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

Optimal. Leaf size=98 \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac{(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.0964343, antiderivative size = 98, 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} \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac{(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.288538, size = 96, normalized size = 0.98 \[ \frac{\frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac{(a d-b c) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.073, size = 309, normalized size = 3.2 \begin{align*} 2\,{\frac{{d}^{2}\tan \left ( 1/2\,fx+e/2 \right ) a}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) c}}-2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) db}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}+2\,{\frac{da}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}-2\,{\frac{cb}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}+2\,{\frac{ca}{f \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\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{bd}{f \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)*d^2/(c^2-d^2)/c*tan(1/2*f*x+1/2*e)*a-2/f/(c*tan(1/2*f*x+
1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)*d/(c^2-d^2)*tan(1/2*f*x+1/2*e)*b+2/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+
1/2*e)*d+c)/(c^2-d^2)*d*a-2/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c^2-d^2)*c*b+2/f/(c^2-d^2)^(3
/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*c*a-2/f/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f
*x+1/2*e)+2*d)/(c^2-d^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((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74316, size = 871, normalized size = 8.89 \begin{align*} \left [-\frac{{\left (a c^{2} - b c d +{\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\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 c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) +{\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}}, -\frac{{\left (a c^{2} - b c d +{\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\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 c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) +{\left (c^{5} - 2 \, c^{3} d^{2} + c 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))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.22791, size = 213, normalized size = 2.17 \begin{align*} \frac{2 \,{\left (\frac{{\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 )}{\left (a c - b d\right )}}{{\left (c^{2} - d^{2}\right )}^{\frac{3}{2}}} - \frac{b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b c^{2} - a c d}{{\left (c^{3} - c d^{2}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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