3.4.10 \(\int \frac {\sec ^3(x)}{a+b \sin ^2(x)} \, dx\) [310]

Optimal. Leaf size=61 \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^2}+\frac {(a+3 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^2}+\frac {\sec (x) \tan (x)}{2 (a+b)} \]

[Out]

1/2*(a+3*b)*arctanh(sin(x))/(a+b)^2+b^(3/2)*arctan(sin(x)*b^(1/2)/a^(1/2))/(a+b)^2/a^(1/2)+1/2*sec(x)*tan(x)/(
a+b)

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Rubi [A]
time = 0.06, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3269, 425, 536, 212, 211} \begin {gather*} \frac {b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^2}+\frac {(a+3 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^2}+\frac {\tan (x) \sec (x)}{2 (a+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Sin[x]^2),x]

[Out]

(b^(3/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2) + ((a + 3*b)*ArcTanh[Sin[x]])/(2*(a + b)^2) + (
Sec[x]*Tan[x])/(2*(a + b))

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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(61)=122\).
time = 0.24, size = 147, normalized size = 2.41 \begin {gather*} \frac {-\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{\sqrt {a}}+\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a}}-2 (a+3 b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 (a+3 b) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {a+b}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}-\frac {a+b}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}}{4 (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + b*Sin[x]^2),x]

[Out]

((-2*b^(3/2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/Sqrt[a] + (2*b^(3/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/Sqrt[a]
- 2*(a + 3*b)*Log[Cos[x/2] - Sin[x/2]] + 2*(a + 3*b)*Log[Cos[x/2] + Sin[x/2]] + (a + b)/(Cos[x/2] - Sin[x/2])^
2 - (a + b)/(Cos[x/2] + Sin[x/2])^2)/(4*(a + b)^2)

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Maple [A]
time = 0.31, size = 96, normalized size = 1.57

method	result	size
default	\(\frac {b^{2} \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{\left (a +b \right )^{2} \sqrt {a b}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (x \right )-1\right )}+\frac {\left (-a -3 b \right ) \ln \left (\sin \left (x \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (1+\sin \left (x \right )\right )}+\frac {\left (a +3 b \right ) \ln \left (1+\sin \left (x \right )\right )}{4 \left (a +b \right )^{2}}\)	\(96\)
risch	\(-\frac {i \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2} \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+i\right ) a}{2 a^{2}+4 a b +2 b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )^{2}}-\frac {\sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )^{2}}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3/(a+b*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

b^2/(a+b)^2/(a*b)^(1/2)*arctan(b*sin(x)/(a*b)^(1/2))-1/(4*a+4*b)/(sin(x)-1)+1/4/(a+b)^2*(-a-3*b)*ln(sin(x)-1)-
1/(4*a+4*b)/(1+sin(x))+1/4*(a+3*b)/(a+b)^2*ln(1+sin(x))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (49) = 98\).
time = 0.48, size = 104, normalized size = 1.70 \begin {gather*} \frac {b^{2} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} + \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {\sin \left (x\right )}{2 \, {\left ({\left (a + b\right )} \sin \left (x\right )^{2} - a - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

b^2*arctan(b*sin(x)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) + 1/4*(a + 3*b)*log(sin(x) + 1)/(a^2 + 2*a*b +
b^2) - 1/4*(a + 3*b)*log(sin(x) - 1)/(a^2 + 2*a*b + b^2) - 1/2*sin(x)/((a + b)*sin(x)^2 - a - b)

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Fricas [A]
time = 0.46, size = 203, normalized size = 3.33 \begin {gather*} \left [\frac {2 \, b \sqrt {-\frac {b}{a}} \cos \left (x\right )^{2} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + {\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (a + b\right )} \sin \left (x\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (x\right )^{2}}, \frac {4 \, b \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \left (x\right )\right ) \cos \left (x\right )^{2} + {\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (a + b\right )} \sin \left (x\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (x\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

[1/4*(2*b*sqrt(-b/a)*cos(x)^2*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + (a + 3
*b)*cos(x)^2*log(sin(x) + 1) - (a + 3*b)*cos(x)^2*log(-sin(x) + 1) + 2*(a + b)*sin(x))/((a^2 + 2*a*b + b^2)*co
s(x)^2), 1/4*(4*b*sqrt(b/a)*arctan(sqrt(b/a)*sin(x))*cos(x)^2 + (a + 3*b)*cos(x)^2*log(sin(x) + 1) - (a + 3*b)
*cos(x)^2*log(-sin(x) + 1) + 2*(a + b)*sin(x))/((a^2 + 2*a*b + b^2)*cos(x)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+b*sin(x)**2),x)

[Out]

Integral(sec(x)**3/(a + b*sin(x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (49) = 98\).
time = 0.42, size = 102, normalized size = 1.67 \begin {gather*} \frac {b^{2} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} + \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 3 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {\sin \left (x\right )}{2 \, {\left (\sin \left (x\right )^{2} - 1\right )} {\left (a + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*sin(x)^2),x, algorithm="giac")

