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

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

[Out]

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

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Rubi [A]  time = 0.167061, 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} \[ \frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 f \sqrt{a^2-b^2}}+\frac{d x (2 b c-a d)}{b^2}-\frac{d^2 \cos (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c - a*d)*x)/b^2 + (2*(b*c - a*d)^2*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b^2*Sqrt[a^2 - b
^2]*f) - (d^2*Cos[e + f*x])/(b*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{(c+d \sin (e+f x))^2}{a+b \sin (e+f x)} \, dx &=-\frac{d^2 \cos (e+f x)}{b f}+\frac{\int \frac{b c^2+d (2 b c-a d) \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{b}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}+\frac{(b c-a d)^2 \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^2}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}+\frac{\left (2 (b c-a d)^2\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^2 f}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}-\frac{\left (4 (b c-a d)^2\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^2 f}\\ &=\frac{d (2 b c-a d) x}{b^2}+\frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2} f}-\frac{d^2 \cos (e+f x)}{b f}\\ \end{align*}

Mathematica [A]  time = 0.159349, size = 90, normalized size = 0.97 \[ \frac{\frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+d (e+f x) (2 b c-a d)-b d^2 \cos (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.065, size = 226, normalized size = 2.4 \begin{align*} -2\,{\frac{{d}^{2}}{bf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) a}{{b}^{2}f}}+4\,{\frac{d\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{bf}}+2\,{\frac{{a}^{2}{d}^{2}}{{b}^{2}f\sqrt{{a}^{2}-{b}^{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{acd}{bf\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{c}^{2}}{f\sqrt{{a}^{2}-{b}^{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((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e)),x)

[Out]

-2/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)-2/f*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))*a+4/f*d/b*arctan(tan(1/2*f*x+1/2*e)
)*c+2/f/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*d^2-4/f/b/(a^2-b^2)^(
1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a*c*d+2/f/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/
2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*c^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((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59945, size = 771, normalized size = 8.29 \begin{align*} \left [-\frac{2 \,{\left (a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right ) - 2 \,{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} f}, -\frac{{\left (a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right ) -{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right )}{{\left (a^{2} b^{2} - b^{4}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.35917, size = 181, normalized size = 1.95 \begin{align*} \frac{\frac{{\left (2 \, b c d - a d^{2}\right )}{\left (f x + e\right )}}{b^{2}} - \frac{2 \, d^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} b} + \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 (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 )}}{\sqrt{a^{2} - b^{2}} b^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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