3.490 \(\int (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=298 \[ \frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{35 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c+3 d) \left (c^2-10 c d-7 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{35 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f} \]

[Out]

(4*a^2*(c^2 - 7*c*d - 10*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(35*d*f) + (4*a^2*(c - 7*d)*Cos[e + f*x]*
(c + d*Sin[e + f*x])^(3/2))/(35*d*f) - (2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(7*d*f) - (4*a^2*(c + 3
*d)*(c^2 - 10*c*d - 7*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(35*d^2*f*Sq
rt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c^2 - 7*c*d - 10*d^2)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(35*d^2*f*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.478873, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{35 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c+3 d) \left (c^2-10 c d-7 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{35 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(c^2 - 7*c*d - 10*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(35*d*f) + (4*a^2*(c - 7*d)*Cos[e + f*x]*
(c + d*Sin[e + f*x])^(3/2))/(35*d*f) - (2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(7*d*f) - (4*a^2*(c + 3
*d)*(c^2 - 10*c*d - 7*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(35*d^2*f*Sq
rt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c^2 - 7*c*d - 10*d^2)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(35*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 (c+d \sin (e+f x))^{3/2} \, dx &=-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}+\frac{2 \int \left (6 a^2 d-a^2 (c-7 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2} \, dx}{7 d}\\ &=\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}+\frac{4 \int \sqrt{c+d \sin (e+f x)} \left (\frac{3}{2} a^2 d (9 c+7 d)-\frac{3}{2} a^2 \left (c^2-7 c d-10 d^2\right ) \sin (e+f x)\right ) \, dx}{35 d}\\ &=\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}+\frac{8 \int \frac{\frac{3}{2} a^2 d \left (13 c^2+14 c d+5 d^2\right )-\frac{3}{4} a^2 (c+3 d) \left (c^2-10 c d-7 d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{105 d}\\ &=\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}-\frac{\left (2 a^2 (c+3 d) \left (c^2-10 c d-7 d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{35 d^2}+\frac{\left (2 a^2 \left (c^2-7 c d-10 d^2\right ) \left (c^2-d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{35 d^2}\\ &=\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}-\frac{\left (2 a^2 (c+3 d) \left (c^2-10 c d-7 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}{35 d^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a^2 \left (c^2-7 c d-10 d^2\right ) \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}{35 d^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{35 d f}+\frac{4 a^2 (c-7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d f}-\frac{2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 d f}-\frac{4 a^2 (c+3 d) \left (c^2-10 c d-7 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)}}{35 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 a^2 \left (c^2-7 c d-10 d^2\right ) \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}}}{35 d^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.10408, size = 262, normalized size = 0.88 \[ \frac{a^2 \left (d \cos (e+f x) \left (-d \left (36 c^2+168 c d+95 d^2\right ) \sin (e+f x)-112 c^2 d-4 c^3+2 d^2 (13 c+14 d) \cos (2 (e+f x))-106 c d^2+5 d^3 \sin (3 (e+f x))-28 d^3\right )-8 \left (-11 c^2 d^2-7 c^3 d+c^4+7 c d^3+10 d^4\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+8 \left (-44 c^2 d^2-6 c^3 d+c^4-58 c d^3-21 d^4\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{70 d^2 f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*(c^4 - 6*c^3*d - 44*c^2*d^2 - 58*c*d^3 - 21*d^4)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[
(c + d*Sin[e + f*x])/(c + d)] - 8*(c^4 - 7*c^3*d - 11*c^2*d^2 + 7*c*d^3 + 10*d^4)*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]*(-4*c^3 - 112*c^2*d - 106*c*d^2 - 28*d
^3 + 2*d^2*(13*c + 14*d)*Cos[2*(e + f*x)] - d*(36*c^2 + 168*c*d + 95*d^2)*Sin[e + f*x] + 5*d^3*Sin[3*(e + f*x)
])))/(70*d^2*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 1.149, size = 1316, normalized size = 4.4 \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))^(3/2),x)

[Out]

-2/35*a^2*(-13*c*d^4*sin(f*x+e)^4+28*c^2*d^3+20*c*d^4-68*((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))*c^3*d
^2-9*c^2*d^3*sin(f*x+e)^3-42*c*d^4*sin(f*x+e)^3-c^3*d^2*sin(f*x+e)^2-28*c^2*d^3*sin(f*x+e)^2-7*c*d^4*sin(f*x+e
)^2+9*c^2*d^3*sin(f*x+e)+42*c*d^4*sin(f*x+e)-5*d^5*sin(f*x+e)^5-14*d^5*sin(f*x+e)^4-15*d^5*sin(f*x+e)^3+14*d^5
*sin(f*x+e)^2+20*d^5*sin(f*x+e)+c^3*d^2+62*((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^5-2*((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^5-42*((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^5-64*((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))*c^2*d^3+68*((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))*c*d^4+14*((c+d*s
in(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*d+76*((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^2+28*(
(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^3-74*((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^
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)*EllipticF(
((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d)/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}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{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))^(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (2 \, a^{2} c + 2 \, a^{2} d -{\left (a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} -{\left (a^{2} d \cos \left (f x + e\right )^{2} - 2 \, a^{2} c - 2 \, a^{2} d\right )} \sin \left (f x + e\right )\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))^(3/2),x, algorithm="fricas")

[Out]

integral((2*a^2*c + 2*a^2*d - (a^2*c + 2*a^2*d)*cos(f*x + e)^2 - (a^2*d*cos(f*x + e)^2 - 2*a^2*c - 2*a^2*d)*si
n(f*x + e))*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 c \sqrt{c + d \sin{\left (e + f x \right )}}\, dx + \int 2 c \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}\, dx + \int c \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int d \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}\, dx + \int 2 d \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int d \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{3}{\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))**(3/2),x)

[Out]

a**2*(Integral(c*sqrt(c + d*sin(e + f*x)), x) + Integral(2*c*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integ
ral(c*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(d*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + In
tegral(2*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3,
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))^(3/2),x, algorithm="giac")

[Out]

Timed out