3.5.28 \(\int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [428]

Optimal. Leaf size=201 \[ \frac {\sqrt {i a-b} (A+i B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(2 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {\sqrt {i a+b} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \]

[Out]

(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2)/d+(2*A*b+B*a)*arctanh(b^(1
/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/b^(1/2)-(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan
(d*x+c))^(1/2))*(I*a+b)^(1/2)/d+B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d

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Rubi [A]
time = 1.02, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3691, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\sqrt {-b+i a} (A+i B) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {\sqrt {b+i a} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((2*A*b + a*
B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[b]*d) - (Sqrt[I*a + b]*(A - I*B)*ArcT
anh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c +
d*x]])/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 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 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 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 3691

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] +
Dist[1/(m + n), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m +
a*d*n) + (A*b*c + a*B*c + a*A*d - b*B*d)*(m + n)*Tan[e + f*x] + (A*b*d*(m + n) + B*(a*d*m + b*c*n))*Tan[e + f
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 +
d^2, 0] && LtQ[0, m, 1] && LtQ[0, 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*} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\int \frac {-\frac {a B}{2}+(a A-b B) \tan (c+d x)+\frac {1}{2} (2 A b+a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \frac {-\frac {a B}{2}+(a A-b B) x+\frac {1}{2} (2 A b+a B) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \left (\frac {2 A b+a B}{2 \sqrt {x} \sqrt {a+b x}}-\frac {A b+a B-(a A-b B) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\text {Subst}\left (\int \frac {A b+a B-(a A-b B) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(2 A b+a B) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\text {Subst}\left (\int \left (\frac {a A-b B+i (A b+a B)}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-a A+b B+i (A b+a B)}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {(2 A b+a B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(2 A b+a B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {(2 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {i a-b} (A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(2 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {\sqrt {i a+b} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\\ \end {align*}

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Mathematica [A]
time = 1.85, size = 239, normalized size = 1.19 \begin {gather*} \frac {\sqrt [4]{-1} \sqrt {-a+i b} (A-i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} \sqrt {a+i b} (A+i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {(2 A b+a B) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+b \tan (c+d x)}}{\sqrt {a} \sqrt {b} \sqrt {1+\frac {b \tan (c+d x)}{a}}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

((-1)^(1/4)*Sqrt[-a + I*b]*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c +
d*x]]] + (-1)^(1/4)*Sqrt[a + I*b]*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Ta
n[c + d*x]]] + B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + ((2*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*
x]])/Sqrt[a]]*Sqrt[a + b*Tan[c + d*x]])/(Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c + d*x])/a]))/d

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 296.35, size = 2178538, normalized size = 10838.50

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2178538\)
default	\(\text {Expression too large to display}\)	\(2178538\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\tan {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))*sqrt(tan(c + d*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)

