3.3.1 \(\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [201]

Optimal. Leaf size=258 \[ \frac {363 (-1)^{3/4} a^{5/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[Out]

363/64*(-1)^(3/4)*a^(5/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+(4+4*I)*a^(5/
2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-149/64*I*a^2*tan(d*x+c)^(1/2)*(a+I*a*tan
(d*x+c))^(1/2)/d+107/96*a^2*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/d+17/24*I*a^2*(a+I*a*tan(d*x+c))^(1/2)*t
an(d*x+c)^(5/2)/d-1/4*a^2*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(7/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.52, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3637, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {363 (-1)^{3/4} a^{5/2} \text {ArcTan}\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} \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 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(363*(-1)^(3/4)*a^(5/2)*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)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (((149*I)/64)*
a^2*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (107*a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
/(96*d) + (((17*I)/24)*a^2*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (a^2*Tan[c + d*x]^(7/2)*Sqrt[a +
I*a*Tan[c + d*x]])/(4*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 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*} \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {1}{4} a \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {15 a}{2}+\frac {17}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {1}{12} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {85 i a^2}{4}+\frac {107}{4} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-\frac {321 a^3}{8}-\frac {447}{8} i a^3 \tan (c+d x)\right ) \, dx}{24 a}\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {447 i a^4}{16}-\frac {1089}{16} a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{24 a^2}\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {1}{128} (363 i a) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\left (4 i a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\right ) \text {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\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 {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\right ) \text {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} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\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 )}{64 d}\\ &=\frac {363 (-1)^{3/4} a^{5/2} \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} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.39, size = 232, normalized size = 0.90 \begin {gather*} \frac {a^2 \left (\frac {6 e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \left (512 \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-363 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-i \sec ^3(c+d x) (1205 \cos (c+d x)+583 \cos (3 (c+d x))+70 i \sin (c+d x)+262 i \sin (3 (c+d x))) \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{768 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((6*Sqrt[-1 + E^((2*I)*(c + d*x))]*(512*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 363*Sqr
t[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^(
(2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) - I*Sec[c + d*x]^3*(1205*Cos[c + d*x] + 583*Cos[3*(c + d*x)] +
(70*I)*Sin[c + d*x] + (262*I)*Sin[3*(c + d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[a + I*a*Tan[c + d*x]])/(768*d)

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (204 ) = 408\).
time = 0.20, size = 494, normalized size = 1.91

method	result	size
derivativedivides	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-272 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+96 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )+384 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +1089 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +894 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+384 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -428 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+1536 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(494\)
default	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-272 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+96 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )+384 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +1089 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +894 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+384 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -428 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+1536 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(494\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-272*I*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+96*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+
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
))/(tan(d*x+c)+I))*a+1089*I*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)+894*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+384*(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))/(tan(d*x+c)+I)
)*a-428*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+1536*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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 769 vs. \(2 (192) = 384\).
time = 0.41, size = 769, normalized size = 2.98 \begin {gather*} \frac {\sqrt {2} {\left (-845 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 1275 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 1135 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 321 i \, a^{2} 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}} + 96 \, \sqrt {\frac {131769 i \, a^{5}}{4096 \, d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (363 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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}} + 128 i \, \sqrt {\frac {131769 i \, a^{5}}{4096 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{363 \, a^{2}}\right ) - 96 \, \sqrt {\frac {131769 i \, a^{5}}{4096 \, d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (363 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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}} - 128 i \, \sqrt {\frac {131769 i \, a^{5}}{4096 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{363 \, a^{2}}\right ) - 96 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 96 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{192 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(sqrt(2)*(-845*I*a^2*e^(7*I*d*x + 7*I*c) - 1275*I*a^2*e^(5*I*d*x + 5*I*c) - 1135*I*a^2*e^(3*I*d*x + 3*I*
c) - 321*I*a^2*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)) + 96*sqrt(131769/4096*I*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(1/363*(363*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + 128*I*sqrt(131769/4096*I*a^5/d^2)*d*e^(I*d*x + I
*c))*e^(-I*d*x - I*c)/a^2) - 96*sqrt(131769/4096*I*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(1/363*(363*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + 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)) - 128*I*sqrt(131769/4096*I*a^5/d^2)*d*e^
(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) - 96*sqrt(32*I*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(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + 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)) + I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e
^(-I*d*x - I*c)/a^2) + 96*sqrt(32*I*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(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + 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)) - I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3877 deep

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 1.26Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{poly1[-128*i,0]:[1,0,-2]
%%},[0]%%%}

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________