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

Optimal. Leaf size=289 \[ -\frac{-7 B+i A}{4 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{2 \sqrt [4]{-1} B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{A+3 i B}{6 a d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{-B+i A}{5 d \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

[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]])

________________________________________________________________________________________

Rubi [A]  time = 1.0316, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.237, Rules used = {4241, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{-7 B+i A}{4 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{2 \sqrt [4]{-1} B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{A+3 i B}{6 a d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{-B+i A}{5 d \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[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 4241

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]

Rule 3595

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 3601

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 3544

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 205

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

Rule 3599

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 63

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 217

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 203

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

Rubi steps

\begin{align*} \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 &=\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 ) \operatorname{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 ) \operatorname{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) \tanh ^{-1}\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 ) \operatorname{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) \tanh ^{-1}\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 ) \operatorname{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 \tan ^{-1}\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) \tanh ^{-1}\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]  time = 9.89122, size = 426, normalized size = 1.47 \[ \frac{e^{-3 i (c+d x)} \sqrt{\cot (c+d x)} \sec ^2(c+d x) \left (15 (A-i B) e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )-14 A e^{2 i (c+d x)}+34 A e^{4 i (c+d x)}-23 A e^{6 i (c+d x)}+3 A-24 i B e^{2 i (c+d x)}+144 i B e^{4 i (c+d x)}-123 i B e^{6 i (c+d x)}+30 i \sqrt{2} B e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-30 i \sqrt{2} B e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+3 i B\right ) (A+B \tan (c+d x))}{120 d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[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]

(Sqrt[Cot[c + d*x]]*(3*A + (3*I)*B - 14*A*E^((2*I)*(c + d*x)) - (24*I)*B*E^((2*I)*(c + d*x)) + 34*A*E^((4*I)*(
c + d*x)) + (144*I)*B*E^((4*I)*(c + d*x)) - 23*A*E^((6*I)*(c + d*x)) - (123*I)*B*E^((6*I)*(c + d*x)) + 15*(A -
I*B)*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]]
+ (30*I)*Sqrt[2]*B*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[
2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - (30*I)*Sqrt[2]*B*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(
c + d*x))]*Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sec[c +
d*x]^2*(A + B*Tan[c + d*x]))/(120*d*E^((3*I)*(c + d*x))*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x
])^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.577, size = 2158, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/120-1/120*I)/d/a^3*(60*I*B*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2*sin(
d*x+c)*2^(1/2)-37*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+147*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^
(1/2)+240*B*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-240*B*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-
30*I*B*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+15*A*2^(1/2
)*sin(d*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))-45*B*2^(1/2)*cos(d*x+c)*arctan((1/2
+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))+15*B*2^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*2^(1/2))-60*B*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+I)+60*B*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)-1)-60*B*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+60*B*sin(d*x+c)*ln(((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)-I)+60*I*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^3*2^(1/2)-
30*I*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)-45*I*A*2^(1/2)*cos(d
*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))-15*I*B*2^(1/2)*sin(d*x+c)*arctan((1/2+1/2*
I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))-60*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2
))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+52*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2+252*I*B*si
n(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2+40*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^3-40
*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)+60*I*B*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-I)-60*I*B*ln(((cos
(d*x+c)-1)/sin(d*x+c))^(1/2)+I)+60*I*B*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)-60*I*B*ln(((cos(d*x+c)-1)/sin(d
*x+c))^(1/2)+1)+52*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-252*B*((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-30*B*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)
^2*2^(1/2)-40*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^3+40*I*A*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)+30*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)-120*I*
B*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+120*I*B*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)+1)-37*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-180*I*B*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)-I)+180*I*B*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+I)-180*I*B*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)-1)+180*I*B*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)-240*I*B*cos(d*x+c)*((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)-147*I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-240*B*sin(d*x+c)*cos(d*x+c)^2*ln(((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)-I)+15*I*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+60*
B*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^3*2^(1/2)+240*B*sin(d*x+c)*cos(d*x+
c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+I)-240*B*sin(d*x+c)*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)-1)+240*B*sin(d*x+c)*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+120*B*cos(d*x+c)*sin(d*x+c)*ln(((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)-I)-120*B*cos(d*x+c)*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+I)+120*B*cos
(d*x+c)*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)-120*B*cos(d*x+c)*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(
d*x+c))^(1/2)+1)+240*I*B*cos(d*x+c)^3*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-I)-240*I*B*cos(d*x+c)^3*ln(((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)+I)+240*I*B*cos(d*x+c)^3*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)-240*I*B*cos(d*x+c)^3
*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+240*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^3-120*I*B*cos(d*
x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-I)+120*I*B*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+I))*c
os(d*x+c)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(4*I*sin(d*x+c)*cos(d*x+c)^2+4*cos(d*x+c)^3-I*sin(d
*x+c)-3*cos(d*x+c))/sin(d*x+c)^3/(cos(d*x+c)/sin(d*x+c))^(5/2)/((cos(d*x+c)-1)/sin(d*x+c))^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [B]  time = 2.8468, size = 2233, normalized size = 7.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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^(6*I*d*x + 6*I*c)*log((2*I*sqrt(1/2)*a^3*d
*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) - I*A -
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^(I*d*x + I*
c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x +
6*I*c)*log((-2*I*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A +
B)*e^(2*I*d*x + 2*I*c) - I*A - 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^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(6*I
*d*x + 6*I*c)*log(1/605*(208*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) - 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^(I*d*x + I*c) + (156*I*a^3*d*e^(2*I*d*x + 2*I*c) - 52*I*a^3
*d)*sqrt(-4*I*B^2/(a^5*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) + 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(6*I*d*x + 6*
I*c)*log(1/605*(208*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) - 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^(I*d*x + I*c) + (-156*I*a^3*d*e^(2*I*d*x + 2*I*c) + 52*I*a^3*d)*sqrt
(-4*I*B^2/(a^5*d^2)))/(B*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^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*
d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)