\(\int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [73]

Optimal result

Integrand size = 36, antiderivative size = 169 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \]

[Out]

1/4*(7*A-4*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/
2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d-1/4*(I*A+4*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/2*A*cot(d*x+c)^2*(
a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3679, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \]

[In]

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

[Out]

(Sqrt[a]*(7*A - (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (Sqrt[2]*Sqrt[a]*(A - I*B)*ArcTa
nh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(4
*d) - (A*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(2*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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

Rule 214

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

Rule 3561

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

Rule 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], 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] && LtQ[n, -1]

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 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (i A+4 B)-\frac {3}{2} a A \tan (c+d x)\right ) \, dx}{2 a} \\ & = -\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (7 A-4 i B)-\frac {1}{4} a^2 (i A+4 B) \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+(-i A-B) \int \sqrt {a+i a \tan (c+d x)} \, dx-\frac {(7 A-4 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{8 a} \\ & = -\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(2 a (A-i B)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}-\frac {(a (7 A-4 i B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d} \\ & = -\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {(7 i A+4 B) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d} \\ & = \frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.94 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.82 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-4 \sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-\cot (c+d x) (i A+4 B+2 A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[In]

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

[Out]

(Sqrt[a]*(7*A - (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - 4*Sqrt[2]*Sqrt[a]*(A - I*B)*ArcTanh[Sqr
t[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] - Cot[c + d*x]*(I*A + 4*B + 2*A*Cot[c + d*x])*Sqrt[a + I*a*Tan[c +
d*x]])/(4*d)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}+\frac {A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {1}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-4 i B +7 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(146\)
default	\(\frac {2 a^{3} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}+\frac {A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {1}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-4 i B +7 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(146\)

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/d*a^3*(-1/2*(A-I*B)/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/a^2*(-((-1/2*I*B
+1/8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2*I*a*B+1/8*a*A)*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+1/8*(7*A-4*I*B
)/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))))

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. 730 vs. \(2 (130) = 260\).

Time = 0.27 (sec) , antiderivative size = 730, normalized size of antiderivative = 4.32 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {8 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 8 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (49 \, A^{2} - 56 i \, A B - 16 \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, {\left (-7 i \, A - 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-7 i \, A - 4 \, B\right )} a^{2} + 2 \, \sqrt {2} {\left (i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {{\left (49 \, A^{2} - 56 i \, A B - 16 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{7 i \, A + 4 \, B}\right ) - {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (49 \, A^{2} - 56 i \, A B - 16 \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, {\left (-7 i \, A - 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-7 i \, A - 4 \, B\right )} a^{2} + 2 \, \sqrt {2} {\left (-i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {{\left (49 \, A^{2} - 56 i \, A B - 16 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{7 i \, A + 4 \, B}\right ) - 4 \, \sqrt {2} {\left ({\left (3 \, A - 4 i \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, A e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (A + 4 i \, B\right )} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(8*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-
4*((-I*A - B)*a*e^(I*d*x + I*c) - (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 8*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I
*c) + d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) - (-I*d*e^(2*I*d*x + 2*I*c) -
I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) + (d*e^(
4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*log(-16*(3*(-7*I*A -
4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-7*I*A - 4*B)*a^2 + 2*sqrt(2)*(I*a*d*e^(3*I*d*x + 3*I*c) + I*a*d*e^(I*d*x + I*
c))*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(7*I*A +
4*B)) - (d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*log(-16
*(3*(-7*I*A - 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-7*I*A - 4*B)*a^2 + 2*sqrt(2)*(-I*a*d*e^(3*I*d*x + 3*I*c) - I*a*
d*e^(I*d*x + I*c))*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2
*I*c)/(7*I*A + 4*B)) - 4*sqrt(2)*((3*A - 4*I*B)*e^(5*I*d*x + 5*I*c) + 4*A*e^(3*I*d*x + 3*I*c) + (A + 4*I*B)*e^
(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {a^{2} {\left (\frac {4 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {{\left (7 \, A - 4 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - 4 i \, B\right )} + \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 4 i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + a^{3}}\right )}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/8*a^2*(4*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*t
an(d*x + c) + a)))/a^(3/2) - (7*A - 4*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) +
a) + sqrt(a)))/a^(3/2) + 2*((I*a*tan(d*x + c) + a)^(3/2)*(A - 4*I*B) + sqrt(I*a*tan(d*x + c) + a)*(A + 4*I*B)
*a)/((I*a*tan(d*x + c) + a)^2*a - 2*(I*a*tan(d*x + c) + a)*a^2 + a^3))/d

