\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [86]

Optimal result

Integrand size = 36, antiderivative size = 158 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {a^{5/2} (5 i A+2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3674, 3675, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {a^{5/2} (2 B+5 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (-2 B+i A) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \]

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

-((a^(5/2)*((5*I)*A + 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d) + (4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcT
anh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (a^2*(I*A - 2*B)*Sqrt[a + I*a*Tan[c + d*x]])/d - (a*A*C
ot[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/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 3674

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c
+ d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
- 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !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 A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+\int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (5 i A+2 B)+\frac {1}{2} a (A+2 i B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+2 \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{4} a^2 (5 i A+2 B)-\frac {3}{4} a^2 (A-2 i B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}-\left (4 a^2 (A-i B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx+\frac {1}{2} (a (5 i A+2 B)) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+\frac {\left (8 a^3 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}+\frac {\left (a^3 (5 i A+2 B)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+\frac {\left (a^2 (5 A-2 i B)\right ) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {a^{5/2} (5 i A+2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-i a^{5/2} (5 A-2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a^2 (-2 B-A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

((-I)*a^(5/2)*(5*A - (2*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + 4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcTan
h[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + a^2*(-2*B - A*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+2 \sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-a \left (-\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (-2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )}{d}\)	\(131\)
default	\(\frac {2 i a^{2} \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+2 \sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-a \left (-\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (-2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )}{d}\)	\(131\)

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2*I/d*a^2*(I*B*(a+I*a*tan(d*x+c))^(1/2)+2*a^(1/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)
/a^(1/2))-a*(-1/2*I*A*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+1/2*(5*A-2*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. 705 vs. \(2 (125) = 250\).

Time = 0.26 (sec) , antiderivative size = 705, normalized size of antiderivative = 4.46 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {8 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 8 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} + 2 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \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 )}}{{\left (5 i \, A + 2 \, B\right )} a}\right ) + \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \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 )}}{{\left (5 i \, A + 2 \, B\right )} a}\right ) + 4 \, \sqrt {2} {\left ({\left (i \, A + 2 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (i \, A - 2 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((-I*A - B)*a^3*e^(I*d*
x + I*c) + sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))
*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) -
d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^
2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3*(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2*B)*a^3 + 2*sqrt
(2)*sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((5*I*A + 2*B)*a)) + sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(2*I
*d*x + 2*I*c) - d)*log(-16*(3*(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2*B)*a^3 - 2*sqrt(2)*sqrt(-(2
5*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
)))*e^(-2*I*d*x - 2*I*c)/((5*I*A + 2*B)*a)) + 4*sqrt(2)*((I*A + 2*B)*a^2*e^(3*I*d*x + 3*I*c) + (I*A - 2*B)*a^2
*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) - d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.03 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (4 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (5 \, A - 2 i \, B\right )} a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - 4 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A a}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*I*(4*sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqr
t(I*a*tan(d*x + c) + a))) - (5*A - 2*I*B)*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x
+ c) + a) + sqrt(a))) - 4*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 2*I*sqrt(I*a*tan(d*x + c) + a)*A*a/tan(d*x + c))
*a/d

Giac [F]

\[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^2, x)

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 2947, normalized size of antiderivative = 18.65 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

int(cot(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

atan((d^4*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (
A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A
*B*a^5*21i)/(2*d^2))^(1/2)*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*246
4i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*6i)/(A^3*a^14*d*126i - 336*B^3*a^14*d - A*B^2*a^14*d*1032i + 876*A^2*B*
a^14*d + A*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4
- (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/
d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)) + (A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((4
9*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^
4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2)*14i)/(A^3*a^11*d*
126i - 336*B^3*a^11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*
2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2
*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) -
(B^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4
+ (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) +
(A*B*a^5*21i)/(2*d^2))^(1/2)*56i)/(A^3*a^11*d*126i - 336*B^3*a^11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^2
2)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4
*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1
/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) + (88*A*B*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d
^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a
^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2))/(A^3*a^11*d*126i - 336*B^3*a^
11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3
*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B
^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d))*(((49*A^4*a^22)/d^
4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^
6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2)*2i - atan((d^4*(a + a*tan(c + d
*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^
2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2)
*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i
)/d^4)^(1/2)*6i)/(A^3*a^14*d*126i - 336*B^3*a^14*d - A*B^2*a^14*d*1032i + 876*A^2*B*a^14*d - A*a^3*d^3*((49*A^
4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(
1/2)*2i + 4*B*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d
^4 - (A^3*B*a^22*616i)/d^4)^(1/2)) - (A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2
*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (
A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2)*14i)/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*
d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*6
16i)/d^4)^(1/2)*2i + 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2
464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) + (B^2*a^8*d^2*(a + a*tan(c
+ d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328
*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1
/2)*56i)/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^2
2)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i + 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22
)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i +
876*A^2*B*a^11*d) - (88*A*B*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2)
- ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*61
6i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2))/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*d^3*((49*A^4*a^22
)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2
i + 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*
B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d))*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2)
- ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616
i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2)*2i - (2*B*a^2*(a + a*tan(c + d*x)*1i)^(1/2))/d - (A*a^2*(
a + a*tan(c + d*x)*1i)^(1/2))/(d*tan(c + d*x))