3.404 \(\int \frac{\tan (x)}{(a+b \tan ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

[Out]

-ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(5/2)) + (a + b*Tan[x]^2)/(6*a*(a + b
)*(a + b*Tan[x]^4)^(3/2)) + (3*a^2 + b*(5*a + 2*b)*Tan[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tan[x]^4])

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Rubi [A]  time = 0.18687, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {3670, 1248, 741, 823, 12, 725, 206} \[ \frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a + b*Tan[x]^4)^(5/2),x]

[Out]

-ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(5/2)) + (a + b*Tan[x]^2)/(6*a*(a + b
)*(a + b*Tan[x]^4)^(3/2)) + (3*a^2 + b*(5*a + 2*b)*Tan[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tan[x]^4])

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\tan (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x}{\left (1+x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(1+x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right )\\ &=\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-3 a-2 b-2 b x}{(1+x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan ^2(x)\right )}{6 a (a+b)}\\ &=\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 b}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )}{6 a^2 b (a+b)^2}\\ &=\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )}{2 (a+b)^2}\\ &=\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\frac{a-b \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^2}\\ &=-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}+\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.765351, size = 113, normalized size = 0.97 \[ \frac{1}{6} \left (\frac{3 a^2 b \tan ^4(x)+a^2 (4 a+b)+b^2 (5 a+2 b) \tan ^6(x)+3 a b (2 a+b) \tan ^2(x)}{a^2 (a+b)^2 \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{(a+b)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(a + b*Tan[x]^4)^(5/2),x]

[Out]

((-3*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/(a + b)^(5/2) + (a^2*(4*a + b) + 3*a*b*(2*a
+ b)*Tan[x]^2 + 3*a^2*b*Tan[x]^4 + b^2*(5*a + 2*b)*Tan[x]^6)/(a^2*(a + b)^2*(a + b*Tan[x]^4)^(3/2)))/6

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Maple [B]  time = 0.065, size = 602, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a+b*tan(x)^4)^(5/2),x)

[Out]

1/24/((-a*b)^(1/2)+b)/a/(-a*b)^(1/2)/(tan(x)^2-(-a*b)^(1/2)/b)^2*(b*(tan(x)^2-(-a*b)^(1/2)/b)^2+2*(-a*b)^(1/2)
*(tan(x)^2-(-a*b)^(1/2)/b))^(1/2)+1/24/((-a*b)^(1/2)+b)/a^2/(tan(x)^2-(-a*b)^(1/2)/b)*(b*(tan(x)^2-(-a*b)^(1/2
)/b)^2+2*(-a*b)^(1/2)*(tan(x)^2-(-a*b)^(1/2)/b))^(1/2)-1/8*(2*(-a*b)^(1/2)-b)/((-a*b)^(1/2)-b)^2/a^2/(tan(x)^2
+(-a*b)^(1/2)/b)*(b*(tan(x)^2+(-a*b)^(1/2)/b)^2-2*(-a*b)^(1/2)*(tan(x)^2+(-a*b)^(1/2)/b))^(1/2)-1/2*b^2/((-a*b
)^(1/2)+b)^2/((-a*b)^(1/2)-b)^2/(a+b)^(1/2)*ln((2*a+2*b-2*b*(1+tan(x)^2)+2*(a+b)^(1/2)*(b*(1+tan(x)^2)^2-2*b*(
1+tan(x)^2)+a+b)^(1/2))/(1+tan(x)^2))+1/8*(2*(-a*b)^(1/2)+b)/((-a*b)^(1/2)+b)^2/a^2/(tan(x)^2-(-a*b)^(1/2)/b)*
(b*(tan(x)^2-(-a*b)^(1/2)/b)^2+2*(-a*b)^(1/2)*(tan(x)^2-(-a*b)^(1/2)/b))^(1/2)+1/24/((-a*b)^(1/2)-b)/a/(-a*b)^
(1/2)/(tan(x)^2+(-a*b)^(1/2)/b)^2*(b*(tan(x)^2+(-a*b)^(1/2)/b)^2-2*(-a*b)^(1/2)*(tan(x)^2+(-a*b)^(1/2)/b))^(1/
2)-1/24/((-a*b)^(1/2)-b)/a^2/(tan(x)^2+(-a*b)^(1/2)/b)*(b*(tan(x)^2+(-a*b)^(1/2)/b)^2-2*(-a*b)^(1/2)*(tan(x)^2
+(-a*b)^(1/2)/b))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )}{{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(x)/(b*tan(x)^4 + a)^(5/2), x)

