3.13.44 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx\) [1244]

Optimal. Leaf size=195 \[ \frac {(c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {(c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f} \]

[Out]

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

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Rubi [A]
time = 0.61, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3647, 3734, 3620, 3618, 65, 214, 3715} \begin {gather*} -\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )}+\frac {(c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a)}-\frac {(c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (-b+i a)}+\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)*f) - ((c + I*d)^(5/2)*ArcTanh[Sqr
t[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)*f) - (2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e +
f*x]])/Sqrt[b*c - a*d]])/(b^(3/2)*(a^2 + b^2)*f) + (2*d^2*Sqrt[c + d*Tan[e + f*x]])/(b*f)

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 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 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*} \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx &=\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {2 \int \frac {\frac {1}{2} \left (b c^3-a d^3\right )+\frac {1}{2} b d \left (3 c^2-d^2\right ) \tan (e+f x)+\frac {1}{2} d^2 (3 b c-a d) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{b}\\ &=\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {2 \int \frac {\frac {1}{2} b \left (a c^3+3 b c^2 d-3 a c d^2-b d^3\right )+\frac {1}{2} b \left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )}+\frac {(b c-a d)^3 \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {(c-i d)^3 \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)}+\frac {(c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)}+\frac {(b c-a d)^3 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{b \left (a^2+b^2\right ) f}\\ &=\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}-\frac {\left (i (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b) f}-\frac {(i c+d)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b) f}+\frac {\left (2 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{b \left (a^2+b^2\right ) d f}\\ &=-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}-\frac {(c-i d)^3 \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a-i b) d f}-\frac {(c+i d)^3 \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(a+i b) d f}\\ &=\frac {(c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {(c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 199, normalized size = 1.02 \begin {gather*} \frac {b^{3/2} (-i a+b) (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+b^{3/2} (i a+b) (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )+2 \sqrt {b} \left (a^2+b^2\right ) d^2 \sqrt {c+d \tan (e+f x)}}{b^{3/2} \left (a^2+b^2\right ) f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1277\) vs. \(2(165)=330\).
