3.8.32 \(\int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx\) [732]

Optimal. Leaf size=254 \[ \frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {2 \left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 b (b c-5 a d) \left (c^2-d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/5*b^2*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f+4/15*b*(-5*a*d+b*c)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f-2/15*
(3*(5*a^2+3*b^2)*d^2-2*b*c*(-5*a*d+b*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)*Ellipti
cE(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*b*(-5*a*d+b*c)*(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 = 0.27, antiderivative size = 254, 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 = {2870, 2832, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\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 {4 b \left (c^2-d^2\right ) (b c-5 a 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 )}{15 d^2 f \sqrt {c+d \sin (e+f x)}}+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(b*c - 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d*f) - (2*b^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^
(3/2))/(5*d*f) + (2*(3*(5*a^2 + 3*b^2)*d^2 - 2*b*c*(b*c - 5*a*d))*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*b*(b*c - 5*a*d)*(c^2 - d^2)*Elli
pticF[(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 2870

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-d^2)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f
*x])^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c
, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx &=-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {2 \int \sqrt {c+d \sin (e+f x)} \left (\frac {1}{2} \left (5 a^2+3 b^2\right ) d-b (b c-5 a d) \sin (e+f x)\right ) \, dx}{5 d}\\ &=\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {4 \int \frac {\frac {1}{4} d \left (15 a^2 c+7 b^2 c+10 a b d\right )+\frac {1}{4} \left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d}\\ &=\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {\left (2 b (b c-5 a d) \left (c^2-d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^2}+\frac {1}{15} \left (15 a^2+9 b^2-\frac {2 b c (b c-5 a d)}{d^2}\right ) \int \sqrt {c+d \sin (e+f x)} \, dx\\ &=\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {\left (\left (15 a^2+9 b^2-\frac {2 b c (b c-5 a d)}{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 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 b (b c-5 a 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 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {2 \left (15 a^2+9 b^2-\frac {2 b c (b c-5 a d)}{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 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 b (b c-5 a 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 = 0.97, size = 214, normalized size = 0.84 \begin {gather*} \frac {2 \left (-d^2 \left (15 a^2 c+7 b^2 c+10 a b d\right ) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )+\left (-10 a b c d-15 a^2 d^2+b^2 \left (2 c^2-9 d^2\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 b d \cos (e+f x) (c+d \sin (e+f x)) (b c+10 a d+3 b d \sin (e+f x))}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(d^2*(15*a^2*c + 7*b^2*c + 10*a*b*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]) + (-10*a*b*c*d - 15
*a^2*d^2 + b^2*(2*c^2 - 9*d^2))*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e +
Pi - 2*f*x)/4, (2*d)/(c + d)]))*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 2*b*d*Cos[e + f*x]*(c + d*Sin[e + f*x])*
(b*c + 10*a*d + 3*b*d*Sin[e + f*x]))/(15*d^2*f*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. \(1099\) vs. \(2(300)=600\).
time = 19.71, size = 1100, normalized size = 4.33

method	result	size
default	\(\text {Expression too large to display}\)	\(1100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b^2*d*(-2/5/d*sin(f*x+e)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+8/15
/d^2*c*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+4/15/d*c*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x
+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(
f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2*(3/5+8/15/d^2*c^2)*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-
sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-1/d*c-1)*E
llipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(
c+d))^(1/2))))+(2*a*b*d+b^2*c)*(-2/3/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2/3*(1/d*c-1)*((c+d*sin(f*x+e))
/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)
^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-4/3/d*c*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d)
)^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)
*((-1/d*c-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^
(1/2),((c-d)/(c+d))^(1/2))))+2*(a^2*d+2*a*b*c)*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d
))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-1/d*c-1)*EllipticE(((c+d*s
in(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))+2*
a^2*c*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/
(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))/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+b*sin(f*x+e))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*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 = 0.18, size = 575, normalized size = 2.26 \begin {gather*} \frac {\sqrt {2} {\left (4 \, b^{2} c^{3} - 20 \, a b c^{2} d + 30 \, a b d^{3} + 3 \, {\left (5 \, a^{2} + b^{2}\right )} c d^{2}\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 (4 \, b^{2} c^{3} - 20 \, a b c^{2} d + 30 \, a b d^{3} + 3 \, {\left (5 \, a^{2} + b^{2}\right )} c d^{2}\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 (-2 i \, b^{2} c^{2} d + 10 i \, a b c d^{2} + 3 i \, {\left (5 \, a^{2} + 3 \, b^{2}\right )} 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 (2 i \, b^{2} c^{2} d - 10 i \, a b c d^{2} - 3 i \, {\left (5 \, a^{2} + 3 \, b^{2}\right )} 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 ) - 6 \, {\left (3 \, b^{2} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (b^{2} c d^{2} + 10 \, a b d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{45 \, d^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/45*(sqrt(2)*(4*b^2*c^3 - 20*a*b*c^2*d + 30*a*b*d^3 + 3*(5*a^2 + b^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)*(4*b^2*c^3 - 20*a*b*c^2*d + 30*a*b*d^3 + 3*(5*a^2 + b^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*sqrt(2)*(-2*I*b^2*c^2*d + 10*I*a*b*c*d^2 + 3*I*(5*a^2 + 3*b^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)*(2*I*b^2*c^2*d - 10*I*a*
b*c*d^2 - 3*I*(5*a^2 + 3*b^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)) - 6*(3*b^2*d^3*cos(f*x + e)*sin(f*x + e) + (b^2*c*d^2 + 10*a*b*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*} \int \left (a + b \sin {\left (e + f x \right )}\right )^{2} \sqrt {c + d \sin {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**2*(c+d*sin(f*x+e))**(1/2),x)

[Out]

Integral((a + b*sin(e + f*x))**2*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+b*sin(f*x+e))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*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+b\,\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 + b*sin(e + f*x))^2*(c + d*sin(e + f*x))^(1/2),x)

[Out]

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

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