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

Optimal. Leaf size=237 \[ -\frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i x}{8 \sqrt [3]{2} a^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}} \]

[Out]

((-I/8)*x)/(2^(1/3)*a^(4/3)) - (Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/
3))])/(4*2^(1/3)*a^(4/3)*d) - Log[Cos[c + d*x]]/(8*2^(1/3)*a^(4/3)*d) - (3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[
c + d*x])^(1/3)])/(8*2^(1/3)*a^(4/3)*d) + 15/(8*d*(a + I*a*Tan[c + d*x])^(4/3)) + (3*Tan[c + d*x]^2)/(2*d*(a +
I*a*Tan[c + d*x])^(4/3)) - 27/(4*a*d*(a + I*a*Tan[c + d*x])^(1/3))

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Rubi [A]  time = 0.307354, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3560, 3590, 3526, 3481, 55, 617, 204, 31} \[ -\frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i x}{8 \sqrt [3]{2} a^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*x)/(2^(1/3)*a^(4/3)) - (Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/
3))])/(4*2^(1/3)*a^(4/3)*d) - Log[Cos[c + d*x]]/(8*2^(1/3)*a^(4/3)*d) - (3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[
c + d*x])^(1/3)])/(8*2^(1/3)*a^(4/3)*d) + 15/(8*d*(a + I*a*Tan[c + d*x])^(4/3)) + (3*Tan[c + d*x]^2)/(2*d*(a +
I*a*Tan[c + d*x])^(4/3)) - 27/(4*a*d*(a + I*a*Tan[c + d*x])^(1/3))

Rule 3560

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

Rule 3590

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((A*b - a*B)*(a*c + b*d)*(a + b*Tan[e + f*x])^m)/(2*a^2*f*m), x] + Dist[
1/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B*d + 2*a*B*d*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[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 ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{3 \int \frac{\tan (c+d x) \left (2 a-\frac{4}{3} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{4/3}} \, dx}{2 a}\\ &=\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}+\frac{(3 i) \int \frac{\frac{10 a^2}{3}-\frac{8}{3} i a^2 \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{4 a^3}\\ &=\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{i \int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{\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{i x}{8 \sqrt [3]{2} a^{4/3}}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3 \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 \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{i x}{8 \sqrt [3]{2} a^{4/3}}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 \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{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3 \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{i x}{8 \sqrt [3]{2} a^{4/3}}-\frac{\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{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 \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{15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac{27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.845419, size = 130, normalized size = 0.55 \[ \frac{3 \sec ^2(c+d x) \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\sin (2 (c+d x))-i \cos (2 (c+d x)))-18 \sin (2 (c+d x))+17 i \cos (2 (c+d x))+9 i\right )}{16 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[c + d*x]^2*(9*I + (17*I)*Cos[2*(c + d*x)] - 18*Sin[2*(c + d*x)] + Hypergeometric2F1[2/3, 1, 5/3, E^((2*
I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*((-I)*Cos[2*(c + d*x)] + 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 = 0.022, size = 198, normalized size = 0.8 \begin{align*} -{\frac{3}{2\,{a}^{2}d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}-{\frac{{2}^{{\frac{2}{3}}}}{8\,d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){a}^{-{\frac{4}{3}}}}+{\frac{{2}^{{\frac{2}{3}}}}{16\,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{\sqrt{3}{2}^{{\frac{2}{3}}}}{8\,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{15}{4\,ad}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{3}{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)^3/(a+I*a*tan(d*x+c))^(4/3),x)

[Out]

-3/2/d/a^2*(a+I*a*tan(d*x+c))^(2/3)-1/8/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/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/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))-15/4/a/d/(a+I*a*tan(d*x
+c))^(1/3)+3/8/d/(a+I*a*tan(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)^3/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.79109, size = 1107, normalized size = 4.67 \begin{align*} \frac{{\left (32 \, \left (\frac{1}{2}\right )^{\frac{1}{3}} a^{2} d \left (-\frac{1}{a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac{1}{2}\right )^{\frac{2}{3}} a^{3} d^{2} \left (-\frac{1}{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 (\frac{1}{2}\right )^{\frac{1}{3}}{\left (16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (-\frac{1}{a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{1}{16} \, \left (\frac{1}{2}\right )^{\frac{2}{3}}{\left (16 i \, \sqrt{3} a^{3} d^{2} + 16 \, a^{3} d^{2}\right )} \left (-\frac{1}{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 (\frac{1}{2}\right )^{\frac{1}{3}}{\left (-16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (-\frac{1}{a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{1}{16} \, \left (\frac{1}{2}\right )^{\frac{2}{3}}{\left (-16 i \, \sqrt{3} a^{3} d^{2} + 16 \, a^{3} d^{2}\right )} \left (-\frac{1}{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 ) - 12 \cdot 2^{\frac{2}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (35 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{128 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(32*(1/2)^(1/3)*a^2*d*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-2*(1/2)^(2/3)*a^3*d^2*(-1/(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)) + (1/2)^(1/3)*(16*I*sqrt(3)*a^2*
d - 16*a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(1/16*(1/2)^(2/3)*(16*I*sqrt(3)*a^3*d^2 + 16*a^3*d^2
)*(-1/(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)) + (1/2)^(1/3)*(-
16*I*sqrt(3)*a^2*d - 16*a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(1/16*(1/2)^(2/3)*(-16*I*sqrt(3)*a^
3*d^2 + 16*a^3*d^2)*(-1/(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)
) - 12*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(35*e^(4*I*d*x + 4*I*c) + 18*e^(2*I*d*x + 2*I*c) - 1)*e^(4/
3*I*d*x + 4/3*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 ^{3}{\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)**3/(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral(tan(c + d*x)**3/(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 )^{3}}{{\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)^3/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

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