\(\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx\) [584]

Optimal result

Integrand size = 23, antiderivative size = 250 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(a+b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}-\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 \sqrt {\tan (c+d x)}}{b d} \]

[Out]

-2*a^(5/2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(3/2)/(a^2+b^2)/d-1/2*(a+b)*arctan(-1+2^(1/2)*tan(d*x+c)
^(1/2))/(a^2+b^2)/d*2^(1/2)-1/2*(a+b)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-1/4*(a-b)*ln(1-2^
(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)/d*2^(1/2)+1/4*(a-b)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^
2+b^2)/d*2^(1/2)+2*tan(d*x+c)^(1/2)/b/d

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3647, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(a+b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {(a+b) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {(a-b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {(a-b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2} d \left (a^2+b^2\right )}+\frac {2 \sqrt {\tan (c+d x)}}{b d} \]

[In]

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

[Out]

((a + b)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - ((a + b)*ArcTan[1 + Sqrt[2]*Sqrt[Ta
n[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*a^(5/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(b^(3/2)*(a^2
+ b^2)*d) - ((a - b)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + ((a - b)
*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + (2*Sqrt[Tan[c + d*x]])/(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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3647

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

Rule 3715

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

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.62 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\sqrt [4]{-1} b^{3/2} (-i a+b) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 a^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt [4]{-1} b^{3/2} (i a+b) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 a^2 \sqrt {b} \sqrt {\tan (c+d x)}+2 b^{5/2} \sqrt {\tan (c+d x)}}{b^{3/2} \left (a^2+b^2\right ) d} \]

[In]

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

[Out]

((-1)^(1/4)*b^(3/2)*((-I)*a + b)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 2*a^(5/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c
+ d*x]])/Sqrt[a]] + (-1)^(1/4)*b^(3/2)*(I*a + b)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*a^2*Sqrt[b]*Sqrt[T
an[c + d*x]] + 2*b^(5/2)*Sqrt[Tan[c + d*x]])/(b^(3/2)*(a^2 + b^2)*d)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b}+\frac {-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{3} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{b \left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\)	\(241\)
default	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b}+\frac {-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{3} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{b \left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\)	\(241\)

[In]

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

[Out]

1/d*(2/b*tan(d*x+c)^(1/2)+2/(a^2+b^2)*(-1/8*b*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*t
an(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))-1/8*a
*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/
2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))-2/b*a^3/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)
^(1/2)/(a*b)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1385 vs. \(2 (210) = 420\).

Time = 0.32 (sec) , antiderivative size = 2796, normalized size of antiderivative = 11.18 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/2*((a^2*b + b^3)*d*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*
a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sq
rt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^2*b - b^3)*d)*sqrt(-((
a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)
) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - (a^2*b + b^3)*d*sqrt(-((a^4 + 2*
a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*
b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(-((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*
a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^2*b - b^3)*d)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4
- 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^
2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - (a^2*b + b^3)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b
^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log((
(a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d
^4)) + (a^2*b - b^3)*d)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*
a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) + (
a^2*b + b^3)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 +
4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(-((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4
- 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + (a^2*b - b^3)*d)*sqrt(((a^4 + 2*a
^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b
)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - 2*a^2*sqrt(-a/b)*log(-(2*b*sqrt(-a/b)*sqr
t(tan(d*x + c)) - b*tan(d*x + c) + a)/(b*tan(d*x + c) + a)) - 4*(a^2 + b^2)*sqrt(tan(d*x + c)))/((a^2*b + b^3)
*d), -1/2*(4*a^2*sqrt(a/b)*arctan(b*sqrt(a/b)*sqrt(tan(d*x + c))/a) + (a^2*b + b^3)*d*sqrt(-((a^4 + 2*a^2*b^2
+ b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4
+ 2*a^2*b^2 + b^4)*d^2))*log(((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 +
6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^2*b - b^3)*d)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b
^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^
2 - b^2)*sqrt(tan(d*x + c))) - (a^2*b + b^3)*d*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4
)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(-((a^5 +
2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) -
(a^2*b - b^3)*d)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^
4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - (a^2*b
+ b^3)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2
*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^
2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + (a^2*b - b^3)*d)*sqrt(((a^4 + 2*a^2*b^2
+ b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4
+ 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) + (a^2*b + b^3)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^
2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*
b^2 + b^4)*d^2))*log(-((a^5 + 2*a^3*b^2 + a*b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b
^4 + 4*a^2*b^6 + b^8)*d^4)) + (a^2*b - b^3)*d)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)
/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*
sqrt(tan(d*x + c))) - 4*(a^2 + b^2)*sqrt(tan(d*x + c)))/((a^2*b + b^3)*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {8 \, a^{3} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{2} + b^{2}} - \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b}}{4 \, d} \]