time = 0.56, size = 1278, normalized size = 6.55 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [B]
time = 13.25, size = 2500, normalized size = 12.82 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^2*(c + d*tan(e + f*x))^(1/2))/(b*f) - atan(((((((32*(4*b^9*c*d^12*f^4 - 4*a*b^8*d^13*f^4 - 8*a^3*b^6*d^13
*f^4 - 4*a^5*b^4*d^13*f^4 + 16*b^9*c^3*d^10*f^4 + 12*b^9*c^5*d^8*f^4 + 40*a^2*b^7*c^3*d^10*f^4 + 24*a^2*b^7*c^
5*d^8*f^4 - 48*a^3*b^6*c^2*d^11*f^4 - 40*a^3*b^6*c^4*d^9*f^4 + 32*a^4*b^5*c^3*d^10*f^4 + 12*a^4*b^5*c^5*d^8*f^
4 - 24*a^5*b^4*c^2*d^11*f^4 - 20*a^5*b^4*c^4*d^9*f^4 + 8*a^6*b^3*c^3*d^10*f^4 - 24*a*b^8*c^2*d^11*f^4 - 20*a*b
^8*c^4*d^9*f^4 + 16*a^2*b^7*c*d^12*f^4 + 20*a^4*b^5*c*d^12*f^4 + 8*a^6*b^3*c*d^12*f^4))/(b*f^5) - (32*(c + d*t
an(e + f*x))^(1/2)*(-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1
i + 2*a*b*f^2)))^(1/2)*(16*b^10*d^10*f^4 + 16*a^2*b^8*d^10*f^4 - 16*a^4*b^6*d^10*f^4 - 16*a^6*b^4*d^10*f^4 + 2
4*b^10*c^2*d^8*f^4 + 40*a^2*b^8*c^2*d^8*f^4 + 8*a^4*b^6*c^2*d^8*f^4 - 8*a^6*b^4*c^2*d^8*f^4 + 8*a*b^9*c*d^9*f^
4 + 24*a^3*b^7*c*d^9*f^4 + 24*a^5*b^5*c*d^9*f^4 + 8*a^7*b^3*c*d^9*f^4))/(b*f^4))*(-(c*d^4*5i + 5*c^4*d + c^5*1
i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2*a*b*f^2)))^(1/2) + (32*(c + d*tan(e + f*x)
)^(1/2)*(16*a^7*b*d^15*f^2 - 14*a*b^7*d^15*f^2 - 8*a^8*c*d^14*f^2 + 38*b^8*c*d^14*f^2 + 4*a^3*b^5*d^15*f^2 + 2
*a^5*b^3*d^15*f^2 - 10*b^8*c^3*d^12*f^2 - 102*b^8*c^5*d^10*f^2 + 18*b^8*c^7*d^8*f^2 + 100*a^2*b^6*c^3*d^12*f^2
+ 36*a^2*b^6*c^5*d^10*f^2 - 12*a^2*b^6*c^7*d^8*f^2 - 60*a^3*b^5*c^2*d^13*f^2 + 140*a^3*b^5*c^4*d^11*f^2 + 44*
a^3*b^5*c^6*d^9*f^2 - 170*a^4*b^4*c^3*d^12*f^2 - 150*a^4*b^4*c^5*d^10*f^2 + 2*a^4*b^4*c^7*d^8*f^2 + 162*a^5*b^
3*c^2*d^13*f^2 + 190*a^5*b^3*c^4*d^11*f^2 - 2*a^5*b^3*c^6*d^9*f^2 - 120*a^6*b^2*c^3*d^12*f^2 + 114*a*b^7*c^2*d
^13*f^2 + 110*a*b^7*c^4*d^11*f^2 - 114*a*b^7*c^6*d^9*f^2 - 44*a^2*b^6*c*d^14*f^2 - 2*a^4*b^4*c*d^14*f^2 - 88*a
^6*b^2*c*d^14*f^2 + 48*a^7*b*c^2*d^13*f^2))/(b*f^4))*(-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d
^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2*a*b*f^2)))^(1/2) + (32*(12*a^6*b*d^18*f^2 + 8*a^7*c*d^17*f^2 + a^2*b^5
*d^18*f^2 - 15*a^4*b^3*d^18*f^2 + 8*a^7*c^3*d^15*f^2 - 3*b^7*c^2*d^16*f^2 - 48*b^7*c^4*d^14*f^2 + 30*b^7*c^6*d
^12*f^2 + 72*b^7*c^8*d^10*f^2 - 3*b^7*c^10*d^8*f^2 - 171*a^2*b^5*c^2*d^16*f^2 + 558*a^2*b^5*c^4*d^14*f^2 + 522
*a^2*b^5*c^6*d^12*f^2 - 207*a^2*b^5*c^8*d^10*f^2 + a^2*b^5*c^10*d^8*f^2 - 640*a^3*b^4*c^3*d^15*f^2 - 372*a^3*b
