\(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) [566]

Optimal result

Integrand size = 38, antiderivative size = 289 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2*(-1)^(1/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c
)^(1/2)/a^(5/2)/d+(1/8+1/8*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x
+c)^(1/2)*tan(d*x+c)^(1/2)/a^(5/2)/d+1/4*(-I*A+7*B)/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B
)/d/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2)+1/6*(A+3*I*B)/a/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {4326, 3676, 3682, 3625, 211, 3680, 65, 223, 209} \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \sqrt [4]{-1} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {-7 B+i A}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {-B+i A}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(2*(-1)^(1/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*
Sqrt[Tan[c + d*x]])/(a^(5/2)*d) + ((1/8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a +
I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(a^(5/2)*d) + (I*A - B)/(5*d*Cot[c + d*x]^(5/2)*(a
+ I*a*Tan[c + d*x])^(5/2)) + (A + (3*I)*B)/(6*a*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) - (I*A - 7*
B)/(4*a^2*d*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 209

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

Rule 211

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3625

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

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]

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 3682

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

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx \\ & = \frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (i A-B)+5 i a B \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {15}{4} a^2 (A+3 i B)-15 a^2 B \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {15}{8} a^3 (i A-7 B)-15 i a^3 B \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{15 a^6} \\ & = \frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^4}+\frac {\left ((i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3} \\ & = \frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {\left (i (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d} \\ & = \frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = \frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^2 d} \\ & = \frac {2 \sqrt [4]{-1} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.97 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{120}+\frac {i}{120}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-120-120 i) \sqrt [4]{-1} \sqrt {a} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-\sec ^2(c+d x) \sqrt {1+i \tan (c+d x)} \left ((-1-i) \sqrt {a} (-11 A-21 i B+2 (13 A+63 i B) \cos (2 (c+d x))+20 i (A+6 i B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}+15 (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )\right )}{a^{5/2} d \sqrt {1+i \tan (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((1/120 + I/120)*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-120 - 120*I)*(-1)^(1/4)*Sqrt[a]*B*ArcSinh[(-1)^(1/4)
*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^3*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - Sec[c + d*x]^2*Sqrt[1 + I*Tan[c
+ d*x]]*((-1 - I)*Sqrt[a]*(-11*A - (21*I)*B + 2*(13*A + (63*I)*B)*Cos[2*(c + d*x)] + (20*I)*(A + (6*I)*B)*Sin[
2*(c + d*x)])*Sqrt[Tan[c + d*x]] + 15*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[
c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])))/(a^(5/2)*d*Sqrt[1 + I*Tan[c +
d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1551 vs. \(2 (230 ) = 460\).

Time = 0.55 (sec) , antiderivative size = 1552, normalized size of antiderivative = 5.37

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1552\)
default	\(\text {Expression too large to display}\)	\(1552\)

[In]

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

[Out]

-1/240/d/(1/tan(d*x+c))^(5/2)/tan(d*x+c)^2*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(240*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta
n(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a-15*A*(I*a)^(1/2)*ln(-(-2*2^(1/2
)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a+60*I*A*(I*a
)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))
*2^(1/2)*a*tan(d*x+c)^3-420*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+240*B*ln(1/2*(2*I
*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^4-
1440*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/
2)*a*tan(d*x+c)^2-220*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+148*A*(I*a)^
(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+1548*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-15*A*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^4+60*B*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(
-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^3+9
0*A*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*
x+c)+I))*2^(1/2)*a*tan(d*x+c)^2-60*B*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)+15*I*B*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a-960*I*B*ln(1/2*(2*I*a*tan
(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^3-60*I*A
*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c
)+I))*2^(1/2)*a*tan(d*x+c)+60*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-1380*I*B*(I*a
)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+960*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)-308*I*A*(I*a)^(1/2)*(-I
*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-90*I*B*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^2+15*I*B*(I*a)^(
1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^
(1/2)*a*tan(d*x+c)^4+588*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3)/(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^4

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. 799 vs. \(2 (217) = 434\).

Time = 0.27 (sec) , antiderivative size = 799, normalized size of antiderivative = 2.76 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 30 \, a^{3} d \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) - 30 \, a^{3} d \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} - \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) + \sqrt {2} {\left ({\left (23 \, A + 123 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (17 \, A + 72 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (7 \, A + 12 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/
2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/
(e^(2*I*d*x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x -
I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(
2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2
*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e
^(-I*d*x - I*c)/(I*A + B)) + 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-16*(3*B*a^2*e^(2*I*d*x
+ 2*I*c) - B*a^2 + sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-4*I*B^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)
/B) - 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 - sqr
t(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-4*I*B^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/B) + sqrt(2)*((23*A +
123*I*B)*e^(6*I*d*x + 6*I*c) - 2*(17*A + 72*I*B)*e^(4*I*d*x + 4*I*c) + 2*(7*A + 12*I*B)*e^(2*I*d*x + 2*I*c) -
3*A - 3*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^
(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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