3.5.91 \(\int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx\) [491]

Optimal. Leaf size=239 \[ \frac {4 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}-\frac {4 a^2 \left (c^2-5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c-5 d) \left (c^2-d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/5*a^2*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f+4/15*a^2*(c-5*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f+4/15*a^2
*(c^2-5*c*d-12*d^2)*(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/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4/15*a^2*(c-5*d)*
(c^2-d^2)*(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/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 239, 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 = {2842, 2832, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \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} \end {gather*}

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 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 2832

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)*Sim
p[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 2842

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

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*}

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Mathematica [A]
time = 0.94, size = 244, normalized size = 1.02 \begin {gather*} -\frac {a^2 (1+\sin (e+f x))^2 \left (-4 \left (c^3-4 c^2 d-17 c d^2-12 d^3\right ) 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}}+4 \left (c^3-5 c^2 d-c d^2+5 d^3\right ) 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}}-d \cos (e+f x) \left (-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)\right )\right )}{15 d^2 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*Sqrt[c + d*Sin[e + f*x]],x]

[Out]

-1/15*(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])))/(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] Leaf count of result is larger than twice the leaf count of optimal. \(1034\) vs. \(2(285)=570\).
time = 5.17, size = 1035, normalized size = 4.33

method	result	size
default	\(-\frac {2 a^{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}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right ) c^{3} d -34 c^{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^{2}-2 c \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}+34 \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^{4}-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^{4}+10 \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} d +26 \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^{2} d^{2}-10 \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^{3}-24 \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 ) d^{4}-3 d^{4} \left (\sin ^{4}\left (f x +e \right )\right )-4 c \,d^{3} \left (\sin ^{3}\left (f x +e \right )\right )-10 d^{4} \left (\sin ^{3}\left (f x +e \right )\right )-c^{2} d^{2} \left (\sin ^{2}\left (f x +e \right )\right )-10 c \,d^{3} \left (\sin ^{2}\left (f x +e \right )\right )+3 d^{4} \left (\sin ^{2}\left (f x +e \right )\right )+4 c \,d^{3} \sin \left (f x +e \right )+10 d^{4} \sin \left (f x +e \right )+c^{2} d^{2}+10 d^{3} c \right )}{15 d^{3} \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(1035\)

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,method=_RETURNVERBOSE)

[Out]

-2/15*a^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)*El
lipticF(((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)*(-(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^2-2*c*((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+34*((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^4-2*((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^4+10*((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*d+26*((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^2*d^2-10*((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^3-24*((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))*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*si
n(f*x+e)+c^2*d^2+10*d^3*c)/d^3/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))^(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 554, normalized size = 2.32 \begin {gather*} \frac {2 \, {\left (\sqrt {2} {\left (2 \, a^{2} c^{3} - 10 \, a^{2} c^{2} d + 9 \, a^{2} c d^{2} + 15 \, a^{2} d^{3}\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 ) + \sqrt {2} {\left (2 \, a^{2} c^{3} - 10 \, a^{2} c^{2} d + 9 \, a^{2} c d^{2} + 15 \, a^{2} d^{3}\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 \, \sqrt {2} {\left (-i \, a^{2} c^{2} d + 5 i \, a^{2} c d^{2} + 12 i \, a^{2} d^{3}\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 \, \sqrt {2} {\left (i \, a^{2} c^{2} d - 5 i \, a^{2} c d^{2} - 12 i \, a^{2} d^{3}\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 (3 \, a^{2} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a^{2} c d^{2} + 10 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}\right )}}{45 \, d^{3} f} \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))^(1/2),x, algorithm="fricas")

[Out]

2/45*(sqrt(2)*(2*a^2*c^3 - 10*a^2*c^2*d + 9*a^2*c*d^2 + 15*a^2*d^3)*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^3 - 10*a^2*c^2*d + 9*a^2*c*d^2 + 15*a^2*d^3)*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^2*d + 5*I*a^2*c*d^2 + 12*I*a^2*d^3)*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*sqrt(2)*(I*a^2*c^2*d - 5*I*a^2*c*d^2 - 12*I*a^2*d^3)*sqrt(-I*d)*we
ierstrassZeta(-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*(3*a^2
*d^3*cos(f*x + e)*sin(f*x + e) + (a^2*c*d^2 + 10*a^2*d^3)*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(d^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {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))**(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]
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))^(1/2),x, algorithm="giac")

[Out]

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

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

[Out]

int((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^(1/2), x)

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