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

Optimal. Leaf size=98 \[ \frac{2 a (c-d) (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{a x (B c-d (A+B))}{d^2}-\frac{a B \cos (e+f x)}{d f} \]

[Out]

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

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Rubi [A]  time = 0.273027, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2968, 3023, 2735, 2660, 618, 204} \[ \frac{2 a (c-d) (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{a x (B c-d (A+B))}{d^2}-\frac{a B \cos (e+f x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.650054, size = 196, normalized size = 2. \[ \frac{a (\sin (e+f x)+1) \left (\frac{2 (c-d) (\cos (e)-i \sin (e)) (B c-A d) \tan ^{-1}\left (\frac{(\cos (e)-i \sin (e)) \sec \left (\frac{f x}{2}\right ) \left (c \sin \left (\frac{f x}{2}\right )+d \cos \left (e+\frac{f x}{2}\right )\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{f \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}+A d x+B x (d-c)+\frac{B d \sin (e) \sin (f x)}{f}-\frac{B d \cos (e) \cos (f x)}{f}\right )}{d^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*d*x + B*(-c + d)*x - (B*d*Cos[e]*Cos[f*x])/f + (2*(c - d)*(B*c - A*d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*S
in[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e
]))/(Sqrt[c^2 - d^2]*f*Sqrt[(Cos[e] - I*Sin[e])^2]) + (B*d*Sin[e]*Sin[f*x])/f)*(1 + Sin[e + f*x]))/(d^2*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])^2)

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Maple [B]  time = 0.117, size = 294, normalized size = 3. \begin{align*} -2\,{\frac{Aac}{df\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{Aa}{f\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{Ba{c}^{2}}{f{d}^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{Bac}{df\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{Ba}{df \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{Aa\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{df}}-2\,{\frac{Ba\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{f{d}^{2}}}+2\,{\frac{Ba\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{df}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*a/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/(c^2-d^2)^(1/2)*ar
ctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A+2/f*a/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/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-2/f*a/d*B/(1+tan(1/2*f*x+1/2*e)^2)+2/f*a/d*A*arctan(tan(1/2*f*x+1/2*e))-2/f*a/d^2*B*arctan(tan(1/2*
f*x+1/2*e))*c+2/f*a/d*B*arctan(tan(1/2*f*x+1/2*e))

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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((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98764, size = 680, normalized size = 6.94 \begin{align*} \left [-\frac{2 \, B a d \cos \left (f x + e\right ) + 2 \,{\left (B a c -{\left (A + B\right )} a d\right )} f x -{\left (B a c - A a d\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 \, d^{2} f}, -\frac{B a d \cos \left (f x + e\right ) +{\left (B a c -{\left (A + B\right )} a d\right )} f x +{\left (B a c - A a d\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 )}{d^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.21573, size = 190, normalized size = 1.94 \begin{align*} -\frac{\frac{{\left (B a c - A a d - B a d\right )}{\left (f x + e\right )}}{d^{2}} + \frac{2 \, B a}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} d} - \frac{2 \,{\left (B a c^{2} - A a c d - B a c d + A a d^{2}\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^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((B*a*c - A*a*d - B*a*d)*(f*x + e)/d^2 + 2*B*a/((tan(1/2*f*x + 1/2*e)^2 + 1)*d) - 2*(B*a*c^2 - A*a*c*d - B*a*
c*d + A*a*d^2)*(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^2))/f