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

Optimal. Leaf size=211 \[ -\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(-9 B+7 i A) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((1/2 + I/2)*(I*A + 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]])/(Sqrt[a]*d) + ((A + I*B)*Cot[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((7*I)
*A - 9*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d) - ((5*A + (3*I)*B)*Cot[c + d*x]^(3/2)*Sqrt[a
+ I*a*Tan[c + d*x]])/(3*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.696978, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4241, 3596, 3598, 12, 3544, 205} \[ -\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(-9 B+7 i A) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/2 + I/2)*(I*A + 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]])/(Sqrt[a]*d) + ((A + I*B)*Cot[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((7*I)
*A - 9*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d) - ((5*A + (3*I)*B)*Cot[c + d*x]^(3/2)*Sqrt[a
+ I*a*Tan[c + d*x]])/(3*a*d)

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{\cot ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \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{1}{2} a (5 A+3 i B)-2 a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (7 i A-9 B)-\frac{1}{2} a^2 (5 A+3 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(7 i A-9 B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int -\frac{3 a^3 (A-i B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(7 i A-9 B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{\left ((A-i 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}{2 a}\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(7 i A-9 B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (i a (A-i 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 )}{d}\\ &=\frac{\left (\frac{1}{2}+\frac{i}{2}\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)}}{\sqrt{a} d}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(7 i A-9 B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{(5 A+3 i B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}\\ \end{align*}

Mathematica [A]  time = 4.05174, size = 166, normalized size = 0.79 \[ \frac{\sqrt{\cot (c+d x)} \csc (c+d x) \sec (c+d x) \left ((5 A+9 i B) \cos (2 (c+d x))+\frac{3}{2} (A-i B) e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+2 i A \sin (2 (c+d x))-9 A-6 B \sin (2 (c+d x))-9 i B\right )}{6 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Csc[c + d*x]*Sec[c + d*x]*(-9*A - (9*I)*B + (3*(A - I*B)*(-1 + E^((2*I)*(c + d*x)))^(3/2)*
ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(2*E^(I*(c + d*x))) + (5*A + (9*I)*B)*Cos[2*(c + d*x)
] + (2*I)*A*Sin[2*(c + d*x)] - 6*B*Sin[2*(c + d*x)]))/(6*d*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 0.709, size = 683, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/6-1/6*I)/d/a*(-6*I*B*sin(d*x+c)*cos(d*x+c)-9*I*B*cos(d*x+c)^2+3*I*A*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/
sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+3*A*cos(d*x+c)^2*((cos
(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+3*B*cos(d*x
+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))
*2^(1/2)-7*I*A+3*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*2^(1/2))*2^(1/2)-3*I*B*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1
)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)-2*I*A*cos(d*x+c)*sin(d*x+c)-3*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(
(1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+5*I*A*cos(d*x+c)^2+9*I*B-5*A*cos(d*x+c)^2-2*A*c
os(d*x+c)*sin(d*x+c)-9*B*cos(d*x+c)^2+6*B*cos(d*x+c)*sin(d*x+c)+3*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan
((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+3*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+7*A+9*B)*sin(d*x+c)*(cos(d*x+c)/sin
(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c)^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(cot(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 1.68108, size = 1353, normalized size = 6.41 \begin{align*} \frac{\sqrt{2}{\left ({\left (14 i \, A - 30 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-36 i \, A + 36 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, A - 6 \, 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}} e^{\left (i \, d x + i \, c\right )} + 3 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} \log \left (\frac{{\left (a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - 3 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} \log \left (-\frac{{\left (a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right )}{12 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*((14*I*A - 30*B)*e^(4*I*d*x + 4*I*c) + (-36*I*A + 36*B)*e^(2*I*d*x + 2*I*c) + 6*I*A - 6*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) + 3*(
a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)/(a*d^2))*log((a*d*sqrt((-
2*I*A^2 - 4*A*B + 2*I*B^2)/(a*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)*sq
rt(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)) - 3*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-2*I*A^2 - 4*A*B
+ 2*I*B^2)/(a*d^2))*log(-(a*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)/(a*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)))/(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x
+ 2*I*c))

________________________________________________________________________________________

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)**(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

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 )^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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