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

Optimal. Leaf size=193 \[ -\frac{\sqrt{b} \left (15 a^2+10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} f (a-b)^4}-\frac{b (11 a+b) \tan (e+f x)}{8 a f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )}-\frac{3 b \tan (e+f x)}{4 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac{x (a+5 b)}{2 (a-b)^4} \]

[Out]

((a + 5*b)*x)/(2*(a - b)^4) - (Sqrt[b]*(15*a^2 + 10*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(3
/2)*(a - b)^4*f) - (Cos[e + f*x]*Sin[e + f*x])/(2*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - (3*b*Tan[e + f*x])/(4*
(a - b)^2*f*(a + b*Tan[e + f*x]^2)^2) - (b*(11*a + b)*Tan[e + f*x])/(8*a*(a - b)^3*f*(a + b*Tan[e + f*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.245847, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{\sqrt{b} \left (15 a^2+10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} f (a-b)^4}-\frac{b (11 a+b) \tan (e+f x)}{8 a f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )}-\frac{3 b \tan (e+f x)}{4 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac{x (a+5 b)}{2 (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 5*b)*x)/(2*(a - b)^4) - (Sqrt[b]*(15*a^2 + 10*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(3
/2)*(a - b)^4*f) - (Cos[e + f*x]*Sin[e + f*x])/(2*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - (3*b*Tan[e + f*x])/(4*
(a - b)^2*f*(a + b*Tan[e + f*x]^2)^2) - (b*(11*a + b)*Tan[e + f*x])/(8*a*(a - b)^3*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 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 )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^3} \, 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 )^2}+\frac{\operatorname{Subst}\left (\int \frac{a-5 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, 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 )^2}-\frac{3 b \tan (e+f x)}{4 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{2 a (2 a+b)-18 a b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 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 )^2}-\frac{3 b \tan (e+f x)}{4 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (11 a+b) \tan (e+f x)}{8 a (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a \left (4 a^2+9 a b-b^2\right )-2 a b (11 a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{16 a^2 (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 )^2}-\frac{3 b \tan (e+f x)}{4 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (11 a+b) \tan (e+f x)}{8 a (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}+\frac{(a+5 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^4 f}-\frac{\left (b \left (15 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 (a-b)^4 f}\\ &=\frac{(a+5 b) x}{2 (a-b)^4}-\frac{\sqrt{b} \left (15 a^2+10 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} (a-b)^4 f}-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{3 b \tan (e+f x)}{4 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b (11 a+b) \tan (e+f x)}{8 a (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.59452, size = 164, normalized size = 0.85 \[ \frac{\frac{\sqrt{b} \left (-15 a^2-10 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{4 b^2 (a-b) \sin (2 (e+f x))}{((a-b) \cos (2 (e+f x))+a+b)^2}+4 (a+5 b) (e+f x)-2 (a-b) \sin (2 (e+f x))-\frac{b (a-b) (9 a+b) \sin (2 (e+f x))}{a ((a-b) \cos (2 (e+f x))+a+b)}}{8 f (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + 5*b)*(e + f*x) + (Sqrt[b]*(-15*a^2 - 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/a^(3/2) - 2
*(a - b)*Sin[2*(e + f*x)] + (4*(a - b)*b^2*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)])^2 - ((a - b)*b
*(9*a + b)*Sin[2*(e + f*x)])/(a*(a + b + (a - b)*Cos[2*(e + f*x)])))/(8*(a - b)^4*f)

________________________________________________________________________________________

Maple [B]  time = 0.081, size = 430, normalized size = 2.2 \begin{align*} -{\frac{7\,a{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{4\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}-{\frac{9\,b\tan \left ( fx+e \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{5\,{b}^{2}\tan \left ( fx+e \right ) a}{4\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{3}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,ab}{8\,f \left ( a-b \right ) ^{4}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{5\,{b}^{2}}{4\,f \left ( a-b \right ) ^{4}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{3}}{8\,f \left ( a-b \right ) ^{4}a}\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 ) ^{4} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{b\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{4} \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 ) ^{4}}}+{\frac{5\,\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{2\,f \left ( a-b \right ) ^{4}}} \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)^3,x)

