3.302 \(\int \frac{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=213 \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \]

[Out]

x/(8*2^(1/3)*a^(4/3)) - ((I/4)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3
))])/(2^(1/3)*a^(4/3)*d) - ((I/8)*Log[Cos[c + d*x]])/(2^(1/3)*a^(4/3)*d) - (((3*I)/8)*Log[2^(1/3)*a^(1/3) - (a
+ I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(4/3)*d) - ((3*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) + ((9*I)/4)/(a*d
*(a + I*a*Tan[c + d*x])^(1/3))

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Rubi [A]  time = 0.202164, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3540, 3526, 3481, 55, 617, 204, 31} \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

x/(8*2^(1/3)*a^(4/3)) - ((I/4)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3
))])/(2^(1/3)*a^(4/3)*d) - ((I/8)*Log[Cos[c + d*x]])/(2^(1/3)*a^(4/3)*d) - (((3*I)/8)*Log[2^(1/3)*a^(1/3) - (a
+ I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(4/3)*d) - ((3*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) + ((9*I)/4)/(a*d
*(a + I*a*Tan[c + d*x])^(1/3))

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rule 3526

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

Rule 3481

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

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 \frac{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{\int \frac{a-2 i a \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.662838, size = 128, normalized size = 0.6 \[ \frac{3 \sec ^2(c+d x) \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+6 i \sin (2 (c+d x))+5 \cos (2 (c+d x))+5\right )}{16 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

(3*Sec[c + d*x]^2*(5 + 5*Cos[2*(c + d*x)] + Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(
c + d*x)))]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + (6*I)*Sin[2*(c + d*x)]))/(16*a*d*(-I + Tan[c + d*x])*(a
+ I*a*Tan[c + d*x])^(1/3))

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Maple [A]  time = 0.02, size = 181, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{8}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){a}^{-{\frac{4}{3}}}}+{\frac{{\frac{i}{16}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ){a}^{-{\frac{4}{3}}}}-{\frac{{\frac{i}{8}}\sqrt{3}{2}^{{\frac{2}{3}}}}{d}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ){a}^{-{\frac{4}{3}}}}+{\frac{{\frac{9\,i}{4}}}{ad}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{{\frac{3\,i}{8}}}{d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x)

[Out]

-1/8*I/d/a^(4/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))+1/16*I/d/a^(4/3)*2^(2/3)*ln((a+I*a*tan(d
*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))-1/8*I/d/a^(4/3)*3^(1/2)*2^(2/3)*arctan(
1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1))+9/4*I/a/d/(a+I*a*tan(d*x+c))^(1/3)-3/8*I/d/(a+I*a*ta
n(d*x+c))^(4/3)

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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(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.80004, size = 1040, normalized size = 4.88 \begin{align*} \frac{{\left (32 \, a^{2} d \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (32 \, a^{3} d^{2} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) + 2^{\frac{2}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (33 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )} +{\left (-16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (16 i \, \sqrt{3} a^{3} d^{2} - 16 \, a^{3} d^{2}\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) +{\left (16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (-16 i \, \sqrt{3} a^{3} d^{2} - 16 \, a^{3} d^{2}\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

1/32*(32*a^2*d*(1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(32*a^3*d^2*(1/128*I/(a^4*d^3))^(2/3) + 2^(1/3
)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + 2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(
33*I*e^(4*I*d*x + 4*I*c) + 30*I*e^(2*I*d*x + 2*I*c) - 3*I)*e^(4/3*I*d*x + 4/3*I*c) + (-16*I*sqrt(3)*a^2*d - 16
*a^2*d)*(1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log((16*I*sqrt(3)*a^3*d^2 - 16*a^3*d^2)*(1/128*I/(a^4*d^
3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + (16*I*sqrt(3)*a^2*d - 16*a^
2*d)*(1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log((-16*I*sqrt(3)*a^3*d^2 - 16*a^3*d^2)*(1/128*I/(a^4*d^3)
)^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral(tan(c + d*x)**2/(a*(I*tan(c + d*x) + 1))**(4/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^2/(I*a*tan(d*x + c) + a)^(4/3), x)