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Fricas [B]  time = 4.12989, size = 1355, normalized size = 11.58 \begin{align*} \left [\frac{3 \,{\left (a^{2} b^{2} \tan \left (x\right )^{8} + 2 \, a^{3} b \tan \left (x\right )^{4} + a^{4}\right )} \sqrt{a + b} \log \left (\frac{{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} + 2 \, \sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) + 2 \,{\left ({\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (x\right )^{6} + 3 \,{\left (a^{3} b + a^{2} b^{2}\right )} \tan \left (x\right )^{4} + 4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2} + 3 \,{\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{2}\right )} \sqrt{b \tan \left (x\right )^{4} + a}}{12 \,{\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} \tan \left (x\right )^{8} + a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (x\right )^{4}\right )}}, -\frac{3 \,{\left (a^{2} b^{2} \tan \left (x\right )^{8} + 2 \, a^{3} b \tan \left (x\right )^{4} + a^{4}\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) -{\left ({\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (x\right )^{6} + 3 \,{\left (a^{3} b + a^{2} b^{2}\right )} \tan \left (x\right )^{4} + 4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2} + 3 \,{\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{2}\right )} \sqrt{b \tan \left (x\right )^{4} + a}}{6 \,{\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} \tan \left (x\right )^{8} + a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (x\right )^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a^2*b^2*tan(x)^8 + 2*a^3*b*tan(x)^4 + a^4)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2
+ 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + 2*((5*a^2*
b^2 + 7*a*b^3 + 2*b^4)*tan(x)^6 + 3*(a^3*b + a^2*b^2)*tan(x)^4 + 4*a^4 + 5*a^3*b + a^2*b^2 + 3*(2*a^3*b + 3*a^
2*b^2 + a*b^3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*tan(x)^8 + a^7 + 3
*a^6*b + 3*a^5*b^2 + a^4*b^3 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^3*b^4)*tan(x)^4), -1/6*(3*(a^2*b^2*tan(x)^
8 + 2*a^3*b*tan(x)^4 + a^4)*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2
)*tan(x)^4 + a^2 + a*b)) - ((5*a^2*b^2 + 7*a*b^3 + 2*b^4)*tan(x)^6 + 3*(a^3*b + a^2*b^2)*tan(x)^4 + 4*a^4 + 5*
a^3*b + a^2*b^2 + 3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))/((a^5*b^2 + 3*a^4*b^3 + 3*a^
3*b^4 + a^2*b^5)*tan(x)^8 + a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^3*b^4)*
tan(x)^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*tan(x)**4)**(5/2),x)

[Out]

Integral(tan(x)/(a + b*tan(x)**4)**(5/2), x)

