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

Optimal. Leaf size=286 \[ -\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{(-317 B+707 i A) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((1/8 + I/8)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (21*A + (11*I)*B)/(30*a*d*Tan[c + d*x]^(3/2
)*(a + I*a*Tan[c + d*x])^(3/2)) + (89*A + (39*I)*B)/(20*a^2*d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) -
((361*A + (151*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Tan[c + d*x]^(3/2)) + (((707*I)*A - 317*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/(60*a^3*d*Sqrt[Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.00663, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ -\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{(-317 B+707 i A) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (21*A + (11*I)*B)/(30*a*d*Tan[c + d*x]^(3/2
)*(a + I*a*Tan[c + d*x])^(3/2)) + (89*A + (39*I)*B)/(20*a^2*d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) -
((361*A + (151*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Tan[c + d*x]^(3/2)) + (((707*I)*A - 317*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/(60*a^3*d*Sqrt[Tan[c + d*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{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (13 A+3 i B)-4 a (i A-B) \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{3}{4} a^2 (47 A+17 i B)-\frac{3}{2} a^2 (21 i A-11 B) \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{8} a^3 (361 A+151 i B)-\frac{3}{2} a^3 (89 i A-39 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{16} a^4 (707 i A-317 B)-\frac{3}{8} a^4 (361 A+151 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{45 a^7}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(707 i A-317 B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{4 \int -\frac{45 a^5 (A-i B) \sqrt{a+i a \tan (c+d x)}}{32 \sqrt{\tan (c+d x)}} \, dx}{45 a^8}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(707 i A-317 B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}-\frac{(A-i B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(707 i A-317 B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{(i A+B) \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 )}{a^{5/2} d}+\frac{A+i B}{5 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{21 A+11 i B}{30 a d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{89 A+39 i B}{20 a^2 d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(361 A+151 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(707 i A-317 B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 9.45376, size = 701, normalized size = 2.45 \[ \frac{\sqrt{\tan (c+d x)} \sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left ((21 A+16 i B) \left (-\frac{\cos (c)}{60}+\frac{1}{60} i \sin (c)\right ) \cos (4 d x)+(11 A+6 i B) \left (-\frac{7 \cos (c)}{20}-\frac{7}{20} i \sin (c)\right ) \cos (2 d x)+(A+i B) \left (-\frac{1}{40} \cos (3 c)+\frac{1}{40} i \sin (3 c)\right ) \cos (6 d x)+(11 A+6 i B) \left (-\frac{7 \sin (c)}{20}+\frac{7}{20} i \cos (c)\right ) \sin (2 d x)+(21 A+16 i B) \left (\frac{\sin (c)}{60}+\frac{1}{60} i \cos (c)\right ) \sin (4 d x)+(A+i B) \left (\frac{1}{40} \sin (3 c)+\frac{1}{40} i \cos (3 c)\right ) \sin (6 d x)+\frac{2}{3} \csc (c) \csc (c+d x) \left (4 i A \sin (3 c-d x)-4 i A \sin (3 c+d x)+4 A \cos (3 c-d x)-4 A \cos (3 c+d x)-\frac{3}{2} B \sin (3 c-d x)+\frac{3}{2} B \sin (3 c+d x)+\frac{3}{2} i B \cos (3 c-d x)-\frac{3}{2} i B \cos (3 c+d x)\right )+i \csc (c) \left (\frac{1}{120} \cos (3 c)+\frac{1}{120} i \sin (3 c)\right ) (343 i A \sin (c)+640 A \cos (c)-223 B \sin (c)+240 i B \cos (c))+\left (-\frac{2}{3} A \cos (3 c)-\frac{2}{3} i A \sin (3 c)\right ) \csc ^2(c+d x)\right )}{d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))}+\frac{(B+i A) \sqrt{e^{i d x}} e^{-i (d x-2 c)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{3}{2}}(c+d x) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) (\cos (d x)+i \sin (d x))^{5/2} (A+B \tan (c+d x))}{4 \sqrt{2} d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I*A + B)*Sqrt[E^(I*d*x)]*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*ArcT
anh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])^(5/2)*(A + B*Ta
n[c + d*x]))/(4*Sqrt[2]*d*E^(I*(-2*c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]
