3.16 \(\int \frac{\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=153 \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \left (a^2-b^2\right )^{3/2}}+\frac{a^2 (A b-a B) \cos (e+f x)}{b^2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{x (A b-2 a B)}{b^3}-\frac{B \cos (e+f x)}{b^2 f} \]

[Out]

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

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Rubi [A]  time = 0.393458, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2988, 3023, 2735, 2660, 618, 204} \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \left (a^2-b^2\right )^{3/2}}+\frac{a^2 (A b-a B) \cos (e+f x)}{b^2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{x (A b-2 a B)}{b^3}-\frac{B \cos (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [A]  time = 0.870006, size = 147, normalized size = 0.96 \[ \frac{\frac{2 a \left (-a^2 A b+2 a^3 B-3 a b^2 B+2 A b^3\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a^2 b (A b-a B) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}+(e+f x) (A b-2 a B)-b B \cos (e+f x)}{b^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - 2*a*B)*(e + f*x) + (2*a*(-(a^2*A*b) + 2*A*b^3 + 2*a^3*B - 3*a*b^2*B)*ArcTan[(b + a*Tan[(e + f*x)/2])/S
qrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - b*B*Cos[e + f*x] + (a^2*b*(A*b - a*B)*Cos[e + f*x])/((a - b)*(a + b)*(a +
b*Sin[e + f*x])))/(b^3*f)

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Maple [B]  time = 0.106, size = 493, normalized size = 3.2 \begin{align*} -2\,{\frac{B}{{b}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{b}^{2}f}}-4\,{\frac{B\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) a}{{b}^{3}f}}+2\,{\frac{a\tan \left ( 1/2\,fx+e/2 \right ) A}{f \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{a}^{2}\tan \left ( 1/2\,fx+e/2 \right ) B}{bf \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{A{a}^{2}}{bf \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{B{a}^{3}}{{b}^{2}f \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{a}^{3}A}{{b}^{2}f \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{aA}{f \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{{a}^{4}B}{{b}^{3}f \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-6\,{\frac{B{a}^{2}}{bf \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f/b^2*B/(1+tan(1/2*f*x+1/2*e)^2)+2/f/b^2*A*arctan(tan(1/2*f*x+1/2*e))-4/f/b^3*B*arctan(tan(1/2*f*x+1/2*e))*
a+2/f*a/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*tan(1/2*f*x+1/2*e)*A-2/f*a^2/b/(tan(1/2*f*
x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*tan(1/2*f*x+1/2*e)*B+2/f*a^2/b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(
1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*A-2/f*a^3/b^2/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*B-2/f*
a^3/b^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*A+4/f*a/(a^2-b^2)^(3/2)*arcta
n(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*A+4/f*a^4/b^3/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+
1/2*e)+2*b)/(a^2-b^2)^(1/2))*B-6/f*a^2/b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/
2))*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*sin(f*x+e))/(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.83895, size = 1729, normalized size = 11.3 \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*sin(f*x+e))/(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*f*x + (2*B*a^5 - A*a^4*b - 3*
B*a^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4)*sin(f*x + e))*sqrt(-a^2 + b^2)*log
(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 + 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x +
e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*(2*B*a^5*b - A*a^4*b^2 - 3*B
*a^3*b^3 + A*a^2*b^4 + B*a*b^5)*cos(f*x + e) + 2*((2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b
^5 - A*b^6)*f*x + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(f*x + e))*sin(f*x + e))/((a^4*b^4 - 2*a^2*b^6 + b^8)*f
*sin(f*x + e) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*f), -((2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b
^4 - A*a*b^5)*f*x + (2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*
A*a*b^4)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + (2*B*a^5
*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b^4 + B*a*b^5)*cos(f*x + e) + ((2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A
*a^2*b^4 + 2*B*a*b^5 - A*b^6)*f*x + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(f*x + e))*sin(f*x + e))/((a^4*b^4 -
2*a^2*b^6 + b^8)*f*sin(f*x + e) + (a^5*b^3 - 2*a^3*b^5 + a*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)**2*(A+B*sin(f*x+e))/(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.21139, size = 501, normalized size = 3.27 \begin{align*} \frac{\frac{2 \,{\left (2 \, B a^{4} - A a^{3} b - 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (B a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - A a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - A a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, B b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, B a^{3} - A a^{2} b - B a b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a\right )}{\left (a^{2} b^{2} - b^{4}\right )}} - \frac{{\left (2 \, B a - A b\right )}{\left (f x + e\right )}}{b^{3}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(2*B*a^4 - A*a^3*b - 3*B*a^2*b^2 + 2*A*a*b^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*
f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/((a^2*b^3 - b^5)*sqrt(a^2 - b^2)) - 2*(B*a^2*b*tan(1/2*f*x + 1/2*e)^3 - A*
a*b^2*tan(1/2*f*x + 1/2*e)^3 + 2*B*a^3*tan(1/2*f*x + 1/2*e)^2 - A*a^2*b*tan(1/2*f*x + 1/2*e)^2 - B*a*b^2*tan(1
/2*f*x + 1/2*e)^2 + 3*B*a^2*b*tan(1/2*f*x + 1/2*e) - A*a*b^2*tan(1/2*f*x + 1/2*e) - 2*B*b^3*tan(1/2*f*x + 1/2*
e) + 2*B*a^3 - A*a^2*b - B*a*b^2)/((a*tan(1/2*f*x + 1/2*e)^4 + 2*b*tan(1/2*f*x + 1/2*e)^3 + 2*a*tan(1/2*f*x +
1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)*(a^2*b^2 - b^4)) - (2*B*a - A*b)*(f*x + e)/b^3)/f