3.3.90 \(\int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [290]

Optimal. Leaf size=282 \[ -\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {i \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d} \]

[Out]

-1/8*x*2^(2/3)/a^(1/3)+1/8*I*ln(cos(d*x+c))*2^(2/3)/a^(1/3)/d+3/8*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3
))*2^(2/3)/a^(1/3)/d+1/4*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^(1/3)/d-15/8*I*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/3)+3/8*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/3)+45/8*
I*(a+I*a*tan(d*x+c))^(2/3)/a/d-39/20*I*(a+I*a*tan(d*x+c))^(5/3)/a^2/d

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Rubi [A]
time = 0.32, 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 = {3641, 3676, 3673, 3608, 3562, 57, 631, 210, 31} \begin {gather*} -\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}+\frac {i \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

-1/4*x/(2^(1/3)*a^(1/3)) + ((I/2)*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^(1/3)*d) + ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) + (((3*I)/4)*Log[2^(1/3)*a^(1/3) -
(a + I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) - (((15*I)/8)*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(1
/3)) + (3*Tan[c + d*x]^3)/(8*d*(a + I*a*Tan[c + d*x])^(1/3)) + (((45*I)/8)*(a + I*a*Tan[c + d*x])^(2/3))/(a*d)
- (((39*I)/20)*(a + I*a*Tan[c + d*x])^(5/3))/(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 57

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

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 3562

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 3608

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

Rule 3641

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 3673

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 3676

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)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \int \frac {\tan ^2(c+d x) \left (3 a-\frac {1}{3} i a \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{8 a}\\ &=-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {9 \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} \left (\frac {20 i a^2}{3}+\frac {52}{9} a^2 \tan (c+d x)\right ) \, dx}{16 a^3}\\ &=-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}+\frac {9 \int (a+i a \tan (c+d x))^{2/3} \left (-\frac {52 a^2}{9}+\frac {20}{3} i a^2 \tan (c+d x)\right ) \, dx}{16 a^3}\\ &=-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}+\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}-\frac {i \text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}+\frac {(3 i) \text {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 )}{4 d}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}-\frac {(3 i) \text {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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.42, size = 125, normalized size = 0.44 \begin {gather*} \frac {3 i \sec ^3(c+d x) (37 \cos (c+d x)+12 \cos (3 (c+d x))+2 i \sin (c+d x)+7 i \sin (3 (c+d x)))+15 \, _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 ) (-i+\tan (c+d x))}{40 d \sqrt [3]{a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

((3*I)*Sec[c + d*x]^3*(37*Cos[c + d*x] + 12*Cos[3*(c + d*x)] + (2*I)*Sin[c + d*x] + (7*I)*Sin[3*(c + d*x)]) +
15*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(-I + Tan[c + d*x]))/(40*d*(a
+ I*a*Tan[c + d*x])^(1/3))

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Maple [A]
time = 0.12, size = 211, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3}}{2}+\frac {a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{3}}\)	\(211\)
default	\(\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3}}{2}+\frac {a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{3}}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)

[Out]

3*I/d/a^3*(1/8*(a+I*a*tan(d*x+c))^(8/3)-2/5*a*(a+I*a*tan(d*x+c))^(5/3)+a^2*(a+I*a*tan(d*x+c))^(2/3)+1/2*(1/6*2
^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(2/3)/a^(1/3)*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))+1/6*3^(1/2)*2^(2/3)/a^(1/3)*arctan(1/3*3^(1/2)*(2^(2/
3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))*a^3+1/2*a^3/(a+I*a*tan(d*x+c))^(1/3))

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Maxima [A]
time = 0.54, size = 208, normalized size = 0.74 \begin {gather*} \frac {i \, {\left (10 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {14}{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 {14}{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 {14}{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 ) + 15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} a^{2} - 48 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{3} + 120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{4} + \frac {60 \, a^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{40 \, a^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*I*(10*sqrt(3)*2^(2/3)*a^(14/3)*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^(14/3)*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^(14/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 15*(I*a*ta
n(d*x + c) + a)^(8/3)*a^2 - 48*(I*a*tan(d*x + c) + a)^(5/3)*a^3 + 120*(I*a*tan(d*x + c) + a)^(2/3)*a^4 + 60*a^
5/(I*a*tan(d*x + c) + a)^(1/3))/(a^5*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (201) = 402\).
time = 1.16, size = 479, normalized size = 1.70 \begin {gather*} -\frac {3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (-19 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 39 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 35 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 20 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} \log \left (8 \, a d^{2} \left (-\frac {i}{16 \, a 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 ) + 10 \, {\left ({\left (-i \, \sqrt {3} a d + a d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (-i \, \sqrt {3} a d + a d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-i \, \sqrt {3} a d + a d\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} \log \left (-4 \, {\left (i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a 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 ) + 10 \, {\left ({\left (i \, \sqrt {3} a d + a d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (i \, \sqrt {3} a d + a d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (i \, \sqrt {3} a d + a d\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} \log \left (-4 \, {\left (-i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a 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 )}{20 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

-1/20*(3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-19*I*e^(6*I*d*x + 6*I*c) - 39*I*e^(4*I*d*x + 4*I*c) - 3
5*I*e^(2*I*d*x + 2*I*c) - 5*I)*e^(4/3*I*d*x + 4/3*I*c) - 20*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*
c) + a*d*e^(2*I*d*x + 2*I*c))*(-1/16*I/(a*d^3))^(1/3)*log(8*a*d^2*(-1/16*I/(a*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)) + 10*((-I*sqrt(3)*a*d + a*d)*e^(6*I*d*x + 6*I*c) + 2*(-I*sq
rt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (-I*sqrt(3)*a*d + a*d)*e^(2*I*d*x + 2*I*c))*(-1/16*I/(a*d^3))^(1/3)*log
(-4*(I*sqrt(3)*a*d^2 + a*d^2)*(-1/16*I/(a*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)) + 10*((I*sqrt(3)*a*d + a*d)*e^(6*I*d*x + 6*I*c) + 2*(I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) +
(I*sqrt(3)*a*d + a*d)*e^(2*I*d*x + 2*I*c))*(-1/16*I/(a*d^3))^(1/3)*log(-4*(-I*sqrt(3)*a*d^2 + a*d^2)*(-1/16*I
/(a*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*d*e^(6*I*d*x + 6*I*
c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^4/(I*a*tan(d*x + c) + a)^(1/3), x)

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Mupad [B]
time = 4.63, size = 266, normalized size = 0.94 \begin {gather*} \frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,6{}\mathrm {i}}{5\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{8/3}\,3{}\mathrm {i}}{8\,a^3\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\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}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\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}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3i/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + ((a + a*tan(c + d*x)*1i)^(2/3)*3i)/(a*d) - ((a + a*tan(c + d*x)*1i)^(
5/3)*6i)/(5*a^2*d) + ((a + a*tan(c + d*x)*1i)^(8/3)*3i)/(8*a^3*d) + ((1i/16)^(1/3)*log((a*(tan(c + d*x)*1i + 1
))^(1/3) - (-1)^(1/3)*2^(1/3)*(-a)^(1/3)))/((-a)^(1/3)*d) - ((1i/16)^(1/3)*log((9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3
)*(3^(1/2)*1i - 1))/(8*d^2) - (9*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2))*((3^(1/2)*1i)/2 + 1/2))/((-a)^(1/3)*d
) + ((1i/16)^(1/3)*log(- (9*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2) - (9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3)*(3^(1/2)
*1i + 1))/(8*d^2))*((3^(1/2)*1i)/2 - 1/2))/((-a)^(1/3)*d)

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