3.5.93 \(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{3/2}} \, dx\) [493]

Optimal. Leaf size=189 \[ \frac {2 a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^2 c E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {4 a^2 (c-d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{d^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(1/2)-4*a^2*c*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+
1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d^2/(c+d)/
f/((c+d*sin(f*x+e))/(c+d))^(1/2)+4*a^2*(c-d)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*Ell
ipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d^2/f/(c+d*sin(f*x+e)
)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2841, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {4 a^2 (c-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 )}{d^2 f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^2 c \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 )}{d^2 f (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]) + (4*a^2*c*EllipticE[(e - Pi/2 + f*x)/2, (
2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(d^2*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (4*a^2*(c - d)*El
lipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(d^2*f*Sqrt[c + d*Sin[e + f*x]]
)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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]))

Rubi steps

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

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Mathematica [A]
time = 0.61, size = 175, normalized size = 0.93 \begin {gather*} -\frac {2 a^2 (1+\sin (e+f x))^2 \left (2 c (c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-(c-d) \left (d \cos (e+f x)+2 (c+d) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )\right )}{d^2 (c+d) f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(1 + Sin[e + f*x])^2*(2*c*(c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e +
f*x])/(c + d)] - (c - d)*(d*Cos[e + f*x] + 2*(c + d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c +
d*Sin[e + f*x])/(c + d)])))/(d^2*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c + d*Sin[e + f*x]])

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Maple [A]
time = 4.61, size = 463, normalized size = 2.45

method	result	size
default	\(-\frac {2 \left (2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c^{3}-2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c \,d^{2}-2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c^{2} d +2 \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {-\frac {\left (\sin \left (f x +e \right )-1\right ) d}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) d^{3}+c \,d^{2} \left (\sin ^{2}\left (f x +e \right )\right )-d^{3} \left (\sin ^{2}\left (f x +e \right )\right )-c \,d^{2}+d^{3}\right ) a^{2}}{d^{3} \left (c +d \right ) \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(463\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*(2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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^3-2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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^
2-2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^2*d+2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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+
c*d^2*sin(f*x+e)^2-d^3*sin(f*x+e)^2-c*d^2+d^3)/d^3*a^2/(c+d)/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 594, normalized size = 3.14 \begin {gather*} \frac {2 \, {\left (3 \, {\left (a^{2} c d^{2} - a^{2} d^{3}\right )} \sqrt {d \sin \left (f x + e\right ) + c} \cos \left (f x + e\right ) - {\left (\sqrt {2} {\left (2 \, a^{2} c^{2} d - 3 \, a^{2} d^{3}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (2 \, a^{2} c^{3} - 3 \, a^{2} c d^{2}\right )}\right )} \sqrt {i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right ) - {\left (\sqrt {2} {\left (2 \, a^{2} c^{2} d - 3 \, a^{2} d^{3}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (2 \, a^{2} c^{3} - 3 \, a^{2} c d^{2}\right )}\right )} \sqrt {-i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right ) - 3 \, {\left (i \, \sqrt {2} a^{2} c d^{2} \sin \left (f x + e\right ) + i \, \sqrt {2} a^{2} c^{2} d\right )} \sqrt {i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right )\right ) - 3 \, {\left (-i \, \sqrt {2} a^{2} c d^{2} \sin \left (f x + e\right ) - i \, \sqrt {2} a^{2} c^{2} d\right )} \sqrt {-i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right )\right )\right )}}{3 \, {\left ({\left (c d^{4} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{2} d^{3} + c d^{4}\right )} f\right )}} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

2/3*(3*(a^2*c*d^2 - a^2*d^3)*sqrt(d*sin(f*x + e) + c)*cos(f*x + e) - (sqrt(2)*(2*a^2*c^2*d - 3*a^2*d^3)*sin(f*
x + e) + sqrt(2)*(2*a^2*c^3 - 3*a^2*c*d^2))*sqrt(I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I
*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) - (sqrt(2)*(2*a^2*c^2*d - 3*a^2*
d^3)*sin(f*x + e) + sqrt(2)*(2*a^2*c^3 - 3*a^2*c*d^2))*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2
, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) - 3*(I*sqrt(2)*a^2*
c*d^2*sin(f*x + e) + I*sqrt(2)*a^2*c^2*d)*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 -
9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f
*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)) - 3*(-I*sqrt(2)*a^2*c*d^2*sin(f*x + e) - I*sqrt(2)*a^2*c^2*d)*sqrt(-
I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c
^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)))/((
c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))**(3/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(3/2),x)

[Out]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(3/2), x)

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