\(\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [193]

Optimal result

Integrand size = 28, antiderivative size = 254 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {23 (-1)^{3/4} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}+\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d} \]

[Out]

23/8*(-1)^(3/4)*a^(3/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+(2+2*I)*a^(3/2)
*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-9/8*I*a*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c)
)^(1/2)/d+7/12*a*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/d+1/3*I*a^2*tan(d*x+c)^(5/2)/d/(a+I*a*tan(d*x+c))^(
1/2)-1/3*a^2*tan(d*x+c)^(7/2)/d/(a+I*a*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3637, 3676, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {23 (-1)^{3/4} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}+\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d} \]

[In]

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

[Out]

(23*(-1)^(3/4)*a^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(8*d) + ((2
+ 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((I/3)*a^2*Tan[c
+ d*x]^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Tan[c + d*x]^(7/2))/(3*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((
(9*I)/8)*a*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (7*a*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x
]])/(12*d)

Rule 65

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 209

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

Rule 211

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

Rule 223

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 3625

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 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3676

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 3678

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 3680

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 3682

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{3} a \int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (\frac {13 a}{2}+\frac {11}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {5 i a^2}{2}-\frac {7}{2} a^2 \tan (c+d x)\right ) \, dx}{3 a} \\ & = \frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {21 a^3}{4}+\frac {27}{4} i a^3 \tan (c+d x)\right ) \, dx}{6 a^2} \\ & = \frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {27 i a^4}{8}+\frac {69}{8} a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a^3} \\ & = \frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {23}{16} i \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+(2 i a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {\left (23 i a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}+\frac {\left (4 a^3\right ) \text {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 {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {\left (23 i a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 d} \\ & = \frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {\left (23 i a^2\right ) \text {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 )}{8 d} \\ & = \frac {23 (-1)^{3/4} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}+\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.39 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.33 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {1+i \tan (c+d x)}}-\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {7 a \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[In]

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

[Out]

(2*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*S
qrt[Tan[c + d*x]]) - (2*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[
c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (7*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c +
d*x]]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d*Sqrt[1 + I*Tan[c + d*x]]) - (((9*I)/8)*a*Sqrt[Tan[c + d*x]]*Sqrt[a +
I*a*Tan[c + d*x]])/d + (7*a*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(12*d) + ((I/3)*a*Tan[c + d*x]^(5/2
)*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (200 ) = 400\).

Time = 1.07 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.76

method	result	size
derivativedivides	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-16 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+24 i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +69 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +24 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +54 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-28 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+96 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{48 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(447\)
default	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-16 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+24 i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +69 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +24 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +54 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-28 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+96 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{48 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(447\)

[In]

int(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-16*I*(-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)^2+24*I*2^(1/2)*(I*a)^(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+69*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a+24*2^(1/2)*(I*a)^(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+54*I*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)-28*(-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)+96*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))/(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (188) = 376\).

Time = 0.27 (sec) , antiderivative size = 673, normalized size of antiderivative = 2.65 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (-49 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} - 38 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 21 i \, a e^{\left (i \, d x + i \, c\right )}\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}} + 12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {529 i \, a^{3}}{64 \, d^{2}}} \log \left (\frac {{\left (23 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} + 16 i \, \sqrt {\frac {529 i \, a^{3}}{64 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{23 \, a}\right ) - 12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {529 i \, a^{3}}{64 \, d^{2}}} \log \left (\frac {{\left (23 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} - 16 i \, \sqrt {\frac {529 i \, a^{3}}{64 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{23 \, a}\right ) - 12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{24 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

1/24*(sqrt(2)*(-49*I*a*e^(5*I*d*x + 5*I*c) - 38*I*a*e^(3*I*d*x + 3*I*c) - 21*I*a*e^(I*d*x + I*c))*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)) + 12*(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(529/64*I*a^3/d^2)*log(1/23*(23*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) + 16*I*sqrt(529/64*I*
a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqr
t(529/64*I*a^3/d^2)*log(1/23*(23*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) - 16*I*sqrt(529/64*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d
*x - I*c)/a) - 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(
a*e^(2*I*d*x + 2*I*c) + a)*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)) + I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(
2*I*d*x + 2*I*c) + d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) - I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c
))*e^(-I*d*x - I*c)/a))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0]Warning, replacing 0 by -35, a substitution variable should perhaps be pu
rged.Warnin

Mupad [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[In]

int(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2), x)