Giac [F]

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

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.85 (sec) , antiderivative size = 702, normalized size of antiderivative = 4.15 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\frac {\left (A\,a^2+B\,a^2\,4{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,d}+\frac {\left (A\,a-B\,a\,4{}\mathrm {i}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2-2\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )+a^2}-\frac {\mathrm {atan}\left (\frac {17\,A^3\,a^4\,d\,\sqrt {-\frac {a}{2}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{17\,d\,A^3\,a^5-9{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}-\frac {B^3\,a^4\,d\,\sqrt {-\frac {a}{2}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,16{}\mathrm {i}}{17\,d\,A^3\,a^5-9{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}+\frac {24\,A\,B^2\,a^4\,d\,\sqrt {-\frac {a}{2}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{17\,d\,A^3\,a^5-9{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}-\frac {A^2\,B\,a^4\,d\,\sqrt {-\frac {a}{2}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,9{}\mathrm {i}}{17\,d\,A^3\,a^5-9{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-\frac {a}{2}}\,2{}\mathrm {i}}{d}+\frac {\sqrt {-a}\,\mathrm {atan}\left (\frac {119\,A^3\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,\left (\frac {119\,d\,A^3\,a^5}{4}-3{}\mathrm {i}\,d\,A^2\,B\,a^5+36\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5\right )}-\frac {B^3\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,16{}\mathrm {i}}{\frac {119\,d\,A^3\,a^5}{4}-3{}\mathrm {i}\,d\,A^2\,B\,a^5+36\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}+\frac {36\,A\,B^2\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\frac {119\,d\,A^3\,a^5}{4}-3{}\mathrm {i}\,d\,A^2\,B\,a^5+36\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}-\frac {A^2\,B\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{\frac {119\,d\,A^3\,a^5}{4}-3{}\mathrm {i}\,d\,A^2\,B\,a^5+36\,d\,A\,B^2\,a^5-16{}\mathrm {i}\,d\,B^3\,a^5}\right )\,\left (4\,B+A\,7{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,d} \]

[In]

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

[Out]

(((A*a^2 + B*a^2*4i)*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d) + ((A*a - B*a*4i)*(a + a*tan(c + d*x)*1i)^(3/2))/(4*
d))/((a + a*tan(c + d*x)*1i)^2 - 2*a*(a + a*tan(c + d*x)*1i) + a^2) - (atan((17*A^3*a^4*d*(-a/2)^(1/2)*(a + a*
tan(c + d*x)*1i)^(1/2))/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a^5*d - A^2*B*a^5*d*9i) - (B^3*a^4*d*(-a/2)^(
1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*16i)/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a^5*d - A^2*B*a^5*d*9i) + (24
*A*B^2*a^4*d*(-a/2)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a^5*d - A^2*
B*a^5*d*9i) - (A^2*B*a^4*d*(-a/2)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*9i)/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A
*B^2*a^5*d - A^2*B*a^5*d*9i))*(A*1i + B)*(-a/2)^(1/2)*2i)/d + ((-a)^(1/2)*atan((119*A^3*(-a)^(9/2)*d*(a + a*ta
n(c + d*x)*1i)^(1/2))/(4*((119*A^3*a^5*d)/4 - B^3*a^5*d*16i + 36*A*B^2*a^5*d - A^2*B*a^5*d*3i)) - (B^3*(-a)^(9
/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*16i)/((119*A^3*a^5*d)/4 - B^3*a^5*d*16i + 36*A*B^2*a^5*d - A^2*B*a^5*d*3i)
+ (36*A*B^2*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/((119*A^3*a^5*d)/4 - B^3*a^5*d*16i + 36*A*B^2*a^5*d -
A^2*B*a^5*d*3i) - (A^2*B*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*3i)/((119*A^3*a^5*d)/4 - B^3*a^5*d*16i +
36*A*B^2*a^5*d - A^2*B*a^5*d*3i))*(A*7i + 4*B)*1i)/(4*d)