3.682 \(\int \frac{(a+b \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=93 \[ \frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{b x (b c-2 a d)}{d^2}-\frac{b^2 \cos (e+f x)}{d f} \]

[Out]

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

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Rubi [A]  time = 0.181709, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2746, 2735, 2660, 618, 204} \[ \frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{b x (b c-2 a d)}{d^2}-\frac{b^2 \cos (e+f x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2746

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

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

Mathematica [A]  time = 0.211975, size = 89, normalized size = 0.96 \[ -\frac{-\frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+b (e+f x) (b c-2 a d)+b^2 d \cos (e+f x)}{d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 226, normalized size = 2.4 \begin{align*} 2\,{\frac{{a}^{2}}{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 ) }-4\,{\frac{abc}{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{{c}^{2}{b}^{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{{b}^{2}}{df \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+4\,{\frac{b\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) a}{df}}-2\,{\frac{{b}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{f{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/(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^2-4/f/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*b*c+2/f/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))*c^2*b^2-2/f*b^2/d/(1+tan(1/2*f*x+1/2*e)^2)+4/f*b/d*arctan(tan(1/2*f*x+1/2*e))*a-2/f
*b^2/d^2*arctan(tan(1/2*f*x+1/2*e))*c

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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+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77071, size = 798, normalized size = 8.58 \begin{align*} \left [-\frac{2 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d - b^{2} c d^{2} + 2 \, a b d^{3}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-c^{2} + d^{2}} \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 (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}}}{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^{2} c^{2} d - b^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left (c^{2} d^{2} - d^{4}\right )} f}, -\frac{{\left (b^{2} c^{3} - 2 \, a b c^{2} d - b^{2} c d^{2} + 2 \, a b d^{3}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{c^{2} - d^{2}} \arctan \left (-\frac{c \sin \left (f x + e\right ) + d}{\sqrt{c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) +{\left (b^{2} c^{2} d - b^{2} d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{2} d^{2} - d^{4}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.42492, size = 181, normalized size = 1.95 \begin{align*} -\frac{\frac{{\left (b^{2} c - 2 \, a b d\right )}{\left (f x + e\right )}}{d^{2}} + \frac{2 \, b^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} d} - \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} 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+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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