3.441 \(\int \frac{(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=138 \[ \frac{3 a^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d)^2 \sqrt{c^2-d^2}}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d f (c+d)^2 (c+d \sin (e+f x))}+\frac{a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

[Out]

(3*a^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^2*Sqrt[c^2 - d^2]*f) + (a^2*(c - d)*Cos[e +
f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (a^2*(c + 4*d)*Cos[e + f*x])/(2*d*(c + d)^2*f*(c + d*Sin[e + f*
x]))

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Rubi [A]  time = 0.181949, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2762, 2754, 12, 2660, 618, 204} \[ \frac{3 a^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d)^2 \sqrt{c^2-d^2}}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d f (c+d)^2 (c+d \sin (e+f x))}+\frac{a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^3,x]

[Out]

(3*a^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^2*Sqrt[c^2 - d^2]*f) + (a^2*(c - d)*Cos[e +
f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (a^2*(c + 4*d)*Cos[e + f*x])/(2*d*(c + d)^2*f*(c + d*Sin[e + f*
x]))

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

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))^2}{(c+d \sin (e+f x))^3} \, dx &=\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \int \frac{-4 a d-a (c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac{a \int \frac{3 a (c-d) d}{c+d \sin (e+f x)} \, dx}{2 (c-d) d (c+d)^2}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac{\left (3 a^2\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 (c+d)^2}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac{\left (3 a^2\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 )}{(c+d)^2 f}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}-\frac{\left (6 a^2\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 )}{(c+d)^2 f}\\ &=\frac{3 a^2 \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{(c+d)^2 \sqrt{c^2-d^2} f}+\frac{a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.663653, size = 140, normalized size = 1.01 \[ \frac{a^2 \cos (e+f x) \left (-\frac{(c+4 d) \sin (e+f x)+4 c+d}{(c+d) (c+d \sin (e+f x))^2}-\frac{6 \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{(-c-d)^{3/2} \sqrt{d-c} \sqrt{\cos ^2(e+f x)}}\right )}{2 f (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^3,x]

[Out]

(a^2*Cos[e + f*x]*((-6*ArcTan[(Sqrt[-c + d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/((
-c - d)^(3/2)*Sqrt[-c + d]*Sqrt[Cos[e + f*x]^2]) - (4*c + d + (c + 4*d)*Sin[e + f*x])/((c + d)*(c + d*Sin[e +
f*x])^2)))/(2*(c + d)*f)

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Maple [B]  time = 0.132, size = 799, normalized size = 5.8 \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))^2/(c+d*sin(f*x+e))^3,x)

[Out]

1/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c*tan(1/2*f*x+1/2*e)^3-4/f*a^2/(c*
tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*d-2/f*a^2/(c*tan(1/2*f*x
+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)^3*d^2-4/f*a^2/(c*tan(1/2*f*x+1/2*e)
^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c*tan(1/2*f*x+1/2*e)^2-1/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/
2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2*d-8/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e
)*d+c)^2/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)^2*d^2-2/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^
2/(c^2+2*c*d+d^2)/c^2*tan(1/2*f*x+1/2*e)^2*d^3-1/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*c/(
c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)-12/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)
*tan(1/2*f*x+1/2*e)*d-2/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/c/(c^2+2*c*d+d^2)*tan(1/2*f*
x+1/2*e)*d^2-4/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c-1/f*a^2/(c*tan(1/2*
f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*d+3/f*a^2/(c^2+2*c*d+d^2)/(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))

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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))^2/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.84876, size = 1449, normalized size = 10.5 \begin{align*} \left [\frac{2 \,{\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \,{\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{4 \,{\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) -{\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}, \frac{{\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \,{\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) -{\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.45219, size = 470, normalized size = 3.41 \begin{align*} \frac{\frac{3 \,{\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 )} a^{2}}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt{c^{2} - d^{2}}} + \frac{a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4 \, a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 8 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 12 \, a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, a^{2} c^{3} - a^{2} c^{2} d}{{\left (c^{4} + 2 \, c^{3} d + c^{2} 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 )}^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

(3*(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^2/((c^2
+ 2*c*d + d^2)*sqrt(c^2 - d^2)) + (a^2*c^3*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 2*a^2
*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^3*tan(1/2*f*x + 1/2*e)^2 - a^2*c^2*d*tan(1/2*f*x + 1/2*e)^2 - 8*a^2*c*
d^2*tan(1/2*f*x + 1/2*e)^2 - 2*a^2*d^3*tan(1/2*f*x + 1/2*e)^2 - a^2*c^3*tan(1/2*f*x + 1/2*e) - 12*a^2*c^2*d*ta
n(1/2*f*x + 1/2*e) - 2*a^2*c*d^2*tan(1/2*f*x + 1/2*e) - 4*a^2*c^3 - a^2*c^2*d)/((c^4 + 2*c^3*d + c^2*d^2)*(c*t
an(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f