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Mupad [B]
time = 118.04, size = 2500, normalized size = 12.44 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*B*a*tan(c + d*x)^(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (2*B*a*b*tan(c + d*x)^(3/2))/((a + b*tan(
c + d*x))^(1/2) - a^(1/2))^3)/(d + (b^2*d*tan(c + d*x)^2)/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^4 - (2*b*d*ta
n(c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2) - atan(((((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a
+ A*B*b*2i)/(4*d^2))^(1/2)*((((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2)*((((A
^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2)*((((274877906944*(1600*a^12*b^34*d^8 -
16640*a^14*b^32*d^8 + 22784*a^16*b^30*d^8 + 106496*a^18*b^28*d^8 + 65536*a^20*b^26*d^8))/d^8 - (274877906944*
tan(c + d*x)*(1600*a^12*b^35*d^8 - 48000*a^14*b^33*d^8 + 155136*a^16*b^31*d^8 + 466944*a^18*b^29*d^8 + 262144*
a^20*b^27*d^8))/(d^8*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2))*((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*
a + A*B*b*2i)/(4*d^2))^(1/2) - (2199023255552*tan(c + d*x)^(1/2)*(2048*A*a^20*b^27*d^6 - 12536*A*a^16*b^31*d^6
- 3328*A*a^18*b^29*d^6 - 7160*A*a^14*b^33*d^6 + 240*B*a^13*b^34*d^6 - 720*B*a^15*b^32*d^6 + 5696*B*a^17*b^30*
d^6 + 14848*B*a^19*b^28*d^6 + 8192*B*a^21*b^26*d^6))/(d^7*((a + b*tan(c + d*x))^(1/2) - a^(1/2))))*((A^2*b - A
^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2) - (274877906944*(1200*A^2*a^12*b^35*d^6 - 1600
*A^2*a^14*b^33*d^6 + 272464*A^2*a^16*b^31*d^6 + 573952*A^2*a^18*b^29*d^6 + 299008*A^2*a^20*b^27*d^6 - 1440*B^2
*a^12*b^35*d^6 + 8352*B^2*a^14*b^33*d^6 - 320*B^2*a^16*b^31*d^6 + 26880*B^2*a^18*b^29*d^6 + 102400*B^2*a^20*b^
27*d^6 + 65536*B^2*a^22*b^25*d^6 - 11200*A*B*a^13*b^34*d^6 - 25856*A*B*a^15*b^32*d^6 + 303168*A*B*a^17*b^30*d^
6 + 661504*A*B*a^19*b^28*d^6 + 344064*A*B*a^21*b^26*d^6))/d^8 + (274877906944*tan(c + d*x)*(1200*A^2*a^12*b^36
*d^6 - 32160*A^2*a^14*b^34*d^6 + 1125488*A^2*a^16*b^32*d^6 + 2445312*A^2*a^18*b^30*d^6 + 1351680*A^2*a^20*b^28
*d^6 + 65536*A^2*a^22*b^26*d^6 - 1440*B^2*a^12*b^36*d^6 + 32256*B^2*a^14*b^34*d^6 - 41440*B^2*a^16*b^32*d^6 +
68096*B^2*a^18*b^30*d^6 + 405504*B^2*a^20*b^28*d^6 + 262144*B^2*a^22*b^26*d^6 - 16320*A*B*a^13*b^35*d^6 - 3411
2*A*B*a^15*b^33*d^6 + 1294592*A*B*a^17*b^31*d^6 + 2656256*A*B*a^19*b^29*d^6 + 1343488*A*B*a^21*b^27*d^6))/(d^8
*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2)) + (2199023255552*tan(c + d*x)^(1/2)*(7600*A^3*a^18*b^30*d^4 - 3902
*A^3*a^16*b^32*d^4 - 4590*A^3*a^14*b^34*d^4 + 6912*A^3*a^20*b^28*d^4 - 168*B^3*a^13*b^35*d^4 + 982*B^3*a^15*b^
