3.5.94 \(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx\) [494]

Optimal. Leaf size=247 \[ \frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+3 d) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+2 d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 d^2 (c+d) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

2/3*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(3/2)-4/3*a^2*(c+3*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+
e))^(1/2)+4/3*a^2*(c+3*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)*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)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4/3
*a^2*(c+2*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)*EllipticF(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/(c+d)/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 247, normalized size of antiderivative = 1.00, 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 = {2841, 2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \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}} \end {gather*}

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 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 2833

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 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))^{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*}

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Mathematica [A]
time = 1.21, size = 207, normalized size = 0.84 \begin {gather*} -\frac {2 a^2 (1+\sin (e+f x))^2 \left (-2 (c+d)^2 (c+3 d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}+2 (c+d)^2 (c+2 d) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}+d \cos (e+f x) \left (c^2+6 c d+d^2+2 d (c+3 d) \sin (e+f x)\right )\right )}{3 d^2 (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c+d \sin (e+f x))^{3/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(1220\) vs. \(2(293)=586\).
time = 4.80, size = 1221, normalized size = 4.94

method	result	size
default	\(\text {Expression too large to display}\)	\(1221\)

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,method=_RETURNVERBOSE)

[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 = 0.00, 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))^(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.19, size = 1013, normalized size = 4.10 \begin {gather*} \frac {2 \, {\left (2 \, {\left (\sqrt {2} {\left (a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3} + 3 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (a^{2} c^{3} d + 3 \, a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (a^{2} c^{4} + 3 \, a^{2} c^{3} d + 4 \, a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3} + 3 \, a^{2} d^{4}\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 ) + 2 \, {\left (\sqrt {2} {\left (a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3} + 3 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (a^{2} c^{3} d + 3 \, a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (a^{2} c^{4} + 3 \, a^{2} c^{3} d + 4 \, a^{2} c^{2} d^{2} + 3 \, a^{2} c d^{3} + 3 \, a^{2} d^{4}\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 (\sqrt {2} {\left (i \, a^{2} c d^{3} + 3 i \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (-i \, a^{2} c^{2} d^{2} - 3 i \, a^{2} c d^{3}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (-i \, a^{2} c^{3} d - 3 i \, a^{2} c^{2} d^{2} - i \, a^{2} c d^{3} - 3 i \, a^{2} d^{4}\right )}\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 (\sqrt {2} {\left (-i \, a^{2} c d^{3} - 3 i \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (i \, a^{2} c^{2} d^{2} + 3 i \, a^{2} c d^{3}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (i \, a^{2} c^{3} d + 3 i \, a^{2} c^{2} d^{2} + i \, a^{2} c d^{3} + 3 i \, a^{2} d^{4}\right )}\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 (2 \, {\left (a^{2} c d^{3} + 3 \, a^{2} d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a^{2} c^{2} d^{2} + 6 \, a^{2} c d^{3} + a^{2} d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}\right )}}{9 \, {\left ({\left (c^{2} d^{5} + 2 \, c d^{6} + d^{7}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d^{4} + 2 \, c^{2} d^{5} + c d^{6}\right )} f \sin \left (f x + e\right ) - {\left (c^{4} d^{3} + 2 \, c^{3} d^{4} + 2 \, c^{2} d^{5} + 2 \, c d^{6} + d^{7}\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))^(5/2),x, algorithm="fricas")

[Out]

2/9*(2*(sqrt(2)*(a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)^2 - 2*sqrt(2)*(a^2*c^3*d + 3*a^2*c^2*d^2
+ 3*a^2*c*d^3)*sin(f*x + e) - sqrt(2)*(a^2*c^4 + 3*a^2*c^3*d + 4*a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4))*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) + 2*(sqrt(2)*(a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)^2 - 2*sqrt(2)*(
a^2*c^3*d + 3*a^2*c^2*d^2 + 3*a^2*c*d^3)*sin(f*x + e) - sqrt(2)*(a^2*c^4 + 3*a^2*c^3*d + 4*a^2*c^2*d^2 + 3*a^2
*c*d^3 + 3*a^2*d^4))*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*(sqrt(2)*(I*a^2*c*d^3 + 3*I*a^2*d^4)*cos(f*x + e)
^2 + 2*sqrt(2)*(-I*a^2*c^2*d^2 - 3*I*a^2*c*d^3)*sin(f*x + e) + sqrt(2)*(-I*a^2*c^3*d - 3*I*a^2*c^2*d^2 - I*a^2
*c*d^3 - 3*I*a^2*d^4))*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, we
ierstrassPInverse(-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*(sqrt(2)*(-I*a^2*c*d^3 - 3*I*a^2*d^4)*cos(f*x + e)^2 + 2*sqrt(2)*(I*a^2*c^2*d^2 + 3
*I*a^2*c*d^3)*sin(f*x + e) + sqrt(2)*(I*a^2*c^3*d + 3*I*a^2*c^2*d^2 + I*a^2*c*d^3 + 3*I*a^2*d^4))*sqrt(-I*d)*w
eierstrassZeta(-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*(2*(a
^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)*sin(f*x + e) + (a^2*c^2*d^2 + 6*a^2*c*d^3 + a^2*d^4)*cos(f*x + e))*sqrt(d*s
in(f*x + e) + c))/((c^2*d^5 + 2*c*d^6 + d^7)*f*cos(f*x + e)^2 - 2*(c^3*d^4 + 2*c^2*d^5 + c*d^6)*f*sin(f*x + e)
- (c^4*d^3 + 2*c^3*d^4 + 2*c^2*d^5 + 2*c*d^6 + d^7)*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \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 )}^{5/2}} \,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))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(5/2), x)

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