\(\int \frac {\sqrt {a+b \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{3/2}} \, dx\) [155]

Optimal result

Integrand size = 49, antiderivative size = 299 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {a-i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}-\frac {\sqrt {a+i b} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2} f}+\frac {2 \sqrt {b} C \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]

[Out]

-(I*A+B-I*C)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a-I*b)^(1/2)/
(c-I*d)^(3/2)/f-(B-I*(A-C))*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))
*(a+I*b)^(1/2)/(c+I*d)^(3/2)/f+2*C*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))*b^(1
/2)/d^(3/2)/f-2*(A*d^2-B*c*d+C*c^2)*(a+b*tan(f*x+e))^(1/2)/d/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 3.84 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3726, 3736, 6857, 65, 223, 212, 95, 214} \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {a-i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac {\sqrt {a+i b} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c+i d)^{3/2}}-\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {2 \sqrt {b} C \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f} \]

[In]

Int[(Sqrt[a + b*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

-((Sqrt[a - I*b]*(I*A + B - I*C)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Ta
n[e + f*x]])])/((c - I*d)^(3/2)*f)) - (Sqrt[a + I*b]*(B - I*(A - C))*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e +
f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((c + I*d)^(3/2)*f) + (2*Sqrt[b]*C*ArcTanh[(Sqrt[d]*Sqrt[a
+ b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(d^(3/2)*f) - (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b*Ta
n[e + f*x]])/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

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

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 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 3726

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

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 6.11 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {-a+i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{3/2}}+\frac {i \sqrt {a+i b} (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2}}+\frac {(B+i (A-C)) \sqrt {a+b \tan (e+f x)}}{(c-i d) \sqrt {c+d \tan (e+f x)}}+\frac {(-i A+B+i C) \sqrt {a+b \tan (e+f x)}}{(c+i d) \sqrt {c+d \tan (e+f x)}}+\frac {2 C \left (-\sqrt {d} \sqrt {a+b \tan (e+f x)}+\sqrt {b c-a d} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}\right )}{d^{3/2} \sqrt {c+d \tan (e+f x)}}}{f} \]

[In]

Integrate[(Sqrt[a + b*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((Sqrt[-a + I*b]*(I*A + B - I*C)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*
Tan[e + f*x]])])/(-c + I*d)^(3/2) + (I*Sqrt[a + I*b]*(A + I*B - C)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f
*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(c + I*d)^(3/2) + ((B + I*(A - C))*Sqrt[a + b*Tan[e + f*x]])/
((c - I*d)*Sqrt[c + d*Tan[e + f*x]]) + (((-I)*A + B + I*C)*Sqrt[a + b*Tan[e + f*x]])/((c + I*d)*Sqrt[c + d*Tan
[e + f*x]]) + (2*C*(-(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]]) + Sqrt[b*c - a*d]*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e +
f*x]])/Sqrt[b*c - a*d]]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]))/(d^(3/2)*Sqrt[c + d*Tan[e + f*x]]))/f

Maple [F(-1)]

Timed out.

\[\int \frac {\sqrt {a +b \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

[In]

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69574 vs. \(2 (237) = 474\).

Time = 246.04 (sec) , antiderivative size = 139175, normalized size of antiderivative = 465.47 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a + b \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((a+b*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral(sqrt(a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(c + d*tan(e + f*x))**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (e+f\,x\right )}\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right )}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(((a + b*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + d*tan(e + f*x))^(3/2),x)

[Out]

int(((a + b*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + d*tan(e + f*x))^(3/2), x)