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

Optimal. Leaf size=138 \[ -\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 \sqrt {a} f (a-b)^3}-\frac {b \tan (e+f x)}{f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {x (a+3 b)}{2 (a-b)^3} \]

[Out]

1/2*(a+3*b)*x/(a-b)^3-1/2*(3*a+b)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/(a-b)^3/f/a^(1/2)-1/2*cos(f*x+e)*
sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)-b*tan(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.16, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3663, 471, 527, 522, 203, 205} \[ -\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 \sqrt {a} f (a-b)^3}-\frac {b \tan (e+f x)}{f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {x (a+3 b)}{2 (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 3*b)*x)/(2*(a - b)^3) - (Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*Sqrt[a]*(a - b)^3*
f) - (Cos[e + f*x]*Sin[e + f*x])/(2*(a - b)*f*(a + b*Tan[e + f*x]^2)) - (b*Tan[e + f*x])/((a - b)^2*f*(a + b*T
an[e + f*x]^2))

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]

Rule 471

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

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 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 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]

Rubi steps

\begin {align*} \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {a-3 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b) f}\\ &=-\frac {\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{(a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 a (a+b)-4 a b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a (a-b)^2 f}\\ &=-\frac {\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{(a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^3 f}+\frac {(a+3 b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^3 f}\\ &=\frac {(a+3 b) x}{2 (a-b)^3}-\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 \sqrt {a} (a-b)^3 f}-\frac {\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{(a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 111, normalized size = 0.80 \[ -\frac {-2 (a+3 b) (e+f x)+(a-b) \sin (2 (e+f x))+\frac {2 \sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {2 b (a-b) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}}{4 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(-2*(a + 3*b)*(e + f*x) + (2*Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/Sqrt[a] + (a - b)*
Sin[2*(e + f*x)] + (2*(a - b)*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)]))/((a - b)^3*f)

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fricas [A]  time = 0.60, size = 568, normalized size = 4.12 \[ \left [\frac {4 \, {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a b + 3 \, b^{2}\right )} f x - {\left ({\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac {2 \, {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 2 \, {\left (a b + 3 \, b^{2}\right )} f x + {\left ({\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*(a^2 + 2*a*b - 3*b^2)*f*x*cos(f*x + e)^2 + 4*(a*b + 3*b^2)*f*x - ((3*a^2 - 2*a*b - b^2)*cos(f*x + e)^2
+ 3*a*b + b^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*((a^2 - 2*a*b + b^2)*cos(f*x + e)^3 + 2*(a*b - b^2)*cos(f*x + e))*si
n(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4
)*f), 1/4*(2*(a^2 + 2*a*b - 3*b^2)*f*x*cos(f*x + e)^2 + 2*(a*b + 3*b^2)*f*x + ((3*a^2 - 2*a*b - b^2)*cos(f*x +
e)^2 + 3*a*b + b^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*((a^2 - 2*a*b + b^2)*cos(f*x + e)^3 + 2*(a*b - b^2)*cos(f*x + e))*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*
b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f)]

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giac [A]  time = 3.09, size = 195, normalized size = 1.41 \[ \frac {\frac {{\left (f x + e\right )} {\left (a + 3 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b + b^{2}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {2 \, b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + b \tan \left (f x + e\right )^{2} + a\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((f*x + e)*(a + 3*b)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan
(f*x + e)/sqrt(a*b)))*(3*a*b + b^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (2*b*tan(f*x + e)^3 + a*tan(
f*x + e) + b*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + b*tan(f*x + e)^2 + a)*(a^2 - 2*a*b + b^2)))
/f

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maple [A]  time = 0.51, size = 240, normalized size = 1.74 \[ -\frac {b \tan \left (f x +e \right ) a}{2 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{2} \tan \left (f x +e \right )}{2 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right ) a}{2 f \left (a -b \right )^{3} \sqrt {a b}}-\frac {b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \left (a -b \right )^{3} \sqrt {a b}}-\frac {\tan \left (f x +e \right ) a}{2 f \left (a -b \right )^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )}+\frac {\tan \left (f x +e \right ) b}{2 f \left (a -b \right )^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a}{2 f \left (a -b \right )^{3}}+\frac {3 \arctan \left (\tan \left (f x +e \right )\right ) b}{2 f \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/(a-b)^3*b*tan(f*x+e)/(a+b*tan(f*x+e)^2)*a+1/2/f/(a-b)^3*b^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)-3/2/f/(a-b)^3
*b/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))*a-1/2/f/(a-b)^3*b^2/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2
))-1/2/f/(a-b)^3*tan(f*x+e)/(1+tan(f*x+e)^2)*a+1/2/f/(a-b)^3*tan(f*x+e)/(1+tan(f*x+e)^2)*b+1/2/f/(a-b)^3*arcta
n(tan(f*x+e))*a+3/2/f/(a-b)^3*arctan(tan(f*x+e))*b

