3.255 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=171 \[ -\frac{2 a^2 (c-d)^2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f \sqrt{c^2-d^2}}-\frac{a^2 x \left (2 A d (c-2 d)-B \left (2 c^2-4 c d+3 d^2\right )\right )}{2 d^3}+\frac{a^2 (-2 A d+2 B c-3 B d) \cos (e+f x)}{2 d^2 f}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{2 d f} \]

[Out]

-(a^2*(2*A*(c - 2*d)*d - B*(2*c^2 - 4*c*d + 3*d^2))*x)/(2*d^3) - (2*a^2*(c - d)^2*(B*c - A*d)*ArcTan[(d + c*Ta
n[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^3*Sqrt[c^2 - d^2]*f) + (a^2*(2*B*c - 2*A*d - 3*B*d)*Cos[e + f*x])/(2*d^2*
f) - (B*Cos[e + f*x]*(a^2 + a^2*Sin[e + f*x]))/(2*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.52178, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2976, 2968, 3023, 2735, 2660, 618, 204} \[ -\frac{2 a^2 (c-d)^2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f \sqrt{c^2-d^2}}-\frac{a^2 x \left (2 A d (c-2 d)-B \left (2 c^2-4 c d+3 d^2\right )\right )}{2 d^3}+\frac{a^2 (-2 A d+2 B c-3 B d) \cos (e+f x)}{2 d^2 f}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(2*A*(c - 2*d)*d - B*(2*c^2 - 4*c*d + 3*d^2))*x)/(2*d^3) - (2*a^2*(c - d)^2*(B*c - A*d)*ArcTan[(d + c*Ta
n[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^3*Sqrt[c^2 - d^2]*f) + (a^2*(2*B*c - 2*A*d - 3*B*d)*Cos[e + f*x])/(2*d^2*
f) - (B*Cos[e + f*x]*(a^2 + a^2*Sin[e + f*x]))/(2*d*f)

Rule 2976

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

Rule 2968

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

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

Mathematica [A]  time = 0.629303, size = 177, normalized size = 1.04 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (2 (e+f x) \left (2 A d (2 d-c)+B \left (2 c^2-4 c d+3 d^2\right )\right )-\frac{8 (c-d)^2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}-4 d (A d-B c+2 B d) \cos (e+f x)-B d^2 \sin (2 (e+f x))\right )}{4 d^3 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[e + f*x])^2*(2*(2*A*d*(-c + 2*d) + B*(2*c^2 - 4*c*d + 3*d^2))*(e + f*x) - (8*(c - d)^2*(B*c - A*
d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] - 4*d*(-(B*c) + A*d + 2*B*d)*Cos[e + f*x]
- B*d^2*Sin[2*(e + f*x)]))/(4*d^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)

________________________________________________________________________________________

Maple [B]  time = 0.143, size = 713, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^2/d^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A*c^2-4/f*a^2/d/(c^2-d^2)
^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A*c+2/f*a^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*ta
n(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A-2/f*a^2/d^3/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(
c^2-d^2)^(1/2))*B*c^3+4/f*a^2/d^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c
^2-2/f*a^2/d/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c+1/f*a^2/d/(1+tan(1/2
*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)^3-2/f*a^2/d/(1+tan(1/2*f*x+1/2*e)^2)^2*A*tan(1/2*f*x+1/2*e)^2+2/f*a^2/d^
2/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)^2*c-4/f*a^2/d/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e
)^2-1/f*a^2/d/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)-2/f*a^2/d/(1+tan(1/2*f*x+1/2*e)^2)^2*A+2/f*a^2/d
^2/(1+tan(1/2*f*x+1/2*e)^2)^2*B*c-4/f*a^2/d/(1+tan(1/2*f*x+1/2*e)^2)^2*B-2/f*a^2/d^2*arctan(tan(1/2*f*x+1/2*e)
)*A*c+4/f*a^2/d*arctan(tan(1/2*f*x+1/2*e))*A+2/f*a^2/d^3*arctan(tan(1/2*f*x+1/2*e))*B*c^2-4/f*a^2/d^2*arctan(t
an(1/2*f*x+1/2*e))*B*c+3/f*a^2/d*arctan(tan(1/2*f*x+1/2*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((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.29636, size = 1029, normalized size = 6.02 \begin{align*} \left [-\frac{B a^{2} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (2 \, B a^{2} c^{2} - 2 \,{\left (A + 2 \, B\right )} a^{2} c d +{\left (4 \, A + 3 \, B\right )} a^{2} d^{2}\right )} f x +{\left (B a^{2} c^{2} -{\left (A + B\right )} a^{2} c d + A a^{2} d^{2}\right )} \sqrt{-\frac{c - d}{c + d}} \log \left (-\frac{{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} - 2 \,{\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 2 \,{\left (B a^{2} c d -{\left (A + 2 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}, -\frac{B a^{2} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (2 \, B a^{2} c^{2} - 2 \,{\left (A + 2 \, B\right )} a^{2} c d +{\left (4 \, A + 3 \, B\right )} a^{2} d^{2}\right )} f x - 2 \,{\left (B a^{2} c^{2} -{\left (A + B\right )} a^{2} c d + A a^{2} d^{2}\right )} \sqrt{\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt{\frac{c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 2 \,{\left (B a^{2} c d -{\left (A + 2 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[-1/2*(B*a^2*d^2*cos(f*x + e)*sin(f*x + e) - (2*B*a^2*c^2 - 2*(A + 2*B)*a^2*c*d + (4*A + 3*B)*a^2*d^2)*f*x + (
B*a^2*c^2 - (A + B)*a^2*c*d + A*a^2*d^2)*sqrt(-(c - d)/(c + d))*log(-((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin
(f*x + e) - c^2 - d^2 - 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c
+ d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) - 2*(B*a^2*c*d - (A + 2*B)*a^2*d^2)*cos(f*x + e)
)/(d^3*f), -1/2*(B*a^2*d^2*cos(f*x + e)*sin(f*x + e) - (2*B*a^2*c^2 - 2*(A + 2*B)*a^2*c*d + (4*A + 3*B)*a^2*d^
2)*f*x - 2*(B*a^2*c^2 - (A + B)*a^2*c*d + A*a^2*d^2)*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + e) + d)*sqrt((
c - d)/(c + d))/((c - d)*cos(f*x + e))) - 2*(B*a^2*c*d - (A + 2*B)*a^2*d^2)*cos(f*x + e))/(d^3*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((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.26906, size = 424, normalized size = 2.48 \begin{align*} \frac{\frac{{\left (2 \, B a^{2} c^{2} - 2 \, A a^{2} c d - 4 \, B a^{2} c d + 4 \, A a^{2} d^{2} + 3 \, B a^{2} d^{2}\right )}{\left (f x + e\right )}}{d^{3}} - \frac{4 \,{\left (B a^{2} c^{3} - A a^{2} c^{2} d - 2 \, B a^{2} c^{2} d + 2 \, A a^{2} c d^{2} + B a^{2} c d^{2} - A a^{2} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{\sqrt{c^{2} - d^{2}} d^{3}} + \frac{2 \,{\left (B a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, B a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, A a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, B a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, B a^{2} c - 2 \, A a^{2} d - 4 \, B a^{2} d\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} d^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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