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

Optimal. Leaf size=282 \[ \frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {x}{8 \sqrt [3]{2} a^{4/3}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

[Out]

-1/16*x*2^(2/3)/a^(4/3)+1/16*I*ln(cos(d*x+c))*2^(2/3)/a^(4/3)/d+3/16*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(
1/3))*2^(2/3)/a^(4/3)/d+1/8*I*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)*2
^(2/3)/a^(4/3)/d-39/40*I*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(4/3)+3/5*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(4/3)-5
1/10*I/a/d/(a+I*a*tan(d*x+c))^(1/3)-87/40*I*(a+I*a*tan(d*x+c))^(2/3)/a^2/d

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Rubi [A]  time = 0.45, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3560, 3595, 3592, 3526, 3481, 55, 617, 204, 31} \[ \frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(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) - (((39*I)/40)*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(4/
3)) + (3*Tan[c + d*x]^3)/(5*d*(a + I*a*Tan[c + d*x])^(4/3)) - ((51*I)/10)/(a*d*(a + I*a*Tan[c + d*x])^(1/3)) -
(((87*I)/40)*(a + I*a*Tan[c + d*x])^(2/3))/(a^2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 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 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 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 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 3592

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

Rule 3595

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

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {3 \int \frac {\tan ^2(c+d x) \left (3 a-\frac {4}{3} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{4/3}} \, dx}{5 a}\\ &=-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 \int \frac {\tan (c+d x) \left (\frac {26 i a^2}{3}+\frac {58}{9} a^2 \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{40 a^3}\\ &=-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}+\frac {9 \int \frac {-\frac {58 a^2}{9}+\frac {26}{3} i a^2 \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{40 a^3}\\ &=-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}+\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=-\frac {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}-\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 {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}-\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 {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}-\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 {39 i \tan ^2(c+d x)}{40 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{4/3}}-\frac {51 i}{10 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {87 i (a+i a \tan (c+d x))^{2/3}}{40 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 1.60, size = 145, normalized size = 0.51 \[ -\frac {3 \sec ^2(c+d x) \left (5 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+113 \cos (2 (c+d x))+75 i \tan (c+d x)+59 i \sin (3 (c+d x)) \sec (c+d x)+81\right )}{80 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sec[c + d*x]^2*(81 + 113*Cos[2*(c + d*x)] + 5*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)]) + (59*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (75*I)*Tan[
c + d*x]))/(80*a*d*(-I + Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(1/3))

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fricas [B]  time = 0.48, size = 442, normalized size = 1.57 \[ \frac {2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (-693 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 1275 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 375 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} + 160 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} \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 ) - 80 \, {\left ({\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} \log \left (-16 \, {\left (i \, \sqrt {3} a^{3} d^{2} + 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 ) - 80 \, {\left ({\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} \log \left (-16 \, {\left (-i \, \sqrt {3} a^{3} d^{2} + 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 )}{160 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*(2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-693*I*e^(6*I*d*x + 6*I*c) - 1275*I*e^(4*I*d*x + 4*I*c) -
375*I*e^(2*I*d*x + 2*I*c) + 15*I)*e^(4/3*I*d*x + 4/3*I*c) + 160*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x
+ 4*I*c))*(-1/128*I/(a^4*d^3))^(1/3)*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)) - 80*((-I*sqrt(3)*a^2*d + a^2*d)*e^(6*I*d*x + 6*I*c) + (-I*sqrt(3)*a^2
*d + a^2*d)*e^(4*I*d*x + 4*I*c))*(-1/128*I/(a^4*d^3))^(1/3)*log(-16*(I*sqrt(3)*a^3*d^2 + 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)) - 80*((I*sqrt(3)*a^2*d +
a^2*d)*e^(6*I*d*x + 6*I*c) + (I*sqrt(3)*a^2*d + a^2*d)*e^(4*I*d*x + 4*I*c))*(-1/128*I/(a^4*d^3))^(1/3)*log(-1
6*(-I*sqrt(3)*a^3*d^2 + 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)))/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 227, normalized size = 0.80 \[ \frac {3 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d \,a^{3}}-\frac {3 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{d \,a^{2}}+\frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {i 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {21 i}{4 d a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {3 i}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*I/d/a^3*(a+I*a*tan(d*x+c))^(5/3)-3*I/d/a^2*(a+I*a*tan(d*x+c))^(2/3)+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))-21/4*I/d/a/(a+I*a*tan(d*x+c))^(1/3)+3/8*I/d/(a+I*a*tan(d*x+c))^(4/3)

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maxima [A]  time = 0.49, size = 209, normalized size = 0.74 \[ \frac {i \, {\left (10 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 5 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 10 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 48 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{2} - 240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{3} - \frac {30 \, {\left (14 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\right )}}{80 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*I*(10*sqrt(3)*2^(2/3)*a^(11/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/
3))/a^(1/3)) - 5*2^(2/3)*a^(11/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*ta
n(d*x + c) + a)^(2/3)) + 10*2^(2/3)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 48*(I*a*ta
n(d*x + c) + a)^(5/3)*a^2 - 240*(I*a*tan(d*x + c) + a)^(2/3)*a^3 - 30*(14*(I*a*tan(d*x + c) + a)*a^4 - a^5)/(I
*a*tan(d*x + c) + a)^(4/3))/(a^5*d)

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mupad [B]  time = 4.54, size = 263, normalized size = 0.93 \[ \frac {\frac {3{}\mathrm {i}}{8\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,21{}\mathrm {i}}{4\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,3{}\mathrm {i}}{5\,a^3\,d}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (9\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{4/3}\,d}+\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(4/3),x)

[Out]

(3i/(8*d) - ((a + a*tan(c + d*x)*1i)*21i)/(4*a*d))/(a + a*tan(c + d*x)*1i)^(4/3) - ((a + a*tan(c + d*x)*1i)^(2
/3)*3i)/(a^2*d) + ((a + a*tan(c + d*x)*1i)^(5/3)*3i)/(5*a^3*d) - ((1i/128)^(1/3)*log(9*(a*(tan(c + d*x)*1i + 1
))^(1/3) + 9*(-1)^(1/3)*2^(1/3)*a^(1/3)))/(a^(4/3)*d) + ((1i/128)^(1/3)*log(- (9*(a + a*tan(c + d*x)*1i)^(1/3)
)/(16*a^2*d^2) - (9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i - 1))/(32*a^(5/3)*d^2))*((3^(1/2)*1i)/2 + 1/2))/(a^(4/3)*d)
- ((1i/128)^(1/3)*log((9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i + 1))/(32*a^(5/3)*d^2) - (9*(a + a*tan(c + d*x)*1i)^(
1/3))/(16*a^2*d^2))*((3^(1/2)*1i)/2 - 1/2))/(a^(4/3)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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