d*x])/(a - I*b)] - (a - I*b)*Hypergeometr
ic2F1[-2/3, 1, 1/3, (a + b*Tan[c + d*x])/(a + I*b)]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^(2/3))

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Maple [C]  time = 0.017, size = 103, normalized size = 0.3 \begin{align*} -{\frac{3\,b}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) d} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}+{\frac{b}{2\,d \left ({a}^{2}+{b}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{-{{\it \_R}}^{3}+2\,a}{{{\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(1/(a+b*tan(d*x+c))^(5/3),x)

[Out]

-3/2*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(2/3)+1/2/d*b/(a^2+b^2)*sum((-_R^3+2*a)/(_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 \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(-5/3), 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(1/(a+b*tan(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))**(5/3),x)

[Out]

Integral((a + b*tan(c + d*x))**(-5/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^(5/3),x, algorithm="giac")

[Out]

Exception raised: TypeError