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

Optimal. Leaf size=129 \[ \frac{2 (b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{d^2 x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.217906, antiderivative size = 129, 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 = {2790, 2735, 2660, 618, 204} \[ \frac{2 (b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{d^2 x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] - Di
st[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(2*b*c*d - a*(c^2 + d^2)) + (a^2
*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1) + c^2*(m + 2)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 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{(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^2} \, dx &=\frac{(b c-a d)^2 \cos (e+f x)}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\int \frac{-b \left (2 b c d-a \left (c^2+d^2\right )\right )+\left (a^2-b^2\right ) d^2 \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{d^2 x}{b^2}+\frac{(b c-a d)^2 \cos (e+f x)}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left ((b c-a d) \left (a b c+a^2 d-2 b^2 d\right )\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{d^2 x}{b^2}+\frac{(b c-a d)^2 \cos (e+f x)}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left (2 (b c-a d) \left (a b c+a^2 d-2 b^2 d\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^2 \left (a^2-b^2\right ) f}\\ &=\frac{d^2 x}{b^2}+\frac{(b c-a d)^2 \cos (e+f x)}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\left (4 (b c-a d) \left (a b c+a^2 d-2 b^2 d\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^2 \left (a^2-b^2\right ) f}\\ &=\frac{d^2 x}{b^2}+\frac{2 (b c-a d) \left (a b c+a^2 d-2 b^2 d\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2} f}+\frac{(b c-a d)^2 \cos (e+f x)}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.549897, size = 133, normalized size = 1.03 \[ \frac{-\frac{2 \left (a^3 d^2-a b^2 \left (c^2+2 d^2\right )+2 b^3 c d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b (b c-a d)^2 \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}+d^2 (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.079, size = 556, normalized size = 4.3 \begin{align*} 2\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{b}^{2}f}}+2\,{\frac{a\tan \left ( 1/2\,fx+e/2 \right ){d}^{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 ) }}-4\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) bcd}{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{{b}^{2}\tan \left ( 1/2\,fx+e/2 \right ){c}^{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 ) a}}+2\,{\frac{{a}^{2}{d}^{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 ) }}-4\,{\frac{acd}{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{{c}^{2}b}{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}{d}^{2}}{{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 ) }+2\,{\frac{{c}^{2}a}{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{d}^{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{bcd}{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 ) } \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))^2,x)

[Out]

2/f*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))+2/f/(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*tan(1
/2*f*x+1/2*e)*d^2-4/f*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)*c*d+2/f
*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)/a*tan(1/2*f*x+1/2*e)*c^2+2/f/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*d^2-4/f/(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*c*d+2/f*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)*c^2-2/f/b^2/(a^2-b^2)^(3/2)*ar
ctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^3*d^2+2/f/(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*c^2+4/f/(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*d^2-4/f*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))*c*d

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.27134, size = 1401, normalized size = 10.86 \begin{align*} \left [\frac{2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d^{2} f x \sin \left (f x + e\right ) + 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d^{2} f x +{\left (a^{2} b^{2} c^{2} - 2 \, a b^{3} c d -{\left (a^{4} - 2 \, a^{2} b^{2}\right )} d^{2} +{\left (a b^{3} c^{2} - 2 \, b^{4} c d -{\left (a^{3} b - 2 \, a b^{3}\right )} d^{2}\right )} \sin \left (f x + e\right )\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 ({\left (a^{2} b^{3} - b^{5}\right )} c^{2} - 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} c d +{\left (a^{4} b - a^{2} b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} f \sin \left (f x + e\right ) +{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} f\right )}}, \frac{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d^{2} f x \sin \left (f x + e\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d^{2} f x -{\left (a^{2} b^{2} c^{2} - 2 \, a b^{3} c d -{\left (a^{4} - 2 \, a^{2} b^{2}\right )} d^{2} +{\left (a b^{3} c^{2} - 2 \, b^{4} c d -{\left (a^{3} b - 2 \, a b^{3}\right )} d^{2}\right )} \sin \left (f x + e\right )\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 ({\left (a^{2} b^{3} - b^{5}\right )} c^{2} - 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} c d +{\left (a^{4} b - a^{2} b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} f \sin \left (f x + e\right ) +{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\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))^2,x, algorithm="fricas")

[Out]

[1/2*(2*(a^4*b - 2*a^2*b^3 + b^5)*d^2*f*x*sin(f*x + e) + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d^2*f*x + (a^2*b^2*c^2 -
2*a*b^3*c*d - (a^4 - 2*a^2*b^2)*d^2 + (a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)*d^2)*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*((a^2*b^3 - b^5)*
c^2 - 2*(a^3*b^2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^4*b^3 - 2*a^2*b^5 + b^7)*f*sin(f*x +
e) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*f), ((a^4*b - 2*a^2*b^3 + b^5)*d^2*f*x*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*
b^4)*d^2*f*x - (a^2*b^2*c^2 - 2*a*b^3*c*d - (a^4 - 2*a^2*b^2)*d^2 + (a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)
*d^2)*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))) + ((a^2*b^3 -
b^5)*c^2 - 2*(a^3*b^2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^4*b^3 - 2*a^2*b^5 + b^7)*f*sin(
f*x + e) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*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))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.39138, size = 336, normalized size = 2.6 \begin{align*} \frac{\frac{{\left (f x + e\right )} d^{2}}{b^{2}} + \frac{2 \,{\left (a b^{2} c^{2} - 2 \, b^{3} c d - a^{3} d^{2} + 2 \, a b^{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 )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (b^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a b^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{2} b d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )}}{{\left (a^{3} b - a b^{3}\right )}{\left (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 )}}}{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))^2,x, algorithm="giac")

[Out]

((f*x + e)*d^2/b^2 + 2*(a*b^2*c^2 - 2*b^3*c*d - a^3*d^2 + 2*a*b^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)))/((a^2*b^2 - b^4)*sqrt(a^2 - b^2)) + 2*(b^3*c^2*tan(1
/2*f*x + 1/2*e) - 2*a*b^2*c*d*tan(1/2*f*x + 1/2*e) + a^2*b*d^2*tan(1/2*f*x + 1/2*e) + a*b^2*c^2 - 2*a^2*b*c*d
+ a^3*d^2)/((a^3*b - a*b^3)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)))/f