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

Optimal. Leaf size=130 \[ -\frac{\sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^3 f \sqrt{a+b}}-\frac{b \tan (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x (a+4 b)}{2 a^3}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.168873, antiderivative size = 130, 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 = {4132, 471, 527, 522, 203, 205} \[ -\frac{\sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^3 f \sqrt{a+b}}-\frac{b \tan (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x (a+4 b)}{2 a^3}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 (a+b) (a+2 b)-4 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a^2 (a+b) f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a+4 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}-\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}\\ &=\frac{(a+4 b) x}{2 a^3}-\frac{\sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^3 \sqrt{a+b} f}-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 11.8335, size = 825, normalized size = 6.35 \[ -\frac{(\cos (2 e+2 f x) a+a+2 b)^2 \left (16 x+\frac{\left (-a^3+6 b a^2+24 b^2 a+16 b^3\right ) \tan ^{-1}\left (\frac{\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt{b (\cos (e)-i \sin (e))^4}}+\frac{\left (a^2+8 b a+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (\cos (2 (e+f x)) a+a+2 b) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right ) \sec ^4(e+f x)}{128 a^2 \left (b \sec ^2(e+f x)+a\right )^2}-\frac{(\cos (2 e+2 f x) a+a+2 b)^2 \left (-64 (a+2 b) x+\frac{\left (a^4-16 b a^3-144 b^2 a^2-256 b^3 a-128 b^4\right ) \tan ^{-1}\left (\frac{\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt{b (\cos (e)-i \sin (e))^4}}+\frac{16 a \cos (2 f x) \sin (2 e)}{f}+\frac{16 a \cos (2 e) \sin (2 f x)}{f}-\frac{\left (a^3+18 b a^2+48 b^2 a+32 b^3\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (\cos (2 (e+f x)) a+a+2 b) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right ) \sec ^4(e+f x)}{256 a^3 \left (b \sec ^2(e+f x)+a\right )^2}+\frac{(\cos (2 e+2 f x) a+a+2 b)^2 \left (\frac{(a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}-\frac{a \sqrt{b} \sin (2 (e+f x))}{(a+b) (\cos (2 (e+f x)) a+a+2 b)}\right ) \sec ^4(e+f x)}{128 b^{3/2} f \left (b \sec ^2(e+f x)+a\right )^2}+\frac{(\cos (2 e+2 f x) a+a+2 b)^2 \left (\frac{\sqrt{b} (a+2 b) \sin (2 (e+f x))}{(a+b) (\cos (2 (e+f x)) a+a+2 b)}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}\right ) \sec ^4(e+f x)}{256 b^{3/2} f \left (b \sec ^2(e+f x)+a\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(Sec[f*
x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e
])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b^2)*((
a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin
[e]))))/(128*a^2*(a + b*Sec[e + f*x]^2)^2) - ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-64*(a + 2*b)*x
+ ((a^4 - 16*a^3*b - 144*a^2*b^2 - 256*a*b^3 - 128*b^4)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)
*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a
+ b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (16*a*Cos[2*f*x]*Sin[2*e])/f + (16*a*Cos[2*e]*Sin[2*f*x])/f - ((
a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e +
f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(256*a^3*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f
*x])^2*Sec[e + f*x]^4*(((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2
*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))))/(128*b^(3/2)*f*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b +
a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-((a*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2)) + (Sqrt
[b]*(a + 2*b)*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))))/(256*b^(3/2)*f*(a + b*Sec[e + f*x]^
2)^2)

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Maple [A]  time = 0.098, size = 155, normalized size = 1.2 \begin{align*} -{\frac{\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{2\,f{a}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{f{a}^{3}}}-{\frac{b\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,b}{2\,f{a}^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-2\,{\frac{{b}^{2}}{f{a}^{3}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{b\tan \left ( fx+e \right ) }{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.626608, size = 1046, normalized size = 8.05 \begin{align*} \left [\frac{4 \,{\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 4 \,{\left (a b + 4 \, b^{2}\right )} f x +{\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt{-\frac{b}{a + b}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \,{\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac{2 \,{\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 2 \,{\left (a b + 4 \, b^{2}\right )} f x +{\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt{\frac{b}{a + b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \,{\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \,{\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 4*(a*b + 4*b^2)*f*x + ((3*a^2 + 4*a*b)*cos(f*x + e)^2 + 3*a*b + 4*b
^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 +
3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x +
e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 4*(a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a^4*f*cos(f*x
+ e)^2 + a^3*b*f), 1/4*(2*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 2*(a*b + 4*b^2)*f*x + ((3*a^2 + 4*a*b)*cos(f*x +
e)^2 + 3*a*b + 4*b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x +
e)*sin(f*x + e))) - 2*(a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*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*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2681, size = 213, normalized size = 1.64 \begin{align*} \frac{\frac{{\left (f x + e\right )}{\left (a + 4 \, b\right )}}{a^{3}} - \frac{{\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 + b^{2}}}\right )\right )}{\left (3 \, a b + 4 \, b^{2}\right )}}{\sqrt{a b + b^{2}} a^{3}} - \frac{2 \, b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + 2 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )} a^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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