[Out]

b^2*arctan(b*sin(x)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) + 1/4*(a + 3*b)*log(sin(x) + 1)/(a^2 + 2*a*b +
b^2) - 1/4*(a + 3*b)*log(-sin(x) + 1)/(a^2 + 2*a*b + b^2) - 1/2*sin(x)/((sin(x)^2 - 1)*(a + b))

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Mupad [B]
time = 15.36, size = 1139, normalized size = 18.67 \begin {gather*} \frac {\sin \left (x\right )}{2\,{\cos \left (x\right )}^2\,\left (a+b\right )}-\ln \left (\sin \left (x\right )-1\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )+\frac {\ln \left (\sin \left (x\right )+1\right )\,\left (a+3\,b\right )}{4\,{\left (a+b\right )}^2}+\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-a\,b^3}\,\left (\frac {\sin \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}+\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\sin \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2}+\frac {\sqrt {-a\,b^3}\,\left (\frac {\sin \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}+\frac {\sin \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2}}{\frac {\frac {3\,b^5}{2}+\frac {a\,b^4}{2}}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}-\frac {\sqrt {-a\,b^3}\,\left (\frac {\sin \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}+\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\sin \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )}{a^3+2\,a^2\,b+a\,b^2}+\frac {\sqrt {-a\,b^3}\,\left (\frac {\sin \left (x\right )\,\left (a^2\,b^3+6\,a\,b^4+13\,b^5\right )}{4\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\left (\frac {2\,a^5\,b^2+12\,a^4\,b^3+28\,a^3\,b^4+32\,a^2\,b^5+18\,a\,b^6+4\,b^7}{2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}+\frac {\sin \left (x\right )\,\sqrt {-a\,b^3}\,\left (-16\,a^5\,b^2-48\,a^4\,b^3-32\,a^3\,b^4+32\,a^2\,b^5+48\,a\,b^6+16\,b^7\right )}{8\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )\,\sqrt {-a\,b^3}}{2\,\left (a^3+2\,a^2\,b+a\,b^2\right )}\right )}{a^3+2\,a^2\,b+a\,b^2}}\right )\,\sqrt {-a\,b^3}\,1{}\mathrm {i}}{a^3+2\,a^2\,b+a\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + b*sin(x)^2)),x)

[Out]

sin(x)/(2*cos(x)^2*(a + b)) - log(sin(x) - 1)*(b/(2*(a + b)^2) + 1/(4*(a + b))) + (log(sin(x) + 1)*(a + 3*b))/
(4*(a + b)^2) + (atan((((-a*b^3)^(1/2)*((sin(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) + (((18*
a*b^6 + 4*b^7 + 32*a^2*b^5 + 28*a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - (sin(x
)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^
2)*(a*b^2 + 2*a^2*b + a^3)))*(-a*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3)))*1i)/(a*b^2 + 2*a^2*b + a^3) + ((-a*b
^3)^(1/2)*((sin(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) - (((18*a*b^6 + 4*b^7 + 32*a^2*b^5 +
28*a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) + (sin(x)*(-a*b^3)^(1/2)*(48*a*b^6 +
16*b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^2)*(a*b^2 + 2*a^2*b + a^3)))*
(-a*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3)))*1i)/(a*b^2 + 2*a^2*b + a^3))/(((a*b^4)/2 + (3*b^5)/2)/(3*a*b^2 +
3*a^2*b + a^3 + b^3) - ((-a*b^3)^(1/2)*((sin(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) + (((18*
a*b^6 + 4*b^7 + 32*a^2*b^5 + 28*a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - (sin(x
)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^
2)*(a*b^2 + 2*a^2*b + a^3)))*(-a*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3))))/(a*b^2 + 2*a^2*b + a^3) + ((-a*b^3)
^(1/2)*((sin(x)*(6*a*b^4 + 13*b^5 + a^2*b^3))/(4*(2*a*b + a^2 + b^2)) - (((18*a*b^6 + 4*b^7 + 32*a^2*b^5 + 28*
a^3*b^4 + 12*a^4*b^3 + 2*a^5*b^2)/(2*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) + (sin(x)*(-a*b^3)^(1/2)*(48*a*b^6 + 16*
b^7 + 32*a^2*b^5 - 32*a^3*b^4 - 48*a^4*b^3 - 16*a^5*b^2))/(8*(2*a*b + a^2 + b^2)*(a*b^2 + 2*a^2*b + a^3)))*(-a
*b^3)^(1/2))/(2*(a*b^2 + 2*a^2*b + a^3))))/(a*b^2 + 2*a^2*b + a^3)))*(-a*b^3)^(1/2)*1i)/(a*b^2 + 2*a^2*b + a^3
)

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