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]

((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))

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Rubi [A]  time = 0.155567, antiderivative size = 138, normalized size of antiderivative = 1., 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 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 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 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 ^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*}

Mathematica [A]  time = 1.58765, size = 111, normalized size = 0.8 \[ -\frac{-2 (a+3 b) (e+f x)+(a-b) \sin (2 (e+f x))+\frac{2 \sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{2 b (a-b) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}}{4 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-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)]))/(4*(a - b)^3*f)

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Maple [A]  time = 0.077, size = 240, normalized size = 1.7 \begin{align*} -{\frac{\tan \left ( fx+e \right ) ab}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{2}\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,ab}{2\,f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{2}}{2\,f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\tan \left ( fx+e \right ) a}{2\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{b\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) a}{2\,f \left ( a-b \right ) ^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{2\,f \left ( a-b \right ) ^{3}}} \end{align*}

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(b*tan(f*x+e)/(a*b)^(1/2))*a-1/2/f/(a-b)^3*b^2/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(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 [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)^2/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2771, size = 1300, normalized size = 9.42 \begin{align*} \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 ] \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.56774, size = 892, normalized size = 6.46 \begin{align*} -\frac{\frac{2 \,{\left (2 \, a^{3} b - 2 \, a^{2} b^{2} - 2 \, a b^{3} + 2 \, b^{4} + b{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (f x + e\right )}{\sqrt{\frac{a^{3} - a^{2} b - a b^{2} + b^{3} + \sqrt{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )}^{2} - 4 \,{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )}}}{a^{2} b - 2 \, a b^{2} + b^{3}}}}\right )\right )}}{a^{3}{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |} - a^{2} b{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |} - a b^{2}{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |} + b^{3}{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |} +{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}^{2}} + \frac{2 \,{\left (2 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sqrt{a b}{\left | b \right |} - \sqrt{a b}{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |}{\left | b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (f x + e\right )}{\sqrt{\frac{a^{3} - a^{2} b - a b^{2} + b^{3} - \sqrt{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )}^{2} - 4 \,{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )}}}{a^{2} b - 2 \, a b^{2} + b^{3}}}}\right )\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}^{2} b -{\left (a^{3} b - a^{2} b^{2} - a b^{3} + b^{4}\right )}{\left | -a^{3} + 3 \, a^{2} b - 3 \, a b^{2} + b^{3} \right |}} + \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} \end{align*}

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