3.677 \(\int \frac{a+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac{\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2}}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac{(b c-a d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]

[Out]

-(((3*b*c*d - a*(2*c^2 + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c^2 - d^2)^(5/2)*f)) - ((b*
c - a*d)*Cos[e + f*x])/(2*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) + ((3*a*c*d - b*(c^2 + 2*d^2))*Cos[e + f*x])/(
2*(c^2 - d^2)^2*f*(c + d*Sin[e + f*x]))

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Rubi [A]  time = 0.197544, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2}}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac{(b c-a d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])/(c + d*Sin[e + f*x])^3,x]

[Out]

-(((3*b*c*d - a*(2*c^2 + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c^2 - d^2)^(5/2)*f)) - ((b*
c - a*d)*Cos[e + f*x])/(2*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) + ((3*a*c*d - b*(c^2 + 2*d^2))*Cos[e + f*x])/(
2*(c^2 - d^2)^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))^3} \, dx &=-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac{\int \frac{-2 (a c-b d)-(b c-a d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )}\\ &=-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\int \frac{-3 b c d+a \left (2 c^2+d^2\right )}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac{\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac{\left (3 b c d-a \left (2 c^2+d^2\right )\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 )}{\left (c^2-d^2\right )^2 f}\\ &=-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac{\left (2 \left (3 b c d-a \left (2 c^2+d^2\right )\right )\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 )}{\left (c^2-d^2\right )^2 f}\\ &=-\frac{\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \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 )^{5/2} f}-\frac{(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac{\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.60345, size = 157, normalized size = 0.96 \[ \frac{\frac{2 \left (a \left (2 c^2+d^2\right )-3 b c d\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}-\frac{\left (b \left (c^2+2 d^2\right )-3 a c d\right ) \cos (e+f x)}{(c-d)^2 (c+d)^2 (c+d \sin (e+f x))}+\frac{(a d-b c) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])/(c + d*Sin[e + f*x])^3,x]

[Out]

((2*(-3*b*c*d + a*(2*c^2 + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(5/2) + ((-(b*c
) + a*d)*Cos[e + f*x])/((c - d)*(c + d)*(c + d*Sin[e + f*x])^2) - ((-3*a*c*d + b*(c^2 + 2*d^2))*Cos[e + f*x])/
((c - d)^2*(c + d)^2*(c + d*Sin[e + f*x])))/(2*f)

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Maple [B]  time = 0.085, size = 1291, normalized size = 7.9 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