^4*c^5*d^13*f^2 + 360*a^3*b^4*c^7*d^11*f^2 + 2*a^3*b^4*c^9*d^9*f^2 + 372*a^4*b^3*c^2*d^16*f^2 + 42*a^4*b^3*c^4
*d^14*f^2 - 348*a^4*b^3*c^6*d^12*f^2 - 3*a^4*b^3*c^8*d^10*f^2 + 88*a^5*b^2*c^3*d^15*f^2 + 192*a^5*b^2*c^5*d^13
*f^2 + 2*a*b^6*c*d^17*f^2 + 144*a*b^6*c^3*d^15*f^2 - 228*a*b^6*c^5*d^13*f^2 - 312*a*b^6*c^7*d^11*f^2 + 58*a*b^
6*c^9*d^9*f^2 + 90*a^3*b^4*c*d^17*f^2 - 104*a^5*b^2*c*d^17*f^2 - 48*a^6*b*c^2*d^16*f^2 - 60*a^6*b*c^4*d^14*f^2
))/(b*f^5))*(-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2*a
*b*f^2)))^(1/2) - (32*(c + d*tan(e + f*x))^(1/2)*(b^6*d^20 - 2*a^6*d^20 + 30*a^6*c^2*d^18 - 30*a^6*c^4*d^16 +
2*a^6*c^6*d^14 + 6*b^6*c^2*d^18 + 15*b^6*c^4*d^16 + 18*b^6*c^6*d^14 + 45*b^6*c^8*d^12 - 24*b^6*c^10*d^10 + 3*b
^6*c^12*d^8 + 12*a*b^5*c^5*d^15 - 180*a*b^5*c^7*d^13 + 180*a*b^5*c^9*d^11 - 12*a*b^5*c^11*d^9 - 180*a^5*b*c^3*
d^17 + 180*a^5*b*c^5*d^15 - 12*a^5*b*c^7*d^13 - 30*a^2*b^4*c^4*d^16 + 450*a^2*b^4*c^6*d^14 - 450*a^2*b^4*c^8*d
^12 + 30*a^2*b^4*c^10*d^10 + 40*a^3*b^3*c^3*d^17 - 600*a^3*b^3*c^5*d^15 + 600*a^3*b^3*c^7*d^13 - 40*a^3*b^3*c^
9*d^11 - 30*a^4*b^2*c^2*d^18 + 450*a^4*b^2*c^4*d^16 - 450*a^4*b^2*c^6*d^14 + 30*a^4*b^2*c^8*d^12 + 12*a^5*b*c*
d^19))/(b*f^4))*(-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i +
2*a*b*f^2)))^(1/2)*1i - (((((32*(4*b^9*c*d^12*f^4 - 4*a*b^8*d^13*f^4 - 8*a^3*b^6*d^13*f^4 - 4*a^5*b^4*d^13*f^
4 + 16*b^9*c^3*d^10*f^4 + 12*b^9*c^5*d^8*f^4 + 40*a^2*b^7*c^3*d^10*f^4 + 24*a^2*b^7*c^5*d^8*f^4 - 48*a^3*b^6*c
^2*d^11*f^4 - 40*a^3*b^6*c^4*d^9*f^4 + 32*a^4*b^5*c^3*d^10*f^4 + 12*a^4*b^5*c^5*d^8*f^4 - 24*a^5*b^4*c^2*d^11*
f^4 - 20*a^5*b^4*c^4*d^9*f^4 + 8*a^6*b^3*c^3*d^10*f^4 - 24*a*b^8*c^2*d^11*f^4 - 20*a*b^8*c^4*d^9*f^4 + 16*a^2*
b^7*c*d^12*f^4 + 20*a^4*b^5*c*d^12*f^4 + 8*a^6*b^3*c*d^12*f^4))/(b*f^5) + (32*(c + d*tan(e + f*x))^(1/2)*(-(c*
d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2*a*b*f^2)))^(1/2)*(
16*b^10*d^10*f^4 + 16*a^2*b^8*d^10*f^4 - 16*a^4*b^6*d^10*f^4 - 16*a^6*b^4*d^10*f^4 + 24*b^10*c^2*d^8*f^4 + 40*
a^2*b^8*c^2*d^8*f^4 + 8*a^4*b^6*c^2*d^8*f^4 - 8*a^6*b^4*c^2*d^8*f^4 + 8*a*b^9*c*d^9*f^4 + 24*a^3*b^7*c*d^9*f^4
+ 24*a^5*b^5*c*d^9*f^4 + 8*a^7*b^3*c*d^9*f^4))/(b*f^4))*(-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c
^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2*a*b*f^2)))^(1/2) - (32*(c + d*tan(e + f*x))^(1/2)*(16*a^7*b*d^15*f
^2 - 14*a*b^7*d^15*f^2 - 8*a^8*c*d^14*f^2 + 38*b^8*c*d^14*f^2 + 4*a^3*b^5*d^15*f^2 + 2*a^5*b^3*d^15*f^2 - 10*b
^8*c^3*d^12*f^2 - 102*b^8*c^5*d^10*f^2 + 18*b^8*c^7*d^8*f^2 + 100*a^2*b^6*c^3*d^12*f^2 + 36*a^2*b^6*c^5*d^10*f
^2 - 12*a^2*b^6*c^7*d^8*f^2 - 60*a^3*b^5*c^2*d^...

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