[In]

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

[Out]

-1/4*(8*a^3*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2*b + b^3)*sqrt(a*b)) + (2*sqrt(2)*(a + b)*arctan(1/2*s
qrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c
)))) - sqrt(2)*(a - b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a - b)*log(-sqrt(2)*sqrt(
tan(d*x + c)) + tan(d*x + c) + 1))/(a^2 + b^2) - 8*sqrt(tan(d*x + c))/b)/d

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 3693, normalized size of antiderivative = 14.77 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(log((32*a^3*(a^2 - b^2))/(b*d^5) - (((((((((128*a*b^3*(a^2 + b^2)^2)/d - 256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^
2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^2))^(1/2))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (64*a*tan(c + d*x)^(1/2)*(8
*a^6 - 7*b^6 + 2*a^2*b^4 + a^4*b^2))/d^2)*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*a^2*(12*a^4 + b^4 - 15*a^2*b^
2))/d^3)*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4))*(1i/(d^2*(a*1i - b)
^2))^(1/2))/2)*(-1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2))/2 - log((32*a^3*(a^2 - b^2))/(b*d^5) - ((((((
(((128*a*b^3*(a^2 + b^2)^2)/d + 256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^2))^(
1/2))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 + (64*a*tan(c + d*x)^(1/2)*(8*a^6 - 7*b^6 + 2*a^2*b^4 + a^4*b^2))/d^2)*
(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*a^2*(12*a^4 + b^4 - 15*a^2*b^2))/d^3)*(1i/(d^2*(a*1i - b)^2))^(1/2))/2
+ (32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2)*(-1/(4*(b^2*d^2*1i - a^2*d^
2*1i + 2*a*b*d^2)))^(1/2) + atan((((-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*((((32*(4*a*b^8*d^4 + 8*a^
3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) - (32*tan(c + d*x)^(1/2)*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*
(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^
2*2i)))^(1/2) + (32*tan(c + d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*b^3*d^2))/(b*d^4))
*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (32*(12*a^6*b*d^2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5)
) + (32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*1i - ((-1i
/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*((((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) + (
32*tan(c + d*x)^(1/2)*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*
b^6*d^4 - 16*a^6*b^4*d^4))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (32*tan(c + d*x)^(1/2)*
(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*b^3*d^2))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2
i)))^(1/2) - (32*(12*a^6*b*d^2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5)) - (32*tan(c + d*x)^(1/2)*(2*a^6 - b^6
))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*1i)/(((-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^
(1/2)*((((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) - (32*tan(c + d*x)^(1/2)*(-1i/(4*(b^2*d^2
- a^2*d^2 + a*b*d^2*2i)))^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b*d^4))*(-1
i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) + (32*tan(c + d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^
5*d^2 + 2*a^5*b^3*d^2))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (32*(12*a^6*b*d^2 + a^2*b^