33*d^4 + 2494*B^3*a^17*b^31*d^4 + 9024*B^3*a^19*b^29*d^4 + 15872*B^3*a^21*b^27*d^4 + 8192*B^3*a^23*b^25*d^4 +
4922*A*B^2*a^14*b^34*d^4 + 21906*A*B^2*a^16*b^32*d^4 + 69720*A*B^2*a^18*b^30*d^4 + 95744*A*B^2*a^20*b^28*d^4 +
43008*A*B^2*a^22*b^26*d^4 - 180*A^2*B*a^13*b^35*d^4 + 4486*A^2*B*a^15*b^33*d^4 + 52154*A^2*B*a^17*b^31*d^4 +
93568*A^2*B*a^19*b^29*d^4 + 46080*A^2*B*a^21*b^27*d^4))/(d^7*((a + b*tan(c + d*x))^(1/2) - a^(1/2))))*((A^2*b
- A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2) + (274877906944*(300*A^4*a^12*b^36*d^4 + 23
60*A^4*a^14*b^34*d^4 + 149228*A^4*a^16*b^32*d^4 + 307152*A^4*a^18*b^30*d^4 + 164352*A^4*a^20*b^28*d^4 + 4096*A
^4*a^22*b^26*d^4 + 484*B^4*a^12*b^36*d^4 + 328*B^4*a^14*b^34*d^4 + 2308*B^4*a^16*b^32*d^4 - 16048*B^4*a^18*b^3
0*d^4 - 43008*B^4*a^20*b^28*d^4 - 24576*B^4*a^22*b^26*d^4 + 7880*A*B^3*a^13*b^35*d^4 + 28592*A*B^3*a^15*b^33*d
^4 - 55832*A*B^3*a^17*b^31*d^4 - 100672*A*B^3*a^19*b^29*d^4 + 57344*A*B^3*a^21*b^27*d^4 + 81920*A*B^3*a^23*b^2
5*d^4 - 5880*A^3*B*a^13*b^35*d^4 + 47408*A^3*B*a^15*b^33*d^4 + 569576*A^3*B*a^17*b^31*d^4 + 1004928*A^3*B*a^19
*b^29*d^4 + 489472*A^3*B*a^21*b^27*d^4 - 320*A^2*B^2*a^12*b^36*d^4 + 1264*A^2*B^2*a^14*b^34*d^4 - 27264*A^2*B^
2*a^16*b^32*d^4 + 294128*A^2*B^2*a^18*b^30*d^4 + 711168*A^2*B^2*a^20*b^28*d^4 + 389120*A^2*B^2*a^22*b^26*d^4))
/d^8 - (274877906944*tan(c + d*x)*(300*A^4*a^12*b^37*d^4 - 6920*A^4*a^14*b^35*d^4 + 738524*A^4*a^16*b^33*d^4 +
1819680*A^4*a^18*b^31*d^4 + 1384960*A^4*a^20*b^29*d^4 + 311296*A^4*a^22*b^27*d^4 + 484*B^4*a^12*b^37*d^4 - 53
68*B^4*a^14*b^35*d^4 + 1236*B^4*a^16*b^33*d^4 - 88064*B^4*a^18*b^31*d^4 - 205824*B^4*a^20*b^29*d^4 - 110592*B^
4*a^22*b^27*d^4 + 11192*A*B^3*a^13*b^36*d^4 + 45712*A*B^3*a^15*b^34*d^4 - 394344*A*B^3*a^17*b^32*d^4 - 599296*
A*B^3*a^19*b^30*d^4 + 124928*A*B^3*a^21*b^28*d^4 + 294912*A*B^3*a^23*b^26*d^4 - 8680*A^3*B*a^13*b^36*d^4 + 163
152*A^3*B*a^15*b^34*d^4 + 2287096*A^3*B*a^17*b^32*d^4 + 4405760*A^3*B*a^19*b^30*d^4 + 2617344*A^3*B*a^21*b^28*
d^4 + 327680*A^3*B*a^23*b^26*d^4 - 320*A^2*B^2*a^12*b^37*d^4 + 23984*A^2*B^2*a^14*b^35*d^4 - 212608*A^2*B^2*a^
16*b^33*d^4 + 865008*A^2*B^2*a^18*b^31*d^4 + 2561024*A^2*B^2*a^20*b^29*d^4 + 1523712*A^2*B^2*a^22*b^27*d^4 + 6
5536*A^2*B^2*a^24*b^25*d^4))/(d^8*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2)) - (2199023255552*tan(c + d*x)^(1/
2)*(3641*A^5*a^18*b^31*d^2 - 55*A^5*a^16*b^33*d^2 - 960*A^5*a^14*b^35*d^2 + 2864*A^5*a^20*b^29*d^2 + 128*A^5*a
^22*b^27*d^2 + 39*B^5*a^13*b^36*d^2 - 341*B^5*a^15*b^34*d^2 - 2362*B^5*a^17*b^32*d^2 - 6942*B^5*a^19*b^30*d^2
- 8544*B^5*a^21*b^28*d^2 - 3584*B^5*a^23*b^26*d...

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