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

Optimal. Leaf size=134 \[ \frac{a (2 c-d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac{a (c-2 d) \cos (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sin (e+f x))}-\frac{a \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.184297, antiderivative size = 134, 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} \[ \frac{a (2 c-d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac{a (c-2 d) \cos (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sin (e+f x))}-\frac{a \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 1.14404, size = 242, normalized size = 1.81 \[ \frac{a (\sin (e+f x)+1) \left (\frac{4 (2 c-d) (\cos (e)-i \sin (e)) \tan ^{-1}\left (\frac{(\cos (e)-i \sin (e)) \sec \left (\frac{f x}{2}\right ) \left (c \sin \left (\frac{f x}{2}\right )+d \cos \left (e+\frac{f x}{2}\right )\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{(c-d) \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{2 (c+d) \csc (e) (c \cos (e)+d \sin (f x))}{d (c+d \sin (e+f x))^2}+\frac{2 (c-2 d) \csc (e) \sin (f x)+(2 d-4 c) \cot (e)}{(c-d) (c+d \sin (e+f x))}\right )}{4 f (c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1 + Sin[e + f*x])*((4*(2*c - d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)
/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e]))/((c - d)*Sqrt[c^2 - d^2]*Sqrt[(Cos[e
] - I*Sin[e])^2]) + (2*(c + d)*Csc[e]*(c*Cos[e] + d*Sin[f*x]))/(d*(c + d*Sin[e + f*x])^2) + ((-4*c + 2*d)*Cot[
e] + 2*(c - 2*d)*Csc[e]*Sin[f*x])/((c - d)*(c + d*Sin[e + f*x]))))/(4*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2])^2)

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Maple [B]  time = 0.122, size = 1104, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x)

[Out]

-3/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*d/(c^3+c^2*d-c*d^2-d^3)*c*tan(1/2*f*x+1/2*e)^3+2/f*
a/(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^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3+2/f*a/(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^3+c^2*d-c*d^2-d^3)/c*tan(1/2*f*x+1/2*e)^3-2/f*a/(c*tan
(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*c^2*tan(1/2*f*x+1/2*e)^2+2/f*a/(c*tan(1/2*
f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*c*tan(1/2*f*x+1/2*e)^2*d-3/f*a/(c*tan(1/2*f*x+1
/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^2*d^2+4/f*a/(c*tan(1/2*f*x+1/2*e)
^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)/c*tan(1/2*f*x+1/2*e)^2*d^3+2/f*a/(c*tan(1/2*f*x+1/2*e)^2+
2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)/c^2*tan(1/2*f*x+1/2*e)^2*d^4-5/f*a/(c*tan(1/2*f*x+1/2*e)^2+2
*tan(1/2*f*x+1/2*e)*d+c)^2*d*c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)+6/f*a/(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^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)+2/f*a/(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/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)-2/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)
*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*c^2+2/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d
^3)*c*d+1/f*a/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^3+c^2*d-c*d^2-d^3)*d^2+2/f*a/(c^3+c^2*d-c
*d^2-d^3)/(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-1/f*a/(c^3+c^2*d-c*d^2-d^
3)/(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))*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+a*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 = 1.71909, size = 1729, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.4063, size = 518, normalized size = 3.87 \begin{align*} \frac{\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 (2 \, a c - a d\right )}}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} \sqrt{c^{2} - d^{2}}} - \frac{3 \, a c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, a c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, a c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, a c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, a c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a c^{4} - 2 \, a c^{3} d - a c^{2} d^{2}}{{\left (c^{5} + c^{4} d - c^{3} d^{2} - c^{2} d^{3}\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+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

((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)))*(2*a*c - a*d
)/((c^3 + c^2*d - c*d^2 - d^3)*sqrt(c^2 - d^2)) - (3*a*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 2*a*c^2*d^2*tan(1/2*f*x
+ 1/2*e)^3 - 2*a*c*d^3*tan(1/2*f*x + 1/2*e)^3 + 2*a*c^4*tan(1/2*f*x + 1/2*e)^2 - 2*a*c^3*d*tan(1/2*f*x + 1/2*e
)^2 + 3*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 - 4*a*c*d^3*tan(1/2*f*x + 1/2*e)^2 - 2*a*d^4*tan(1/2*f*x + 1/2*e)^2 +
5*a*c^3*d*tan(1/2*f*x + 1/2*e) - 6*a*c^2*d^2*tan(1/2*f*x + 1/2*e) - 2*a*c*d^3*tan(1/2*f*x + 1/2*e) + 2*a*c^4
- 2*a*c^3*d - a*c^2*d^2)/((c^5 + c^4*d - c^3*d^2 - c^2*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*
e) + c)^2))/f