5*d^2 - 15*a^4*b^3*d^2))/(b*d^5)) + (32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2
+ a*b*d^2*2i)))^(1/2) + ((-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*((((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^4
+ 4*a^5*b^4*d^4))/(b*d^5) + (32*tan(c + d*x)^(1/2)*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*(16*b^10*d
^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1
/2) - (32*tan(c + d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*b^3*d^2))/(b*d^4))*(-1i/(4*(
b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (32*(12*a^6*b*d^2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5)) - (32*ta
n(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (64*(a^5 - a^3*b^2
))/(b*d^5)))*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*2i + (2*tan(c + d*x)^(1/2))/(b*d) - (atan((((-a^
5*b^3)^(1/2)*((32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4) + ((-a^5*b^3)^(1/2)*((32*(12*a^6*b*d^2 + a^2*b^5*d
^2 - 15*a^4*b^3*d^2))/(b*d^5) + (((32*tan(c + d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*
b^3*d^2))/(b*d^4) - (((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) + (32*tan(c + d*x)^(1/2)*(-a^
5*b^3)^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b^4*d^5*(a^2 + b^2)))*(-a^5*b^
3)^(1/2))/(b^3*d*(a^2 + b^2)))*(-a^5*b^3)^(1/2))/(b^3*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2)))*1i)/(b^3*d*(a^2 +
b^2)) + ((-a^5*b^3)^(1/2)*((32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(b*d^4) - ((-a^5*b^3)^(1/2)*((32*(12*a^6*b*d^
2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5) - (((32*tan(c + d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5
*d^2 + 2*a^5*b^3*d^2))/(b*d^4) + (((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) - (32*tan(c + d*
x)^(1/2)*(-a^5*b^3)^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b^4*d^5*(a^2 + b^
2)))*(-a^5*b^3)^(1/2))/(b^3*d*(a^2 + b^2)))*(-a^5*b^3)^(1/2))/(b^3*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2)))*1i)/(
b^3*d*(a^2 + b^2)))/((64*(a^5 - a^3*b^2))/(b*d^5) + ((-a^5*b^3)^(1/2)*((32*tan(c + d*x)^(1/2)*(2*a^6 - b^6))/(
b*d^4) + ((-a^5*b^3)^(1/2)*((32*(12*a^6*b*d^2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5) + (((32*tan(c + d*x)^(1
/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*b^3*d^2))/(b*d^4) - (((32*(4*a*b^8*d^4 + 8*a^3*b^6*d^
4 + 4*a^5*b^4*d^4))/(b*d^5) + (32*tan(c + d*x)^(1/2)*(-a^5*b^3)^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b
^6*d^4 - 16*a^6*b^4*d^4))/(b^4*d^5*(a^2 + b^2)))*(-a^5*b^3)^(1/2))/(b^3*d*(a^2 + b^2)))*(-a^5*b^3)^(1/2))/(b^3
*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2)) - ((-a^5*b^3)^(1/2)*((32*tan(c + d*x)^(1/2)*(2*a^6
- b^6))/(b*d^4) - ((-a^5*b^3)^(1/2)*((32*(12*a^6*b*d^2 + a^2*b^5*d^2 - 15*a^4*b^3*d^2))/(b*d^5) - (((32*tan(c
+ d*x)^(1/2)*(16*a^7*b*d^2 - 14*a*b^7*d^2 + 4*a^3*b^5*d^2 + 2*a^5*b^3*d^2))/(b*d^4) + (((32*(4*a*b^8*d^4 + 8*
a^3*b^6*d^4 + 4*a^5*b^4*d^4))/(b*d^5) - (32*tan(c + d*x)^(1/2)*(-a^5*b^3)^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4
- 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b^4*d^5*(a^2 + b^2)))*(-a^5*b^3)^(1/2))/(b^3*d*(a^2 + b^2)))*(-a^5*b^3)^(
1/2))/(b^3*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2))))/(b^3*d*(a^2 + b^2))))*(-a^5*b^3)^(1/2)*2i)/(b^3*d*(a^2 + b^2
))