3.111 \(\int \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=312 \[ \frac{(43 A+20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} d}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (-B+2 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((43*A + (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*a^(5/2)*d) - ((A - I*B)*ArcTanh[Sqrt[a + I*
a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((A + I*B)*Cot[c + d*x]^2)/(5*d*(a + I*a*Tan[c + d
*x])^(5/2)) + ((23*A + (13*I)*B)*Cot[c + d*x]^2)/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((337*A + (167*I)*B)*
Cot[c + d*x]^2)/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (21*((2*I)*A - B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*
x]])/(4*a^3*d) - ((85*A + (41*I)*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(12*a^3*d)

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Rubi [A]  time = 1.23671, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3596, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac{(43 A+20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} d}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (-B+2 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((43*A + (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*a^(5/2)*d) - ((A - I*B)*ArcTanh[Sqrt[a + I*
a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((A + I*B)*Cot[c + d*x]^2)/(5*d*(a + I*a*Tan[c + d
*x])^(5/2)) + ((23*A + (13*I)*B)*Cot[c + d*x]^2)/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((337*A + (167*I)*B)*
Cot[c + d*x]^2)/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (21*((2*I)*A - B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*
x]])/(4*a^3*d) - ((85*A + (41*I)*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(12*a^3*d)

Rule 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3598

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 3600

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]

Rule 3480

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 206

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

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 208

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

Rubi steps

\begin{align*} \int \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\cot ^3(c+d x) \left (a (7 A+2 i B)-\frac{9}{2} a (i A-B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\cot ^3(c+d x) \left (a^2 (44 A+19 i B)-\frac{7}{4} a^2 (23 i A-13 B) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \cot ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{5}{2} a^3 (85 A+41 i B)-\frac{5}{8} a^3 (337 i A-167 B) \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}+\frac{\int \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{315}{2} a^4 (2 i A-B)-\frac{15}{4} a^4 (85 A+41 i B) \tan (c+d x)\right ) \, dx}{30 a^7}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (2 i A-B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}+\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{15}{4} a^5 (43 A+20 i B)+\frac{315}{4} a^5 (2 i A-B) \tan (c+d x)\right ) \, dx}{30 a^8}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (2 i A-B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}-\frac{(43 A+20 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^4}-\frac{(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (2 i A-B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 a^2 d}-\frac{(43 A+20 i B) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 a^2 d}\\ &=-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (2 i A-B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}+\frac{(43 i A-20 B) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 a^3 d}\\ &=\frac{(43 A+20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} d}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{21 (2 i A-B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 a^3 d}-\frac{(85 A+41 i B) \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 a^3 d}\\ \end{align*}

Mathematica [A]  time = 9.09369, size = 317, normalized size = 1.02 \[ \frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x)) \left (\sqrt{2} e^{2 i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\sqrt{2} (43 A+20 i B) \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )-(A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )+\frac{\csc ^2(c+d x) (-15 (44 A+21 i B) \cos (2 (c+d x))+(388 A+203 i B) \cos (4 (c+d x))-695 i A \sin (2 (c+d x))+385 i A \sin (4 (c+d x))+212 A+340 B \sin (2 (c+d x))-200 B \sin (4 (c+d x))+112 i B)}{15 \sqrt{\sec (c+d x)}}\right )}{8 d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(Sec[c + d*x]^(3/2)*(Sqrt[2]*E^((2*I)*(c + d*x))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((
2*I)*(c + d*x))]*(-((A - I*B)*ArcSinh[E^(I*(c + d*x))]) + Sqrt[2]*(43*A + (20*I)*B)*ArcTanh[(Sqrt[2]*E^(I*(c +
d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]]) + (Csc[c + d*x]^2*(212*A + (112*I)*B - 15*(44*A + (21*I)*B)*Cos[2*(c +
d*x)] + (388*A + (203*I)*B)*Cos[4*(c + d*x)] - (695*I)*A*Sin[2*(c + d*x)] + 340*B*Sin[2*(c + d*x)] + (385*I)*
A*Sin[4*(c + d*x)] - 200*B*Sin[4*(c + d*x)]))/(15*Sqrt[Sec[c + d*x]]))*(A + B*Tan[c + d*x]))/(8*d*(A*Cos[c + d
*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2))

