\(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [520]

Optimal result

Integrand size = 23, antiderivative size = 158 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d} \]

[Out]

I*(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-I*(a+I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)
/(a+I*b)^(1/2))/d-4*a*b*(a+b*tan(d*x+c))^(1/2)/d-2/3*b*(a+b*tan(d*x+c))^(3/2)/d+2/7*(a+b*tan(d*x+c))^(7/2)/b/d

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3624, 3563, 3609, 3620, 3618, 65, 214} \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d} \]

[In]

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

[Out]

(I*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d - (I*(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*
Tan[c + d*x]]/Sqrt[a + I*b]])/d - (4*a*b*Sqrt[a + b*Tan[c + d*x]])/d - (2*b*(a + b*Tan[c + d*x])^(3/2))/(3*d)
+ (2*(a + b*Tan[c + d*x])^(7/2))/(7*b*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 214

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

Rule 3563

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\int (a+b \tan (c+d x))^{5/2} \, dx \\ & = -\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\int \sqrt {a+b \tan (c+d x)} \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx \\ & = -\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {1}{2} (a-i b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {(i a-b)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac {(i a+b)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d} \\ & = -\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {(a-i b)^3 \text {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {(a+i b)^3 \text {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d} \\ & = \frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.30 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {21 i (a-i b)^{5/2} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-21 i (a+i b)^{5/2} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {1}{2} \sec ^3(c+d x) \left (3 a \left (3 a^2-46 b^2\right ) \cos (c+d x)+\left (3 a^3-58 a b^2\right ) \cos (3 (c+d x))-2 b \left (-9 a^2+4 b^2+\left (-9 a^2+10 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right ) \sqrt {a+b \tan (c+d x)}}{21 b d} \]

[In]

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

[Out]

((21*I)*(a - I*b)^(5/2)*b*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - (21*I)*(a + I*b)^(5/2)*b*ArcTanh[S
qrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + (Sec[c + d*x]^3*(3*a*(3*a^2 - 46*b^2)*Cos[c + d*x] + (3*a^3 - 58*a*b^
2)*Cos[3*(c + d*x)] - 2*b*(-9*a^2 + 4*b^2 + (-9*a^2 + 10*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])*Sqrt[a + b*Tan[c
+ d*x]])/2)/(21*b*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1193\) vs. \(2(130)=260\).

Time = 0.11 (sec) , antiderivative size = 1194, normalized size of antiderivative = 7.56

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1194\)
default	\(\text {Expression too large to display}\)	\(1194\)

[In]

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

[Out]

2/7*(a+b*tan(d*x+c))^(7/2)/b/d-2/3*b*(a+b*tan(d*x+c))^(3/2)/d-4*a*b*(a+b*tan(d*x+c))^(1/2)/d-1/4/d/b*ln((a+b*t
an(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(
a^2+b^2)^(1/2)*a^2+1/4/d*b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1
/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(
1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-3/4/d*b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a
^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/d*b/(2*(a^2+b^2)^(1
/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*
(a^2+b^2)^(1/2)*a+3/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))
^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*
a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(
1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2-1/4/d*b*
ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a
)^(1/2)*(a^2+b^2)^(1/2)-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^
2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+3/4/d*b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2
)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a
+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)*a-3/d*b/(2*
(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-
2*a)^(1/2))*a^2+1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)
^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1175 vs. \(2 (124) = 248\).

Time = 0.26 (sec) , antiderivative size = 1175, normalized size of antiderivative = 7.44 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/42*(21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^
8 + b^10)/d^4))/d^2)*log((5*a^8*b - 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a^2 - b^2)*d^3*
sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + 2*(5*a^5*b^2 - 10*a^3*b^4 + a*b^6)*d
)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d
^4))/d^2)) - 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20
*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8*b - 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) - ((a^2 - b^
2)*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + 2*(5*a^5*b^2 - 10*a^3*b^4 + a
*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 +
b^10)/d^4))/d^2)) - 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b
^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8*b - 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a
^2 - b^2)*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - 2*(5*a^5*b^2 - 10*a^3*
b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2
*b^8 + b^10)/d^4))/d^2)) + 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 11
0*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8*b - 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a
) - ((a^2 - b^2)*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - 2*(5*a^5*b^2 -
10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 -
20*a^2*b^8 + b^10)/d^4))/d^2)) + 4*(3*b^3*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^2 + 3*a^3 - 49*a*b^2 + (9*a^2
*b - 7*b^3)*tan(d*x + c))*sqrt(b*tan(d*x + c) + a))/(b*d)

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*tan(c + d*x))**(5/2)*tan(c + d*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 27.91 (sec) , antiderivative size = 2165, normalized size of antiderivative = 13.70 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

atan(((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^
5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10
*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5
*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(8*a*b^5*d^2 + 8*a^3*b^3
*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b
^2)/(4*d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a +
b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i +
a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c
+ d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 - a^
4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^
2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^
(1/2) - (16*(6*a^4*b^7 - b^11 + 8*a^6*b^5 + 3*a^8*b^3))/d^3 + (((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 + 64*a*b
^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))
*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1
/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^
3*b^2)/(4*d^2))^(1/2)))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*2i + a
tan(((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 + a^4*b*5i + a^5
+ b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*
a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*
a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(8*a*b^5*d^2 + 8*a^3*b^3*
d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^
2)/(4*d^2))^(1/2))*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a +
b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a
^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c +
d*x))^(1/2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 + a^4
*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2
*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(
1/2) - (16*(6*a^4*b^7 - b^11 + 8*a^6*b^5 + 3*a^8*b^3))/d^3 + (((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 + 64*a*b^
2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*
(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/
2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3
*b^2)/(4*d^2))^(1/2)))*(-(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*2i + ((
2*a^2)/(3*b*d) - (2*(a^2 + b^2))/(3*b*d))*(a + b*tan(c + d*x))^(3/2) + (2*(a + b*tan(c + d*x))^(7/2))/(7*b*d)
+ 2*a*((2*a^2)/(b*d) - (2*(a^2 + b^2))/(b*d))*(a + b*tan(c + d*x))^(1/2)