3.86 \(\int \frac{\sin ^4(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=250 \[ -\frac{3 \sqrt{b} \left (5 a^2+10 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 \sqrt{a} f (a-b)^5}+\frac{3 x \left (a^2+10 a b+5 b^2\right )}{8 (a-b)^5}-\frac{3 b (a+b) \tan (e+f x)}{2 f (a-b)^4 \left (a+b \tan ^2(e+f x)\right )}-\frac{b (7 a+5 b) \tan (e+f x)}{8 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(5 a+3 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^2} \]

[Out]

(3*(a^2 + 10*a*b + 5*b^2)*x)/(8*(a - b)^5) - (3*Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/S
qrt[a]])/(8*Sqrt[a]*(a - b)^5*f) - ((5*a + 3*b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f*(a + b*Tan[e + f*x]^
2)^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - (b*(7*a + 5*b)*Tan[e + f*x])/(8
*(a - b)^3*f*(a + b*Tan[e + f*x]^2)^2) - (3*b*(a + b)*Tan[e + f*x])/(2*(a - b)^4*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.330496, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3663, 470, 527, 522, 203, 205} \[ -\frac{3 \sqrt{b} \left (5 a^2+10 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 \sqrt{a} f (a-b)^5}+\frac{3 x \left (a^2+10 a b+5 b^2\right )}{8 (a-b)^5}-\frac{3 b (a+b) \tan (e+f x)}{2 f (a-b)^4 \left (a+b \tan ^2(e+f x)\right )}-\frac{b (7 a+5 b) \tan (e+f x)}{8 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(5 a+3 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(3*(a^2 + 10*a*b + 5*b^2)*x)/(8*(a - b)^5) - (3*Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/S
qrt[a]])/(8*Sqrt[a]*(a - b)^5*f) - ((5*a + 3*b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f*(a + b*Tan[e + f*x]^
2)^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - (b*(7*a + 5*b)*Tan[e + f*x])/(8
*(a - b)^3*f*(a + b*Tan[e + f*x]^2)^2) - (3*b*(a + b)*Tan[e + f*x])/(2*(a - b)^4*f*(a + b*Tan[e + f*x]^2))

Rule 3663

Int[sin[(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^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

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

Rule 522

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 203

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{a+(-4 a-3 b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac{(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+5 b)-5 b (5 a+3 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac{(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{12 a^2 (a+3 b)-12 a b (7 a+5 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{32 a (a-b)^3 f}\\ &=-\frac{(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{24 a^2 \left (a^2+6 a b+b^2\right )-96 a^2 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{64 a^2 (a-b)^4 f}\\ &=-\frac{(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\left (3 b \left (5 a^2+10 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^5 f}+\frac{\left (3 \left (a^2+10 a b+5 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^5 f}\\ &=\frac{3 \left (a^2+10 a b+5 b^2\right ) x}{8 (a-b)^5}-\frac{3 \sqrt{b} \left (5 a^2+10 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 \sqrt{a} (a-b)^5 f}-\frac{(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.921413, size = 194, normalized size = 0.78 \[ \frac{12 \left (a^2+10 a b+5 b^2\right ) (e+f x)-\frac{12 \sqrt{b} \left (5 a^2+10 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{16 a b^2 (a-b) \sin (2 (e+f x))}{((a-b) \cos (2 (e+f x))+a+b)^2}+(a-b)^2 \sin (4 (e+f x))-8 (a+2 b) (a-b) \sin (2 (e+f x))-\frac{4 b (9 a+5 b) (a-b) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}}{32 f (a-b)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(12*(a^2 + 10*a*b + 5*b^2)*(e + f*x) - (12*Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a
]])/Sqrt[a] - 8*(a - b)*(a + 2*b)*Sin[2*(e + f*x)] + (16*a*(a - b)*b^2*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[
2*(e + f*x)])^2 - (4*(a - b)*b*(9*a + 5*b)*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)]) + (a - b)^2*Si
n[4*(e + f*x)])/(32*(a - b)^5*f)

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Maple [B]  time = 0.082, size = 598, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x)

[Out]