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Maple [B]  time = 0.484, size = 2876, normalized size = 9.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/240/d/a^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(645*A*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-300*B*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*l
n(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-645*A*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*cos(d*x+c)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-15*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+15*I
*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)-300*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+15*I*B*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*
2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-192*B*cos(d*x+c)^7*sin(d*
x+c)-192*I*B*cos(d*x+c)^8-64*I*B*cos(d*x+c)^6-564*I*B*cos(d*x+c)^4+820*I*B*cos(d*x+c)^2+1700*A*cos(d*x+c)^2-11
64*A*cos(d*x+c)^4-224*A*cos(d*x+c)^6-160*B*cos(d*x+c)^5*sin(d*x+c)+1260*B*cos(d*x+c)*sin(d*x+c)-668*B*cos(d*x+
c)^3*sin(d*x+c)-645*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+300*
B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin
(d*x+c))-15*I*A*cos(d*x+c)^2*sin(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)+645*A*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-300*B*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+300*B*cos(d*x+c)*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+
c))-15*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*
x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)-645*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-15*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2
)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2))-192*A*cos(d*x+c)^8+15*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^
(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+15*I*B*2^(1/2)*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2))+15*A*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*
x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+15*A*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2))+645*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*
x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+300*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-645*I*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+
cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-300*I*B*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+192*I*A*cos(d*x+c)^7*sin(d*x+c)+320*I*A*cos(d*x+c)^5*sin(d*x+c)+1
348*I*A*sin(d*x+c)*cos(d*x+c)^3-2520*I*A*sin(d*x+c)*cos(d*x+c)+300*I*B*cos(d*x+c)^3*arctan(1/(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+645*I*A*cos(d*x+c)^2*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+300*I*B*cos(d*x+c)^2*arc
tan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-645*I*A*cos(d*x+c)*ln(-(-(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-300
*I*B*cos(d*x+c)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+645*I*A*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-300*I*B*sin(d*x+c
)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)+645*I*A*cos(d*x+c)^3*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-15*I*B*cos(d*x+c)^3*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*
x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)-15*I*B*cos(d*x+c)^2*ar
ctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*2^(1/2)-645*I*A*cos(d*x+c)^2*sin(d*x+c)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+300*I*B*cos(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(-1+cos(d*x+c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.494, size = 2677, normalized size = 8.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/240*(2*sqrt(2)*((773*A + 403*I*B)*e^(10*I*d*x + 10*I*c) - 6*(97*A + 42*I*B)*e^(8*I*d*x + 8*I*c) - (931*A +
431*I*B)*e^(6*I*d*x + 6*I*c) + 3*(153*A + 83*I*B)*e^(4*I*d*x + 4*I*c) + 2*(19*A + 14*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))*e^(I*d*x + I*c) + 30*sqrt(1/2)*(a^3*d*e^(10*I*d*x + 10*I*c) -
2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log((2*I*sqrt(
1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - 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))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 30*sqrt(1/2)*(
a^3*d*e^(10*I*d*x + 10*I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((A^2 - 2*I*A*B - B
^2)/(a^5*d^2))*log((-2*I*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - 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))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(
4*I*A + 4*B)) - 15*(a^3*d*e^(10*I*d*x + 10*I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqr
t((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2))*log(1/4*(4*sqrt(2)*((7568*I*A - 3520*B)*e^(2*I*d*x + 2*I*c) + 7
568*I*A - 3520*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + (1056*I*a^3*d*e^(2*I*d*x + 2*I*c) + 352*
I*a^3*d)*sqrt((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2)))/((-21801*I*A + 10140*B)*e^(2*I*d*x + 2*I*c) + 2180
1*I*A - 10140*B)) + 15*(a^3*d*e^(10*I*d*x + 10*I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))
*sqrt((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2))*log(1/4*(4*sqrt(2)*((7568*I*A - 3520*B)*e^(2*I*d*x + 2*I*c)
+ 7568*I*A - 3520*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + (-1056*I*a^3*d*e^(2*I*d*x + 2*I*c) -
352*I*a^3*d)*sqrt((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2)))/((-21801*I*A + 10140*B)*e^(2*I*d*x + 2*I*c) +
21801*I*A - 10140*B)))/(a^3*d*e^(10*I*d*x + 10*I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c)
)

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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(cot(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*cot(d*x + c)^3/(I*a*tan(d*x + c) + a)^(5/2), x)