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Giac [B]  time = 1.37004, size = 834, normalized size = 7.13 \begin{align*} \frac{{\left ({\left (\frac{{\left (5 \, a^{7} b^{3} + 32 \, a^{6} b^{4} + 87 \, a^{5} b^{5} + 130 \, a^{4} b^{6} + 115 \, a^{3} b^{7} + 60 \, a^{2} b^{8} + 17 \, a b^{9} + 2 \, b^{10}\right )} \tan \left (x\right )^{2}}{a^{10} b + 8 \, a^{9} b^{2} + 28 \, a^{8} b^{3} + 56 \, a^{7} b^{4} + 70 \, a^{6} b^{5} + 56 \, a^{5} b^{6} + 28 \, a^{4} b^{7} + 8 \, a^{3} b^{8} + a^{2} b^{9}} + \frac{3 \,{\left (a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}\right )}}{a^{10} b + 8 \, a^{9} b^{2} + 28 \, a^{8} b^{3} + 56 \, a^{7} b^{4} + 70 \, a^{6} b^{5} + 56 \, a^{5} b^{6} + 28 \, a^{4} b^{7} + 8 \, a^{3} b^{8} + a^{2} b^{9}}\right )} \tan \left (x\right )^{2} + \frac{3 \,{\left (2 \, a^{8} b^{2} + 13 \, a^{7} b^{3} + 36 \, a^{6} b^{4} + 55 \, a^{5} b^{5} + 50 \, a^{4} b^{6} + 27 \, a^{3} b^{7} + 8 \, a^{2} b^{8} + a b^{9}\right )}}{a^{10} b + 8 \, a^{9} b^{2} + 28 \, a^{8} b^{3} + 56 \, a^{7} b^{4} + 70 \, a^{6} b^{5} + 56 \, a^{5} b^{6} + 28 \, a^{4} b^{7} + 8 \, a^{3} b^{8} + a^{2} b^{9}}\right )} \tan \left (x\right )^{2} + \frac{4 \, a^{9} b + 25 \, a^{8} b^{2} + 66 \, a^{7} b^{3} + 95 \, a^{6} b^{4} + 80 \, a^{5} b^{5} + 39 \, a^{4} b^{6} + 10 \, a^{3} b^{7} + a^{2} b^{8}}{a^{10} b + 8 \, a^{9} b^{2} + 28 \, a^{8} b^{3} + 56 \, a^{7} b^{4} + 70 \, a^{6} b^{5} + 56 \, a^{5} b^{6} + 28 \, a^{4} b^{7} + 8 \, a^{3} b^{8} + a^{2} b^{9}}}{6 \,{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac{3}{2}}} - \frac{\arctan \left (\frac{\sqrt{b} \tan \left (x\right )^{2} - \sqrt{b \tan \left (x\right )^{4} + a} + \sqrt{b}}{\sqrt{-a - b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a - b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="giac")

[Out]

1/6*((((5*a^7*b^3 + 32*a^6*b^4 + 87*a^5*b^5 + 130*a^4*b^6 + 115*a^3*b^7 + 60*a^2*b^8 + 17*a*b^9 + 2*b^10)*tan(
x)^2/(a^10*b + 8*a^9*b^2 + 28*a^8*b^3 + 56*a^7*b^4 + 70*a^6*b^5 + 56*a^5*b^6 + 28*a^4*b^7 + 8*a^3*b^8 + a^2*b^
9) + 3*(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8)/(a^10*b + 8*a^9*b^2
+ 28*a^8*b^3 + 56*a^7*b^4 + 70*a^6*b^5 + 56*a^5*b^6 + 28*a^4*b^7 + 8*a^3*b^8 + a^2*b^9))*tan(x)^2 + 3*(2*a^8*b
^2 + 13*a^7*b^3 + 36*a^6*b^4 + 55*a^5*b^5 + 50*a^4*b^6 + 27*a^3*b^7 + 8*a^2*b^8 + a*b^9)/(a^10*b + 8*a^9*b^2 +
28*a^8*b^3 + 56*a^7*b^4 + 70*a^6*b^5 + 56*a^5*b^6 + 28*a^4*b^7 + 8*a^3*b^8 + a^2*b^9))*tan(x)^2 + (4*a^9*b +
25*a^8*b^2 + 66*a^7*b^3 + 95*a^6*b^4 + 80*a^5*b^5 + 39*a^4*b^6 + 10*a^3*b^7 + a^2*b^8)/(a^10*b + 8*a^9*b^2 + 2
8*a^8*b^3 + 56*a^7*b^4 + 70*a^6*b^5 + 56*a^5*b^6 + 28*a^4*b^7 + 8*a^3*b^8 + a^2*b^9))/(b*tan(x)^4 + a)^(3/2) -
arctan((sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a) + sqrt(b))/sqrt(-a - b))/((a^2 + 2*a*b + b^2)*sqrt(-a - b))