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

Optimal. Leaf size=298 \[ \frac{3 (-1)^{3/4} a^{5/2} (121 B+120 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{a^2 (107 B+104 i A) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{(4+4 i) a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d} \]

[Out]

(3*(-1)^(3/4)*a^(5/2)*((120*I)*A + 121*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c +
d*x]]])/(64*d) + ((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c
+ d*x]]])/d + (a^2*(152*A - (149*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(64*d) + (a^2*((104*I)*A
+ 107*B)*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(96*d) - (a^2*(8*A - (11*I)*B)*Tan[c + d*x]^(5/2)*Sqr
t[a + I*a*Tan[c + d*x]])/(24*d) + ((I/4)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2))/d

________________________________________________________________________________________

Rubi [A]  time = 1.13219, antiderivative size = 298, 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 = {3594, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{3 (-1)^{3/4} a^{5/2} (121 B+120 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{a^2 (107 B+104 i A) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{(4+4 i) a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-1)^(3/4)*a^(5/2)*((120*I)*A + 121*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c +
d*x]]])/(64*d) + ((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c
+ d*x]]])/d + (a^2*(152*A - (149*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(64*d) + (a^2*((104*I)*A
+ 107*B)*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(96*d) - (a^2*(8*A - (11*I)*B)*Tan[c + d*x]^(5/2)*Sqr
t[a + I*a*Tan[c + d*x]])/(24*d) + ((I/4)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2))/d

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && 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 \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{1}{4} \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (8 A-5 i B)+\frac{1}{2} a (8 i A+11 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{1}{12} \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{4} a^2 (88 A-85 i B)+\frac{1}{4} a^2 (104 i A+107 B) \tan (c+d x)\right ) \, dx\\ &=\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{8} a^3 (104 i A+107 B)+\frac{3}{8} a^3 (152 A-149 i B) \tan (c+d x)\right ) \, dx}{24 a}\\ &=\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{16} a^4 (152 A-149 i B)-\frac{9}{16} a^4 (120 i A+121 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{24 a^2}\\ &=\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}-\left (4 a^2 (A-i B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+\frac{1}{128} (3 a (120 A-121 i B)) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{\left (3 a^3 (120 A-121 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{128 d}+\frac{\left (8 a^4 (i A+B)\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{(4+4 i) a^{5/2} (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 )}{d}+\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{\left (3 a^3 (120 A-121 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{64 d}\\ &=\frac{(4+4 i) a^{5/2} (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 )}{d}+\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}+\frac{\left (3 a^3 (120 A-121 i B)\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 )}{64 d}\\ &=-\frac{3 \sqrt [4]{-1} a^{5/2} (120 A-121 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}+\frac{(4+4 i) a^{5/2} (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 )}{d}+\frac{a^2 (152 A-149 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{a^2 (104 i A+107 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}-\frac{a^2 (8 A-11 i B) \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\\ \end{align*}

Mathematica [A]  time = 9.69973, size = 581, normalized size = 1.95 \[ \frac{\cos ^3(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left ((8 A-23 i B) \left (-\frac{1}{24} \cos (2 c)+\frac{1}{24} i \sin (2 c)\right ) \sec ^2(c+d x)+(104 A-131 i B) \sec (c+d x) \left (-\frac{1}{96} \cos (3 c+d x)+\frac{1}{96} i \sin (3 c+d x)\right )+(56 A-65 i B) \left (\frac{13}{192} \cos (2 c)-\frac{13}{192} i \sin (2 c)\right )+\sec ^3(c+d x) \left (-\frac{1}{4} B \sin (3 c+d x)-\frac{1}{4} i B \cos (3 c+d x)\right )\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (3 \sqrt{2} (120 A-121 i B) \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-2048 (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{256 \sqrt{2} d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-2048*(A - I*B)*Log[E^(I*(
c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + 3*Sqrt[2]*(120*A - (121*I)*B)*(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))]] - 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))]]))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(256*Sqrt[2]*d*E^(
(2*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Co
s[d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*((8*A - (23*I)*B)*Sec[c + d*x]
^2*(-Cos[2*c]/24 + (I/24)*Sin[2*c]) + (56*A - (65*I)*B)*((13*Cos[2*c])/192 - ((13*I)/192)*Sin[2*c]) + (104*A -
(131*I)*B)*Sec[c + d*x]*(-Cos[3*c + d*x]/96 + (I/96)*Sin[3*c + d*x]) + Sec[c + d*x]^3*((-I/4)*B*Cos[3*c + d*x
] - (B*Sin[3*c + d*x])/4))*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(d*(Cos[d*x]
+ I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.041, size = 742, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-96*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)-128*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
+272*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+416*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)+447*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-894*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+428*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)-384*I*
(I*a)^(1/2)*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))/(ta
n(d*x+c)+I))*a-456*A*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+912*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)-768*I*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))*(-I*a)^(1/2)*a+384*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))*a*(I*a)^(1/2)
-768*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))*a*(-I*a)^(1/
2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 2.01838, size = 2928, normalized size = 9.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*sqrt(2)*(13*(56*A - 65*I*B)*a^2*e^(6*I*d*x + 6*I*c) + 3*(504*A - 425*I*B)*a^2*e^(4*I*d*x + 4*I*c) + (
1096*A - 1135*I*B)*a^2*e^(2*I*d*x + 2*I*c) + 3*(104*A - 107*I*B)*a^2)*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*sqrt((-129600*I*A^2 - 261360*A*B +
131769*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log((sq
rt(2)*((360*I*A + 363*B)*a^2*e^(2*I*d*x + 2*I*c) + (360*I*A + 363*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq
rt((-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) + 2*I*sqrt((-129600*I*A^2 - 261360*
A*B + 131769*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((360*I*A + 363*B)*a^2)) - 3*sqrt((-1
29600*I*A^2 - 261360*A*B + 131769*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) + d)*log((sqrt(2)*((360*I*A + 363*B)*a^2*e^(2*I*d*x + 2*I*c) + (360*I*A + 363*B)*a^2)*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) - 2*I*sq
rt((-129600*I*A^2 - 261360*A*B + 131769*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((360*I*A
+ 363*B)*a^2)) - 192*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4
*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*
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)
+ I*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*
a^2)) + 192*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3
*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*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) - I*sqrt
((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(d
*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + 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(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.59308, size = 416, normalized size = 1.4 \begin{align*} \frac{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} + 4 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - 2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2}\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{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} +{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a -{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2}\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 )}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )}}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 2 \, a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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