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

Optimal. Leaf size=97 \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.0923569, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2754, 12, 2660, 618, 204} \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2754

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.303997, size = 96, normalized size = 0.99 \[ \frac{\frac{2 (a c-b d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.072, size = 309, normalized size = 3.2 \begin{align*} -2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) bd}{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}{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{da}{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{cb}{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{ca}{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{bd}{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))/(a+b*sin(f*x+e))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06916, size = 871, normalized size = 8.98 \begin{align*} \left [-\frac{{\left (a^{2} c - a b d +{\left (a b c - b^{2} d\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 - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}}, -\frac{{\left (a^{2} c - a b d +{\left (a b c - b^{2} d\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 - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a 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))/(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/2*((a^2*c - a*b*d + (a*b*c - b^2*d)*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 - b^3)*c - (a^3 - a*b^2)*d)*cos(f*x + e))/((a^4*b - 2*a^
2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f), -((a^2*c - a*b*d + (a*b*c - b^2*d)*sin(f*x + e))*s
qrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) - ((a^2*b - b^3)*c - (a^3 - a*b^2)
*d)*cos(f*x + e))/((a^4*b - 2*a^2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*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))/(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.43228, size = 212, normalized size = 2.19 \begin{align*} \frac{2 \,{\left (\frac{{\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 c - b d\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{b^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a b d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a b c - a^{2} d}{{\left (a^{3} - a b^{2}\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 )}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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