3.494 \(\int \frac{(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=247 \[ \frac{4 a^2 (c+2 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt{c+d \sin (e+f x)}}+\frac{2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]
[Out]
(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c + 3*d)*Cos[e + f*x])/(3*d*
(c + d)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^2*(c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c +
d*Sin[e + f*x]])/(3*d^2*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c + 2*d)*EllipticF[(e - Pi/
2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d^2*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.369142, antiderivative size = 247, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used =
{2762, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a^2 (c+2 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt{c+d \sin (e+f x)}}+\frac{2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(5/2),x]
[Out]
(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c + 3*d)*Cos[e + f*x])/(3*d*
(c + d)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^2*(c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c +
d*Sin[e + f*x]])/(3*d^2*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c + 2*d)*EllipticF[(e - Pi/
2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d^2*(c + d)*f*Sqrt[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 2752
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{(2 a) \int \frac{-3 a d-a (c+2 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt{c+d \sin (e+f x)}}+\frac{(4 a) \int \frac{a (c-d) d-\frac{1}{2} a (c-d) (c+3 d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 (c-d) d (c+d)^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt{c+d \sin (e+f x)}}+\frac{\left (2 a^2 (c+2 d)\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d^2 (c+d)}-\frac{\left (2 a^2 (c+3 d)\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{3 d^2 (c+d)^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 a^2 (c+3 d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{3 d^2 (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a^2 (c+2 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{3 d^2 (c+d) \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c+3 d) E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{3 d^2 (c+d)^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 a^2 (c+2 d) F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{3 d^2 (c+d) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.76248, size = 207, normalized size = 0.84 \[ -\frac{2 a^2 (\sin (e+f x)+1)^2 \left (d \cos (e+f x) \left (c^2+2 d (c+3 d) \sin (e+f x)+6 c d+d^2\right )+2 (c+2 d) (c+d)^2 \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{3/2} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-2 (c+3 d) (c+d)^2 \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{3/2} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{3 d^2 f (c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 (c+d \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(5/2),x]
[Out]
(-2*a^2*(1 + Sin[e + f*x])^2*(-2*(c + d)^2*(c + 3*d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*((c + d*S
in[e + f*x])/(c + d))^(3/2) + 2*(c + d)^2*(c + 2*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*((c + d*Si
n[e + f*x])/(c + d))^(3/2) + d*Cos[e + f*x]*(c^2 + 6*c*d + d^2 + 2*d*(c + 3*d)*Sin[e + f*x])))/(3*d^2*(c + d)^
2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c + d*Sin[e + f*x])^(3/2))
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Maple [B] time = 1.007, size = 1221, normalized size = 4.9 \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))^(5/2),x)
[Out]
-2/3*a^2*((2*c*d^3+6*d^4)*sin(f*x+e)*cos(f*x+e)^2-2*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d
/(c-d))^(1/2)*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*d*(EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c
+d))^(1/2))*c^3+3*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-EllipticE((d/(c-d)
*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d^2-3*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d
)/(c+d))^(1/2))*d^3-EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d+EllipticF((d/(c-
d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3)*sin(f*x+e)+(c^2*d^2+6*c*d^3+d^4)*cos(f*x+e)^2+2*(d/(c
-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*Ellipt
icF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3*d-2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d
/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^
(1/2),((c-d)/(c+d))^(1/2))*c*d^3-2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-
d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^4-6*(d
/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*Ell
ipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3*d+2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*
(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*
c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2+6*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/
2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d
^3)/(c+d)^2/(c+d*sin(f*x+e))^(3/2)/d^3/cos(f*x+e)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \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))^(5/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} +{\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\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))^(5/2),x, algorithm="fricas")
[Out]
integral((a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*sqrt(d*sin(f*x + e) + c)/(3*c*d^2*cos(f*x + e)^2 -
c^3 - 3*c*d^2 + (d^3*cos(f*x + e)^2 - 3*c^2*d - d^3)*sin(f*x + e)), x)
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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))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \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))^(5/2),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(5/2), x)