3.13.43 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\) [1243]

Optimal. Leaf size=188 \[ -\frac {(i a+b) (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (b c+a d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.30, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3609, 3620, 3618, 65, 214} \begin {gather*} \frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {(b+i a) (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 1.13, size = 233, normalized size = 1.24 \begin {gather*} \frac {i \left ((a-i b) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c-i d) \left (-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )\right )-(a+i b) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c+i d) \left (-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )\right )\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1250\) vs. \(2(160)=320\).
time = 0.45, size = 1251, normalized size = 6.65 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2/5*b*(c+d*tan(f*x+e))^(5/2)+2/3*a*d*(c+d*tan(f*x+e))^(3/2)+2/3*b*c*(c+d*tan(f*x+e))^(3/2)+4*a*c*d*(c+d*t
an(f*x+e))^(1/2)+2*b*c^2*(c+d*tan(f*x+e))^(1/2)-2*b*d^2*(c+d*tan(f*x+e))^(1/2)+1/2/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)))+1/2/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))))

________________________________________________________________________________________

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((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2),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(d-c>0)', see `assume?` for mor
e details)Is

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

Mupad [B]
time = 34.41, size = 2500, normalized size = 13.30 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(- ((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^
2)/f^4)^(1/2)*(((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c
^3*d^2*f^2)/f^4)^(1/2)*(32*b*d^6 - 32*b*c^4*d^2 + 32*c*d^2*f*(((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/
2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*b
^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*
(c^2 + d^2)^3)/f^3)*((20*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6*d^4*f^4 - 25*b^4*c
^8*d^2*f^4)^(1/2)/(4*f^4) + (b^2*c^5)/(4*f^2) + (5*b^2*c*d^4)/(4*f^2) - (5*b^2*c^3*d^2)/(2*f^2))^(1/2) - log((
(((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)
^(1/2)*(((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*
f^2)/f^4)^(1/2)*(32*b*c^4*d^2 - 32*b*d^6 + 32*c*d^2*f*(((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^
2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*b^2*d^2*
(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*(c^2 +
d^2)^3)/f^3)*(((20*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6*d^4*f^4 - 25*b^4*c^8*d^2
*f^4)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/(4*f^4))^(1/2) - log(((-((-b^4*d^2*f^4*(5*c^
4 + d^4 - 10*c^2*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(((-((-b^4*d^2
*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(32*
b*c^4*d^2 - 32*b*d^6 + 32*c*d^2*f*(-((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c
*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*b^2*d^2*(c + d*tan(e + f*x)
)^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3)*(-((20
*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6*d^4*f^4 - 25*b^4*c^8*d^2*f^4)^(1/2) - b^2*
c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/(4*f^4))^(1/2) + log(- ((-((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2
*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(((-((-b^4*d^2*f^4*(5*c^4 + d^
4 - 10*c^2*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(32*b*d^6 - 32*b*c^4
*d^2 + 32*c*d^2*f*(-((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - b^2*c^5*f^2 - 5*b^2*c*d^4*f^2 + 10*b^
2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*b^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d
^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3)*((b^2*c^5)/(4*f^2) - (
20*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6*d^4*f^4 - 25*b^4*c^8*d^2*f^4)^(1/2)/(4*f
^4) + (5*b^2*c*d^4)/(4*f^2) - (5*b^2*c^3*d^2)/(2*f^2))^(1/2) - log(((-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2
)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((-((-a^4*d^2*f^4*(5*c^4 + d^4 -
10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d
^5 + 32*c*d^2*f*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*
c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6
+ 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*(-((20*a^4*c^2*d^8*f^4 - a
^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^
4*f^2 - 10*a^2*c^3*d^2*f^2)/(4*f^4))^(1/2) - log(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^
5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2)
- a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 + 32*c*d^2*f*(((-
a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/
2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^
4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*(((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*
c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2
*f^2)/(4*f^4))^(1/2) + log(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^
2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2
*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 - 32*c*d^2*f*(((-a^4*d^2*f^4*(5*c^4 + d
^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x)
)^(1/2)))/(2*f) - (16*a^2*d^2*(c + d*tan(e + f*...

________________________________________________________________________________________