*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^3*(
(21*A + (16*I)*B)*Cos[4*d*x]*(-Cos[c]/60 + (I/60)*Sin[c]) + (11*A + (6*I)*B)*Cos[2*d*x]*((-7*Cos[c])/20 - ((7*
I)/20)*Sin[c]) + I*Csc[c]*(640*A*Cos[c] + (240*I)*B*Cos[c] + (343*I)*A*Sin[c] - 223*B*Sin[c])*(Cos[3*c]/120 +
(I/120)*Sin[3*c]) + (A + I*B)*Cos[6*d*x]*(-Cos[3*c]/40 + (I/40)*Sin[3*c]) + Csc[c + d*x]^2*((-2*A*Cos[3*c])/3
- ((2*I)/3)*A*Sin[3*c]) + (11*A + (6*I)*B)*(((7*I)/20)*Cos[c] - (7*Sin[c])/20)*Sin[2*d*x] + (21*A + (16*I)*B)*
((I/60)*Cos[c] + Sin[c]/60)*Sin[4*d*x] + (A + I*B)*((I/40)*Cos[3*c] + Sin[3*c]/40)*Sin[6*d*x] + (2*Csc[c]*Csc[
c + d*x]*(4*A*Cos[3*c - d*x] + ((3*I)/2)*B*Cos[3*c - d*x] - 4*A*Cos[3*c + d*x] - ((3*I)/2)*B*Cos[3*c + d*x] +
(4*I)*A*Sin[3*c - d*x] - (3*B*Sin[3*c - d*x])/2 - (4*I)*A*Sin[3*c + d*x] + (3*B*Sin[3*c + d*x])/2))/3)*Sqrt[Ta
n[c + d*x]]*(A + B*Tan[c + d*x]))/(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 = 0.058, size = 1239, normalized size = 4.3 \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))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)*(-2940*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+
2828*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^5+15*B*2^(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))*tan(d*x+c)^6*a+4468*I*B*(a*tan(
d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-90*I*A*2^(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))*tan(d*x+c)^4*a+60*A*2^(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))*tan(d*x+c)^5*a-60*I*B*2^
(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))*t
an(d*x+c)^5*a+15*I*A*2^(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))*tan(d*x+c)^6*a-90*B*2^(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))*tan(d*x+c)^4*a-1268*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^
(1/2)*tan(d*x+c)^5+640*I*A*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)+15*I*A*2^(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))*tan(d*x+c)^2*
a-60*A*2^(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))*tan(d*x+c)^3*a+9868*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-12260*I*A*(a*tan
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+15*B*2^(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))*tan(d*x+c)^2*a+5660*B*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60*I*B*2^(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))*tan(d*x+c)^3*a-6020*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-
I*a)^(1/2)*tan(d*x+c)^2-480*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-160*A*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a^3/tan(d*x+c)^(3/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c
)+I)^4/(-I*a)^(1/2)

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.4034, size = 1740, normalized size = 6.08 \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))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(sqrt(2)*((983*A + 463*I*B)*e^(10*I*d*x + 10*I*c) - 2*(272*A + 97*I*B)*e^(8*I*d*x + 8*I*c) - 3*(393*A +
163*I*B)*e^(6*I*d*x + 6*I*c) + (381*A + 191*I*B)*e^(4*I*d*x + 4*I*c) + 2*(18*A + 13*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) + 15*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((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*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*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(
(-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*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)))/(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((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.56839, size = 261, normalized size = 0.91 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + \left (2 i - 2\right ) \, a^{4}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} a - 5 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a^{2} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{3} - 7 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{4} + 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^4 + (-(2*I - 2)*(I*a*tan(d*x +
c) + a)*a^3 + (2*I - 2)*a^4)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/(((I*a*t
an(d*x + c) + a)^7*a - 5*(I*a*tan(d*x + c) + a)^6*a^2 + 9*(I*a*tan(d*x + c) + a)^5*a^3 - 7*(I*a*tan(d*x + c) +
a)^4*a^4 + 2*(I*a*tan(d*x + c) + a)^3*a^5)*d)