-7/8/f*b^2/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3*a^2+1/4/f*b^3/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3*a
+5/8/f*b^4/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8/f*b/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)*a^3+3/4/f
*b^2/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)*a^2+3/8/f*b^3/(a-b)^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)*a-15/8/f*b/
(a-b)^5/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))*a^2-15/4/f*b^2/(a-b)^5/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a
*b)^(1/2))*a-3/8/f*b^3/(a-b)^5/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/4/f/(a-b)^5/(1+tan(f*x+e)^2)^2*t
an(f*x+e)^3*a*b+7/8/f/(a-b)^5/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*b^2-5/8/f/(a-b)^5/(1+tan(f*x+e)^2)^2*tan(f*x+e)^
3*a^2-3/8/f/(a-b)^5/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a^2-3/4/f/(a-b)^5/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a*b+9/8/f/(a
-b)^5/(1+tan(f*x+e)^2)^2*tan(f*x+e)*b^2+15/4/f/(a-b)^5*arctan(tan(f*x+e))*a*b+15/8/f/(a-b)^5*arctan(tan(f*x+e)
)*b^2+3/8/f/(a-b)^5*arctan(tan(f*x+e))*a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.26895, size = 2664, normalized size = 10.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(12*(a^4 + 8*a^3*b - 14*a^2*b^2 + 5*b^4)*f*x*cos(f*x + e)^4 + 24*(a^3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f
*x*cos(f*x + e)^2 + 12*(a^2*b^2 + 10*a*b^3 + 5*b^4)*f*x - 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)
^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^3*b + 5*a^2*b^2 - 9*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b/a)*log(((a^2
+ 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 - 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x
+ e))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)
) + 4*(2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^7 - (5*a^4 - 12*a^3*b + 6*a^2*b^2 + 4*a*b^3
- 3*b^4)*cos(f*x + e)^5 - (19*a^3*b - 21*a^2*b^2 - 15*a*b^3 + 17*b^4)*cos(f*x + e)^3 - 12*(a^2*b^2 - b^4)*cos(
f*x + e))*sin(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f
*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^
2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f), 1/16*(6*(a^4 + 8*a^3*b - 14*a^2*b^2 +
5*b^4)*f*x*cos(f*x + e)^4 + 12*(a^3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f*x*cos(f*x + e)^2 + 6*(a^2*b^2 + 10*a*b^
3 + 5*b^4)*f*x + 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^
3*b + 5*a^2*b^2 - 9*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(
b*cos(f*x + e)*sin(f*x + e))) + 2*(2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^7 - (5*a^4 - 12*
a^3*b + 6*a^2*b^2 + 4*a*b^3 - 3*b^4)*cos(f*x + e)^5 - (19*a^3*b - 21*a^2*b^2 - 15*a*b^3 + 17*b^4)*cos(f*x + e)
^3 - 12*(a^2*b^2 - b^4)*cos(f*x + e))*sin(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 2
1*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*
a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 3.3016, size = 1793, normalized size = 7.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/8*(6*(3*a^6*b - 2*a^5*b^2 - 19*a^4*b^3 + 36*a^3*b^4 - 19*a^2*b^5 - 2*a*b^6 + 3*b^7 + 2*a*b*abs(-a^5 + 5*a^4
*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5) + 2*b^2*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 +
b^5))*(pi*floor((f*x + e)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(f*x + e)/sqrt((a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2
*b^3 - 3*a*b^4 + b^5 + sqrt((a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)^2 - 4*(a^5 - 4*a^4*b + 6*a
^3*b^2 - 4*a^2*b^3 + a*b^4)*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)))/(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 -
4*a*b^4 + b^5))))/(a^5*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5) - 3*a^4*b*abs(-a^5 + 5*a
^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5) + 2*a^3*b^2*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a
*b^4 + b^5) + 2*a^2*b^3*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5) - 3*a*b^4*abs(-a^5 + 5*a
^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5) + b^5*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 +
b^5) + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)^2) - 6*(2*sqrt(a*b)*(a + b)*abs(-a^5 + 5*a^4
*b - 10*a^3*b^2 + 10*a^2*b^3 - 5*a*b^4 + b^5)*abs(b) - (3*a^6 - 2*a^5*b - 19*a^4*b^2 + 36*a^3*b^3 - 19*a^2*b^4
- 2*a*b^5 + 3*b^6)*sqrt(a*b)*abs(b))*(pi*floor((f*x + e)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(f*x + e)/sqrt((a^
5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5 - sqrt((a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 +
b^5)^2 - 4*(a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)))/(
a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5))))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)^2
*b - (a^5*b - 3*a^4*b^2 + 2*a^3*b^3 + 2*a^2*b^4 - 3*a*b^5 + b^6)*abs(-a^5 + 5*a^4*b - 10*a^3*b^2 + 10*a^2*b^3
- 5*a*b^4 + b^5)) + (12*a*b^2*tan(f*x + e)^7 + 12*b^3*tan(f*x + e)^7 + 19*a^2*b*tan(f*x + e)^5 + 34*a*b^2*tan(
f*x + e)^5 + 19*b^3*tan(f*x + e)^5 + 5*a^3*tan(f*x + e)^3 + 31*a^2*b*tan(f*x + e)^3 + 31*a*b^2*tan(f*x + e)^3
+ 5*b^3*tan(f*x + e)^3 + 3*a^3*tan(f*x + e) + 18*a^2*b*tan(f*x + e) + 3*a*b^2*tan(f*x + e))/((b*tan(f*x + e)^4
+ a*tan(f*x + e)^2 + b*tan(f*x + e)^2 + a)^2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)))/f