\(\int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) [200]

Optimal result

Integrand size = 34, antiderivative size = 289 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {\sqrt {3} a^{2/3} A \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}-\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}+\frac {3 a^{2/3} A \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}-\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d} \]

[Out]

-1/4*a^(2/3)*(I*A+B)*x*2^(2/3)-1/4*a^(2/3)*(A-I*B)*ln(cos(d*x+c))*2^(2/3)/d-1/2*a^(2/3)*A*ln(tan(d*x+c))/d+3/2
*a^(2/3)*A*ln(a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))/d-3/4*a^(2/3)*(A-I*B)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1
/3))*2^(2/3)/d+a^(2/3)*A*arctan(1/3*(a^(1/3)+2*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)/d-1/2*a^(2/3
)*(A-I*B)*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)/d

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3681, 3562, 57, 631, 210, 31, 3680} \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac {\sqrt {3} a^{2/3} A \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}-\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} x (B+i A)}{2 \sqrt [3]{2}}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}+\frac {3 a^{2/3} A \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d} \]

[In]

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

[Out]

-1/2*(a^(2/3)*(I*A + B)*x)/2^(1/3) + (Sqrt[3]*a^(2/3)*A*ArcTan[(a^(1/3) + 2*(a + I*a*Tan[c + d*x])^(1/3))/(Sqr
t[3]*a^(1/3))])/d - (Sqrt[3]*a^(2/3)*(A - I*B)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3
]*a^(1/3))])/(2^(1/3)*d) - (a^(2/3)*(A - I*B)*Log[Cos[c + d*x]])/(2*2^(1/3)*d) - (a^(2/3)*A*Log[Tan[c + d*x]])
/(2*d) + (3*a^(2/3)*A*Log[a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*d) - (3*a^(2/3)*(A - I*B)*Log[2^(1/3)*a^
(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*2^(1/3)*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 3680

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

Rule 3681

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

Rubi steps \begin{align*} \text {integral}& = \frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{2/3} \, dx}{a}+(i A+B) \int (a+i a \tan (c+d x))^{2/3} \, dx \\ & = \frac {(a A) \text {Subst}\left (\int \frac {1}{x \sqrt [3]{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(a (A-i B)) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}-\frac {\left (3 a^{2/3} A\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}+\frac {(3 a A) \text {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}+\frac {\left (3 a^{2/3} (A-i B)\right ) \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 )}{2 \sqrt [3]{2} d}-\frac {(3 a (A-i B)) \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 )}{2 d} \\ & = -\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}+\frac {3 a^{2/3} A \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}-\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {\left (3 a^{2/3} A\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{d}+\frac {\left (3 a^{2/3} (A-i B)\right ) \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 )}{\sqrt [3]{2} d} \\ & = -\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {\sqrt {3} a^{2/3} A \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{d}-\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{2} d}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}+\frac {3 a^{2/3} A \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}-\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.44 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {3 \left (\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left ((A-i B) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 A \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {2 e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{2 \sqrt [3]{2} d} \]

[In]

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

[Out]

(3*((a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(2/3)*((A - I*B)*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I
)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))] - 2*A*Hypergeometric2F1[2/3, 1, 5/3, (2*E^((2*I)*(c + d*x)))/(1 + E^((
2*I)*(c + d*x)))]))/(2*2^(1/3)*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {3 a \left (\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (i B -A \right )+\left (\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}}}-\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{6 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {1}{3}}}\right ) A \right )}{d}\)	\(248\)
default	\(\frac {3 a \left (\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (i B -A \right )+\left (\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}}}-\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{6 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {1}{3}}}\right ) A \right )}{d}\)	\(248\)

[In]

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

3/d*a*((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+I*B)+(1/3/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-a^(1/3
))-1/6/a^(1/3)*ln((a+I*a*tan(d*x+c))^(2/3)+a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+a^(2/3))+1/3*3^(1/2)/a^(1/3)*arcta
n(1/3*3^(1/2)*(2/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))*A)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (211) = 422\).

