3.491 \(\int (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)} \, dx\)
Optimal. Leaf size=239 \[ \frac{4 a^2 (c-5 d) \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f} \]
[Out]
(4*a^2*(c - 5*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d*f) - (2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3
/2))/(5*d*f) - (4*a^2*(c^2 - 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f
*x]])/(15*d^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c - 5*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)/
2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*f*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.336093, antiderivative size = 239, 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 =
{2763, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a^2 (c-5 d) \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
(4*a^2*(c - 5*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d*f) - (2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3
/2))/(5*d*f) - (4*a^2*(c^2 - 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f
*x]])/(15*d^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c - 5*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)/
2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*f*Sqrt[c + d*Sin[e + f*x]])
Rule 2763
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 2753
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*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] && GtQ[m, 0] && 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 (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \int \left (4 a^2 d-a^2 (c-5 d) \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)} \, dx}{5 d}\\ &=\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{4 \int \frac{\frac{1}{2} a^2 d (11 c+5 d)-\frac{1}{2} a^2 \left (c^2-5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d}\\ &=\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}-\frac{\left (2 a^2 \left (c^2-5 c d-12 d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{15 d^2}+\frac{\left (2 a^2 (c-5 d) \left (c^2-d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d^2}\\ &=\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}-\frac{\left (2 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{15 d^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a^2 (c-5 d) \left (c^2-d^2\right ) \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}{15 d^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{4 a^2 (c-5 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}-\frac{4 a^2 \left (c^2-5 c d-12 d^2\right ) 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)}}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 a^2 (c-5 d) \left (c^2-d^2\right ) 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}}}{15 d^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.42886, size = 244, normalized size = 1.02 \[ -\frac{a^2 (\sin (e+f x)+1)^2 \left (-d \cos (e+f x) \left (-2 c^2-4 d (2 c+5 d) \sin (e+f x)-20 c d+3 d^2 \cos (2 (e+f x))-3 d^2\right )+4 \left (-5 c^2 d+c^3-c d^2+5 d^3\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-4 \left (-4 c^2 d+c^3-17 c d^2-12 d^3\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{15 d^2 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
-(a^2*(1 + Sin[e + f*x])^2*(-4*(c^3 - 4*c^2*d - 17*c*d^2 - 12*d^3)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c +
d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + 4*(c^3 - 5*c^2*d - c*d^2 + 5*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/4, (
2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - d*Cos[e + f*x]*(-2*c^2 - 20*c*d - 3*d^2 + 3*d^2*Cos[2*(e +
f*x)] - 4*d*(2*c + 5*d)*Sin[e + f*x])))/(15*d^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c + d*Sin[e + f
*x]])
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Maple [B] time = 1.029, size = 1035, normalized size = 4.3 \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))^(1/2),x)
[Out]
-2/15*a^2*(2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*E
llipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^3*d-34*c^2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1
+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c
+d))^(1/2))*d^2-2*c*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^
(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^3+34*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-
1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(
c+d))^(1/2))*d^4-2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(
1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4+10*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1
+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c
+d))^(1/2))*c^3*d+26*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))
^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2-10*((c+d*sin(f*x+e))/(c-d))^(1/2)
*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c
-d)/(c+d))^(1/2))*c*d^3-24*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/
(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^4-3*d^4*sin(f*x+e)^4-4*c*d^3*sin(
f*x+e)^3-10*d^4*sin(f*x+e)^3-c^2*d^2*sin(f*x+e)^2-10*c*d^3*sin(f*x+e)^2+3*d^4*sin(f*x+e)^2+4*c*d^3*sin(f*x+e)+
10*d^4*sin(f*x+e)+c^2*d^2+10*c*d^3)/d^3/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \sqrt{d \sin \left (f x + e\right ) + c}\,{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))^(1/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e) + c), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\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}, 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))^(1/2),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), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}\, dx + \int \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int \sqrt{c + d \sin{\left (e + f x \right )}}\, dx\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))**(1/2),x)
[Out]
a**2*(Integral(2*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2
, x) + Integral(sqrt(c + d*sin(e + f*x)), x))
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Giac [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))^(1/2),x, algorithm="giac")
[Out]
Timed out