5/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d^2*c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^3*a-2/f/(
c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d^4/c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^3*a-3/f/(c*tan
(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d*c^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^3*b+4/f/(c*tan(1/2*
f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*c^2*tan(1/2*f*x+1/2*e)^2*a*d+7/f/(c*tan(1/2*f*x+1
/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^2*a*d^3-2/f/(c*tan(1/2*f*x+1/2*e)^2
+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)/c^2*tan(1/2*f*x+1/2*e)^2*a*d^5-2/f/(c*tan(1/2*f*x+1/2*e)^2+2*
tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*c^3*tan(1/2*f*x+1/2*e)^2*b-5/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2
*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*c*tan(1/2*f*x+1/2*e)^2*b*d^2-2/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+
1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)/c*tan(1/2*f*x+1/2*e)^2*b*d^4+11/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*
e)*d+c)^2*d^2*c/(c^4-2*c^2*d^2+d^4)*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)
^2*d^4/c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*a-5/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d*c^
2/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*b-4/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d^3/(c^4-2*
c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)*b+4/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*
a*c^2*d-1/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*a*d^3-2/f/(c*tan(1/2*f*x+1
/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^4-2*c^2*d^2+d^4)*c^3*b-1/f/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)
*d+c)^2/(c^4-2*c^2*d^2+d^4)*c*d^2*b+2/f/(c^4-2*c^2*d^2+d^4)/(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*a+1/f/(c^4-2*c^2*d^2+d^4)/(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*d^2-3/f/(c^4-2*c^2*d^2+d^4)/(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))*b*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((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.00247, size = 1724, normalized size = 10.51 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(b*c^4*d - 3*a*c^3*d^2 + b*c^2*d^3 + 3*a*c*d^4 - 2*b*d^5)*cos(f*x + e)*sin(f*x + e) + (2*a*c^4 - 3*b*c
^3*d + 3*a*c^2*d^2 - 3*b*c*d^3 + a*d^4 - (2*a*c^2*d^2 - 3*b*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(2*a*c^3*d - 3*b
*c^2*d^2 + a*c*d^3)*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*(2*b*c^5 - 4*a*c^4*d - b*c^3*d^2 + 5*a*c^2*d^3 - b*c*d^4 - a*d^5)*cos(f*x + e))/((c^
6*d^2 - 3*c^4*d^4 + 3*c^2*d^6 - d^8)*f*cos(f*x + e)^2 - 2*(c^7*d - 3*c^5*d^3 + 3*c^3*d^5 - c*d^7)*f*sin(f*x +
e) - (c^8 - 2*c^6*d^2 + 2*c^2*d^6 - d^8)*f), 1/2*((b*c^4*d - 3*a*c^3*d^2 + b*c^2*d^3 + 3*a*c*d^4 - 2*b*d^5)*co
s(f*x + e)*sin(f*x + e) + (2*a*c^4 - 3*b*c^3*d + 3*a*c^2*d^2 - 3*b*c*d^3 + a*d^4 - (2*a*c^2*d^2 - 3*b*c*d^3 +
a*d^4)*cos(f*x + e)^2 + 2*(2*a*c^3*d - 3*b*c^2*d^2 + a*c*d^3)*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x
+ e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (2*b*c^5 - 4*a*c^4*d - b*c^3*d^2 + 5*a*c^2*d^3 - b*c*d^4 - a*d^5)
*cos(f*x + e))/((c^6*d^2 - 3*c^4*d^4 + 3*c^2*d^6 - d^8)*f*cos(f*x + e)^2 - 2*(c^7*d - 3*c^5*d^3 + 3*c^3*d^5 -
c*d^7)*f*sin(f*x + e) - (c^8 - 2*c^6*d^2 + 2*c^2*d^6 - d^8)*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.44484, size = 578, normalized size = 3.52 \begin{align*} \frac{\frac{{\left (2 \, a c^{2} - 3 \, b c d + a 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 )}}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} \sqrt{c^{2} - d^{2}}} - \frac{3 \, b c^{4} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 5 \, a c^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, a c d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, b c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, a c^{4} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, b c^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7 \, a c^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b c d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, a d^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, b c^{4} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 11 \, a c^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, b c^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a c d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, b c^{5} - 4 \, a c^{4} d + b c^{3} d^{2} + a c^{2} d^{3}}{{\left (c^{6} - 2 \, c^{4} d^{2} + c^{2} d^{4}\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 )}^{2}}}{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))^3,x, algorithm="giac")

[Out]

((2*a*c^2 - 3*b*c*d + a*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)/sq
rt(c^2 - d^2)))/((c^4 - 2*c^2*d^2 + d^4)*sqrt(c^2 - d^2)) - (3*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 5*a*c^3*d^2*ta
n(1/2*f*x + 1/2*e)^3 + 2*a*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*b*c^5*tan(1/2*f*x + 1/2*e)^2 - 4*a*c^4*d*tan(1/2*f
*x + 1/2*e)^2 + 5*b*c^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 7*a*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*b*c*d^4*tan(1/2*f*
x + 1/2*e)^2 + 2*a*d^5*tan(1/2*f*x + 1/2*e)^2 + 5*b*c^4*d*tan(1/2*f*x + 1/2*e) - 11*a*c^3*d^2*tan(1/2*f*x + 1/
2*e) + 4*b*c^2*d^3*tan(1/2*f*x + 1/2*e) + 2*a*c*d^4*tan(1/2*f*x + 1/2*e) + 2*b*c^5 - 4*a*c^4*d + b*c^3*d^2 + a
*c^2*d^3)/((c^6 - 2*c^4*d^2 + c^2*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f