3.5.29 \(\int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [429]

Optimal. Leaf size=169 \[ -\frac {\sqrt {i a-b} (i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.42, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3695, 3697, 3696, 95, 209, 212, 3715, 65, 223} \begin {gather*} -\frac {\sqrt {-b+i a} (-B+i A) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {b+i a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[I*a - b]*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (2*Sqrt[b
]*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (Sqrt[I*a + b]*(I*A + B)*ArcTanh[(Sqrt
[I*a + 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 209

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

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 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 3695

Int[(Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/Sqrt[(c_.) + (d_.)*t
an[(e_.) + (f_.)*(x_)]], x_Symbol] :> Int[Simp[a*A - b*B + (A*b + a*B)*Tan[e + f*x], x]/(Sqrt[a + b*Tan[e + f*
x]]*Sqrt[c + d*Tan[e + f*x]]), x] + Dist[b*B, Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Ta
n[e + f*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]

Rule 3696

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

Rule 3697

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

Rubi steps

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

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Mathematica [A]
time = 0.56, size = 204, normalized size = 1.21 \begin {gather*} \frac {\sqrt [4]{-1} \left (\sqrt {-a+i b} (i A+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 {a+i b} (-i A+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 )\right )+\frac {2 \sqrt {a} \sqrt {b} B \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.78, size = 2175963, normalized size = 12875.52 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Timed out

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Mupad [B]
time = 17.99, size = 1141, normalized size = 6.75 \begin {gather*} \mathrm {atanh}\left (\frac {a^{3/2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^2\,d^4}-A^2\,b\,d^2}{d^4}}\,\sqrt {-A^4\,a^2\,d^4}-a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^2\,d^4}-A^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,a^2\,d^4}+A^2\,a^{3/2}\,b\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^2\,d^4}-A^2\,b\,d^2}{d^4}}-A^2\,a\,b\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^2\,d^4}-A^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{A^3\,a^3\,d^2-A\,a\,b\,\sqrt {-A^4\,a^2\,d^4}-A\,b^2\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-A^4\,a^2\,d^4}-A^3\,a^{5/2}\,d^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+A^3\,a^2\,b\,d^2\,\mathrm {tan}\left (c+d\,x\right )+A\,\sqrt {a}\,b\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,a^2\,d^4}}\right )\,\sqrt {\frac {\sqrt {-A^4\,a^2\,d^4}-A^2\,b\,d^2}{d^4}}-\mathrm {atanh}\left (\frac {a^{3/2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^2\,d^4}+A^2\,b\,d^2}{d^4}}\,\sqrt {-A^4\,a^2\,d^4}-a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^2\,d^4}+A^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,a^2\,d^4}-A^2\,a^{3/2}\,b\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^2\,d^4}+A^2\,b\,d^2}{d^4}}+A^2\,a\,b\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^2\,d^4}+A^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{A^3\,a^3\,d^2+A\,a\,b\,\sqrt {-A^4\,a^2\,d^4}+A\,b^2\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-A^4\,a^2\,d^4}-A^3\,a^{5/2}\,d^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+A^3\,a^2\,b\,d^2\,\mathrm {tan}\left (c+d\,x\right )-A\,\sqrt {a}\,b\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,a^2\,d^4}}\right )\,\sqrt {-\frac {\sqrt {-A^4\,a^2\,d^4}+A^2\,b\,d^2}{d^4}}+\mathrm {atanh}\left (\frac {2\,\left (\frac {\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^2\,d^4}+B^2\,b\,d^2}{d^4}}}{2}-\frac {d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^2\,d^4}+B^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{2}\right )}{B\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\right )}\right )\,\sqrt {\frac {\sqrt {-B^4\,a^2\,d^4}+B^2\,b\,d^2}{d^4}}+\mathrm {atanh}\left (\frac {2\,\left (\frac {\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^2\,d^4}-B^2\,b\,d^2}{d^4}}}{2}-\frac {d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^2\,d^4}-B^2\,b\,d^2}{d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{2}\right )}{B\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\right )}\right )\,\sqrt {-\frac {\sqrt {-B^4\,a^2\,d^4}-B^2\,b\,d^2}{d^4}}+\frac {4\,B\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}-\sqrt {a}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh((a^(3/2)*d*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^2)/d^4)^(1/2)*(-A^4*a^2*d^4)^(1/2) - a*d*
tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*d^4)^(1
/2) + A^2*a^(3/2)*b*d^3*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^2)/d^4)^(1/2) - A^2*a*b*d^3*tan(c
+ d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2))/(A^3*a^3*d^2 - A*a*b*(
-A^4*a^2*d^4)^(1/2) - A*b^2*tan(c + d*x)*(-A^4*a^2*d^4)^(1/2) - A^3*a^(5/2)*d^2*(a + b*tan(c + d*x))^(1/2) + A
^3*a^2*b*d^2*tan(c + d*x) + A*a^(1/2)*b*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*d^4)^(1/2)))*(((-A^4*a^2*d^4)^(1/
2) - A^2*b*d^2)/d^4)^(1/2) - atanh((a^(3/2)*d*tan(c + d*x)^(1/2)*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/
2)*(-A^4*a^2*d^4)^(1/2) - a*d*tan(c + d*x)^(1/2)*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c
+ d*x))^(1/2)*(-A^4*a^2*d^4)^(1/2) - A^2*a^(3/2)*b*d^3*tan(c + d*x)^(1/2)*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)
/d^4)^(1/2) + A^2*a*b*d^3*tan(c + d*x)^(1/2)*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*
x))^(1/2))/(A^3*a^3*d^2 + A*a*b*(-A^4*a^2*d^4)^(1/2) + A*b^2*tan(c + d*x)*(-A^4*a^2*d^4)^(1/2) - A^3*a^(5/2)*d
^2*(a + b*tan(c + d*x))^(1/2) + A^3*a^2*b*d^2*tan(c + d*x) - A*a^(1/2)*b*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*
d^4)^(1/2)))*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2) + atanh((2*((a^(1/2)*d*tan(c + d*x)^(1/2)*(((-B^4
*a^2*d^4)^(1/2) + B^2*b*d^2)/d^4)^(1/2))/2 - (d*tan(c + d*x)^(1/2)*(((-B^4*a^2*d^4)^(1/2) + B^2*b*d^2)/d^4)^(1
/2)*(a + b*tan(c + d*x))^(1/2))/2))/(B*(a + b*tan(c + d*x) - a^(1/2)*(a + b*tan(c + d*x))^(1/2))))*(((-B^4*a^2
*d^4)^(1/2) + B^2*b*d^2)/d^4)^(1/2) + atanh((2*((a^(1/2)*d*tan(c + d*x)^(1/2)*(-((-B^4*a^2*d^4)^(1/2) - B^2*b*
d^2)/d^4)^(1/2))/2 - (d*tan(c + d*x)^(1/2)*(-((-B^4*a^2*d^4)^(1/2) - B^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x)
)^(1/2))/2))/(B*(a + b*tan(c + d*x) - a^(1/2)*(a + b*tan(c + d*x))^(1/2))))*(-((-B^4*a^2*d^4)^(1/2) - B^2*b*d^
2)/d^4)^(1/2) + (4*B*b^(1/2)*atanh((b^(1/2)*tan(c + d*x)^(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))))/d

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