[Out]

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

________________________________________________________________________________________

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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.87567, size = 2402, normalized size = 12.45 \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)^2/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(16*(a^4 + 3*a^3*b - 9*a^2*b^2 + 5*a*b^3)*f*x*cos(f*x + e)^4 + 32*(a^3*b + 4*a^2*b^2 - 5*a*b^3)*f*x*cos(
f*x + e)^2 + 16*(a^2*b^2 + 5*a*b^3)*f*x - ((15*a^4 - 20*a^3*b - 6*a^2*b^2 + 12*a*b^3 - b^4)*cos(f*x + e)^4 + 1
5*a^2*b^2 + 10*a*b^3 - b^4 + 2*(15*a^3*b - 5*a^2*b^2 - 11*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*(4*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + (17*a^3*b - 33*a^2*b^2 + 15*a*b^3 + b^4)*cos(f*x +
e)^3 + (11*a^2*b^2 - 10*a*b^3 - b^4)*cos(f*x + e))*sin(f*x + e))/((a^7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 1
5*a^3*b^4 - 6*a^2*b^5 + a*b^6)*f*cos(f*x + e)^4 + 2*(a^6*b - 5*a^5*b^2 + 10*a^4*b^3 - 10*a^3*b^4 + 5*a^2*b^5 -
a*b^6)*f*cos(f*x + e)^2 + (a^5*b^2 - 4*a^4*b^3 + 6*a^3*b^4 - 4*a^2*b^5 + a*b^6)*f), 1/16*(8*(a^4 + 3*a^3*b -
9*a^2*b^2 + 5*a*b^3)*f*x*cos(f*x + e)^4 + 16*(a^3*b + 4*a^2*b^2 - 5*a*b^3)*f*x*cos(f*x + e)^2 + 8*(a^2*b^2 + 5
*a*b^3)*f*x + ((15*a^4 - 20*a^3*b - 6*a^2*b^2 + 12*a*b^3 - b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 10*a*b^3 - b^4 +
2*(15*a^3*b - 5*a^2*b^2 - 11*a*b^3 + b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*s
qrt(b/a)/(b*cos(f*x + e)*sin(f*x + e))) - 2*(4*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + (17*a^3*b
- 33*a^2*b^2 + 15*a*b^3 + b^4)*cos(f*x + e)^3 + (11*a^2*b^2 - 10*a*b^3 - b^4)*cos(f*x + e))*sin(f*x + e))/((a^
7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 15*a^3*b^4 - 6*a^2*b^5 + a*b^6)*f*cos(f*x + e)^4 + 2*(a^6*b - 5*a^5*b^
2 + 10*a^4*b^3 - 10*a^3*b^4 + 5*a^2*b^5 - a*b^6)*f*cos(f*x + e)^2 + (a^5*b^2 - 4*a^4*b^3 + 6*a^3*b^4 - 4*a^2*b
^5 + a*b^6)*f)]

________________________________________________________________________________________

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.61142, size = 381, normalized size = 1.97 \begin{align*} \frac{\frac{4 \,{\left (f x + e\right )}{\left (a + 5 \, b\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (15 \, a^{2} b + 10 \, a b^{2} - b^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )}}{{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt{a b}} - \frac{4 \, \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (\tan \left (f x + e\right )^{2} + 1\right )}} - \frac{7 \, a b^{2} \tan \left (f x + e\right )^{3} + b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) - a b^{2} \tan \left (f x + e\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}}}{8 \, 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)^3,x, algorithm="giac")

[Out]

1/8*(4*(f*x + e)*(a + 5*b)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (15*a^2*b + 10*a*b^2 - b^3)*(pi*floor
((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^
4)*sqrt(a*b)) - 4*tan(f*x + e)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(f*x + e)^2 + 1)) - (7*a*b^2*tan(f*x + e)^
3 + b^3*tan(f*x + e)^3 + 9*a^2*b*tan(f*x + e) - a*b^2*tan(f*x + e))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(b*ta
n(f*x + e)^2 + a)^2))/f