Time = 0.27 (sec) , antiderivative size = 711, normalized size of antiderivative = 2.46 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(1/2)^(1/3)*(-I*sqrt(3) - 1)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*
A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(I*sqrt(3)*d^2 - d^2)*(
-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) + 1/2*(1/2)^(1/3)*(I*sqrt(3) -
1)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x +
2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(-I*sqrt(3)*d^2 - d^2)*(-(A^3 - 3*I*A^2*B - 3*A*B^2
+ I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) + 1/2*(A^3*a^2/d^3)^(1/3)*(I*sqrt(3) - 1)*log(1/2*(2*2^(1/
3)*A^2*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (I*sqrt(3)*d^2 + d^2)*(A^3*a^2/d^3)^(2/
3))/(A^2*a)) + 1/2*(A^3*a^2/d^3)^(1/3)*(-I*sqrt(3) - 1)*log(1/2*(2*2^(1/3)*A^2*a*(a/(e^(2*I*d*x + 2*I*c) + 1))
^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (-I*sqrt(3)*d^2 + d^2)*(A^3*a^2/d^3)^(2/3))/(A^2*a)) + (1/2)^(1/3)*(-(A^3 - 3
*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^
(1/3)*e^(2/3*I*d*x + 2/3*I*c) - 2*(1/2)^(2/3)*d^2*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2
- 2*I*A*B - B^2)*a)) + (A^3*a^2/d^3)^(1/3)*log((2^(1/3)*A^2*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x
+ 2/3*I*c) - (A^3*a^2/d^3)^(2/3)*d^2)/(A^2*a))

Sympy [F]

\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {2}{3}} \left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**(2/3)*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(2/3)*(A + B*tan(c + d*x))*cot(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 4 \, \sqrt {3} A a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, A a^{\frac {2}{3}} \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, A a^{\frac {2}{3}} \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{4 \, d} \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(3)*2^(2/3)*(A - I*B)*a^(2/3)*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)) - 2^(2/3)*(A - I*B)*a^(2/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1
/3) + (I*a*tan(d*x + c) + a)^(2/3)) + 2*2^(2/3)*(A - I*B)*a^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a
)^(1/3)) - 4*sqrt(3)*A*a^(2/3)*arctan(1/3*sqrt(3)*(2*(I*a*tan(d*x + c) + a)^(1/3) + a^(1/3))/a^(1/3)) + 2*A*a^
(2/3)*log((I*a*tan(d*x + c) + a)^(2/3) + (I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + a^(2/3)) - 4*A*a^(2/3)*log((I*
a*tan(d*x + c) + a)^(1/3) - a^(1/3)))/d

Giac [F]

\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \cot \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.45 (sec) , antiderivative size = 1761, normalized size of antiderivative = 6.09 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(2/3),x)

[Out]

log(- (486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - (1458*a^7*d^6*((A^3*a^2)/d^3)^(2/3) + 243*d*(a + a*
tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i))*((A^3*a^2)/d^3)^(1/3))*((A^3*a^2)/d^3)^
(2/3) - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*((A
^3*a^2)/d^3)^(1/3) + log(- (486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - (1458*a^7*d^6*(-(A^3*a^2 + B^3
*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3) + 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^
2*a^8*d^3 + A*B*a^8*d^3*2i))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3))*(-(A^3*a^2
+ B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3) - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^
4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3)
)^(1/3) + (log(- ((3^(1/2)*1i - 1)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - ((3^(1/2)*1i - 1)*(2
43*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i) + (729*a^7*d^6*(3^(1/2)*1i -
1)^2*((A^3*a^2)/d^3)^(2/3))/2)*((A^3*a^2)/d^3)^(1/3))/2)*((A^3*a^2)/d^3)^(2/3))/4 - 243*d*(a + a*tan(c + d*x)
*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*(3^(1/2)*1i - 1)*((A^3*a^2)/d^3)^(1/
3))/2 - (log(- ((3^(1/2)*1i + 1)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) + ((3^(1/2)*1i + 1)*(243
*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i) + (729*a^7*d^6*(3^(1/2)*1i + 1
)^2*((A^3*a^2)/d^3)^(2/3))/2)*((A^3*a^2)/d^3)^(1/3))/2)*((A^3*a^2)/d^3)^(2/3))/4 - 243*d*(a + a*tan(c + d*x)*1
i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*(3^(1/2)*1i + 1)*((A^3*a^2)/d^3)^(1/3)
)/2 - log(- ((3^(1/2)*1i)/2 + 1/2)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) + (243*d*(a + a*tan(c
+ d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i) + 1458*a^7*d^6*((3^(1/2)*1i)/2 + 1/2)^2*(-(A^3
*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3))*((3^(1/2)*1i)/2 + 1/2)*(-(A^3*a^2 + B^3*a^2*1i
- 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^
(2/3) - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*((3
^(1/2)*1i)/2 + 1/2)*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3) + log(- ((3^(1/2)*1i)
/2 - 1/2)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - (243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8
*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i) + 1458*a^7*d^6*((3^(1/2)*1i)/2 - 1/2)^2*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B
^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^
2*3i)/(2*d^3))^(1/3))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3) - 243*d*(a + a*tan(
c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^10))*((3^(1/2)*1i)/2 - 1/2)*(-(A^
3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3)