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maxima [A]  time = 0.73, size = 185, normalized size = 1.34 \[ \frac {\frac {{\left (f x + e\right )} {\left (a + 3 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {2 \, b \tan \left (f x + e\right )^{3} + {\left (a + b\right )} \tan \left (f x + e\right )}{{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{4} + a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((f*x + e)*(a + 3*b)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a*b + b^2)*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^
3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (2*b*tan(f*x + e)^3 + (a + b)*tan(f*x + e))/((a^2*b - 2*a*b^2 + b^3)
*tan(f*x + e)^4 + a^3 - 2*a^2*b + a*b^2 + (a^3 - a^2*b - a*b^2 + b^3)*tan(f*x + e)^2))/f

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mupad [B]  time = 14.94, size = 3301, normalized size = 23.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((-a*b)^(1/2)*((tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)
+ ((-a*b)^(1/2)*(3*a + b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^4 + 10*a^6*b^3
- 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) - (tan(e + f*x)*(-a*b)^(1
/2)*(3*a + b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2))
/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^3 + 3*a^3*b - a
^4 - 3*a^2*b^2)))*(3*a + b)*1i)/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)) + ((-a*b)^(1/2)*((tan(e + f*x)*(6*a*b^
4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2) - ((-a*b)^(1/2)*(3*a + b)*((10*a*b^8 - 2*b
^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^
6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) + (tan(e + f*x)*(-a*b)^(1/2)*(3*a + b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^
7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^
4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)))*(3*a + b)*1i)/(4*(a*b^3 +
3*a^3*b - a^4 - 3*a^2*b^2)))/((5*a*b^4 + (3*b^5)/2 + (3*a^2*b^3)/2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b
^4 - 20*a^3*b^3 + 15*a^4*b^2) - ((-a*b)^(1/2)*((tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4
*a*b^3 + b^4 + 6*a^2*b^2) + ((-a*b)^(1/2)*(3*a + b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5
- 18*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)
- (tan(e + f*x)*(-a*b)^(1/2)*(3*a + b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 +
40*a^6*b^3 - 8*a^7*b^2))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))
)/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)))*(3*a + b))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)) + ((-a*b)^(1/2)*
((tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2) - ((-a*b)^(1/2)*(3*a
+ b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 -
6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) + (tan(e + f*x)*(-a*b)^(1/2)*(3*a + b)*(40*a*b
^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2))/(4*(a*b^3 + 3*a^3*b
- a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)))*(3*
a + b))/(4*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2))))*(-a*b)^(1/2)*(3*a + b)*1i)/(2*f*(a*b^3 + 3*a^3*b - a^4 - 3*a
^2*b^2)) - (atan((((a + 3*b)*(((a + 3*b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^
4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) - (tan(e +
f*x)*(a + 3*b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2)
)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^2*3i - a^2
*b*3i + a^3*1i - b^3*1i)) + (tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^
2*b^2))*1i)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - ((a + 3*b)*(((a + 3*b)*((10*a*b^8 - 2*b^9 - 18*a^2*b
^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^
4 - 20*a^3*b^3 + 15*a^4*b^2) + (tan(e + f*x)*(a + 3*b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^
5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3
+ b^4 + 6*a^2*b^2))))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3
))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))*1i)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))/(((a + 3*b)*(
((a + 3*b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 18*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a
^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) - (tan(e + f*x)*(a + 3*b)*(40*a*b^8 - 8*b
^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*a^7*b^2))/(4*(a*b^2*3i - a^2*b*3i + a^
3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (t
an(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))/(4*(a*b^2*3i - a^2*b*
3i + a^3*1i - b^3*1i)) - (5*a*b^4 + (3*b^5)/2 + (3*a^2*b^3)/2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 2
0*a^3*b^3 + 15*a^4*b^2) + ((a + 3*b)*(((a + 3*b)*((10*a*b^8 - 2*b^9 - 18*a^2*b^7 + 10*a^3*b^6 + 10*a^4*b^5 - 1
8*a^5*b^4 + 10*a^6*b^3 - 2*a^7*b^2)/(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) + (
tan(e + f*x)*(a + 3*b)*(40*a*b^8 - 8*b^9 - 72*a^2*b^7 + 40*a^3*b^6 + 40*a^4*b^5 - 72*a^5*b^4 + 40*a^6*b^3 - 8*
a^7*b^2))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(4*(a*b^2*
3i - a^2*b*3i + a^3*1i - b^3*1i)) - (tan(e + f*x)*(6*a*b^4 + 5*b^5 + 5*a^2*b^3))/(a^4 - 4*a^3*b - 4*a*b^3 + b^
4 + 6*a^2*b^2)))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))))*(a + 3*b)*1i)/(2*f*(a*b^2*3i - a^2*b*3i + a^3*1
i - b^3*1i)) - ((tan(e + f*x)*(a + b))/(2*(a^2 - 2*a*b + b^2)) + (b*tan(e + f*x)^3)/(a^2 - 2*a*b + b^2))/(f*(a
+ tan(e + f*x)^2*(a + b) + b*tan(e + f*x)^4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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