3.586 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{9/2}} \, dx\)
Optimal. Leaf size=254 \[ -\frac{4 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^4 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^3 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{6 a^3 (c-d) (c+5 d) \cos (e+f x)}{35 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{7 d f (c+d) (c+d \sin (e+f x))^{7/2}} \]
[Out]
(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(7*d*(c + d)*f*(c + d*Sin[e + f*x])^(7/2)) + (6*a^3*(c -
d)*(c + 5*d)*Cos[e + f*x])/(35*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (2*a^3*
(3*c^2 + 22*c*d + 115*d^2)*Cos[e + f*x])/(105*d^2*(c + d)^3*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3
/2)) - (4*a^3*(3*c^2 + 22*c*d + 115*d^2)*Cos[e + f*x])/(105*d^2*(c + d)^4*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c +
d*Sin[e + f*x]])
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Rubi [A] time = 0.625349, antiderivative size = 254, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used =
{2762, 2980, 2772, 2771} \[ -\frac{4 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^4 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^3 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}+\frac{6 a^3 (c-d) (c+5 d) \cos (e+f x)}{35 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}+\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{7 d f (c+d) (c+d \sin (e+f x))^{7/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(9/2),x]
[Out]
(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(7*d*(c + d)*f*(c + d*Sin[e + f*x])^(7/2)) + (6*a^3*(c -
d)*(c + 5*d)*Cos[e + f*x])/(35*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)) - (2*a^3*
(3*c^2 + 22*c*d + 115*d^2)*Cos[e + f*x])/(105*d^2*(c + d)^3*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3
/2)) - (4*a^3*(3*c^2 + 22*c*d + 115*d^2)*Cos[e + f*x])/(105*d^2*(c + d)^4*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c +
d*Sin[e + f*x]])
Rule 2762
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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]))
Rule 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Rule 2772
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Rule 2771
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{9/2}} \, dx &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}-\frac{(2 a) \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-15 d)-\frac{1}{2} a (3 c+11 d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{7/2}} \, dx}{7 d (c+d)}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac{6 a^3 (c-d) (c+5 d) \cos (e+f x)}{35 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}+\frac{\left (a^2 \left (3 c^2+22 c d+115 d^2\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx}{35 d^2 (c+d)^2}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac{6 a^3 (c-d) (c+5 d) \cos (e+f x)}{35 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{2 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{\left (2 a^2 \left (3 c^2+22 c d+115 d^2\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{105 d^2 (c+d)^3}\\ &=\frac{2 a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac{6 a^3 (c-d) (c+5 d) \cos (e+f x)}{35 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac{2 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{4 a^3 \left (3 c^2+22 c d+115 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^4 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 4.20083, size = 216, normalized size = 0.85 \[ \frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-\left (1865 c^2 d+196 c^3+694 c d^2+465 d^3\right ) \sin (e+f x)+\left (157 c^2 d+21 c^3+827 c d^2+115 d^3\right ) \cos (2 (e+f x))+3 c^2 d \sin (3 (e+f x))-495 c^2 d-623 c^3+22 c d^2 \sin (3 (e+f x))-977 c d^2+115 d^3 \sin (3 (e+f x))-145 d^3\right )}{105 f (c+d)^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{7/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(9/2),x]
[Out]
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(-623*c^3 - 495*c^2*d - 977*c*d^2 - 145*
d^3 + (21*c^3 + 157*c^2*d + 827*c*d^2 + 115*d^3)*Cos[2*(e + f*x)] - (196*c^3 + 1865*c^2*d + 694*c*d^2 + 465*d^
3)*Sin[e + f*x] + 3*c^2*d*Sin[3*(e + f*x)] + 22*c*d^2*Sin[3*(e + f*x)] + 115*d^3*Sin[3*(e + f*x)]))/(105*(c +
d)^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c + d*Sin[e + f*x])^(7/2))
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Maple [B] time = 0.374, size = 1223, normalized size = 4.8 \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))^(5/2)/(c+d*sin(f*x+e))^(9/2),x)
[Out]
-2/105/f/(c+d)^4*(c+d*sin(f*x+e))^(1/2)*(a*(1+sin(f*x+e)))^(5/2)*(-485*cos(f*x+e)^8*c*d^6-78*cos(f*x+e)^6*c^6*
d+575*sin(f*x+e)*cos(f*x+e)^8*d^7+225*sin(f*x+e)*cos(f*x+e)^8*c*d^6+1408*sin(f*x+e)*cos(f*x+e)^6*c^3*d^4+2392*
sin(f*x+e)*cos(f*x+e)^6*c^2*d^5-950*sin(f*x+e)*cos(f*x+e)^6*c*d^6+479*sin(f*x+e)*cos(f*x+e)^4*c^5*d^2+1261*sin
(f*x+e)*cos(f*x+e)^4*c^4*d^3-4367*sin(f*x+e)*cos(f*x+e)^4*c^3*d^4-7117*sin(f*x+e)*cos(f*x+e)^4*c^2*d^5+1669*si
n(f*x+e)*cos(f*x+e)^4*c*d^6+368*sin(f*x+e)*cos(f*x+e)^2*c^6*d-3344*sin(f*x+e)*cos(f*x+e)^2*c^5*d^2-5008*sin(f*
x+e)*cos(f*x+e)^2*c^4*d^3+4560*sin(f*x+e)*cos(f*x+e)^2*c^3*d^4+7120*sin(f*x+e)*cos(f*x+e)^2*c^2*d^5-1328*sin(f
*x+e)*cos(f*x+e)^2*c*d^6+896*c^7+1152*c^6*d+2432*c^2*d^5-384*c*d^6-2944*c^4*d^3-640*d^7-2176*c^5*d^2+1664*c^3*
d^4-2350*sin(f*x+e)*cos(f*x+e)^6*d^7-172*cos(f*x+e)^6*c^4*d^3-2968*cos(f*x+e)^6*c^3*d^4-5370*cos(f*x+e)^6*c^2*
d^5+1590*cos(f*x+e)^6*c*d^6+3615*sin(f*x+e)*cos(f*x+e)^4*d^7+39*cos(f*x+e)^4*c^6*d-1879*cos(f*x+e)^4*c^5*d^2-3
397*cos(f*x+e)^4*c^4*d^3+895*cos(f*x+e)^8*c^2*d^5+6439*cos(f*x+e)^4*c^3*d^4+10373*cos(f*x+e)^4*c^2*d^5-2285*co
s(f*x+e)^4*c*d^6+112*sin(f*x+e)*cos(f*x+e)^2*c^7-2480*sin(f*x+e)*cos(f*x+e)^2*d^7-944*cos(f*x+e)^2*c^6*d+4432*
cos(f*x+e)^2*c^5*d^2+6480*cos(f*x+e)^2*c^4*d^3-5392*cos(f*x+e)^2*c^3*d^4-8336*cos(f*x+e)^2*c^2*d^5+1520*cos(f*
x+e)^2*c*d^6-1152*sin(f*x+e)*c^6*d+2176*sin(f*x+e)*c^5*d^2+2944*sin(f*x+e)*c^4*d^3-1664*sin(f*x+e)*c^3*d^4-243
2*sin(f*x+e)*c^2*d^5+384*sin(f*x+e)*c*d^6-1555*cos(f*x+e)^8*d^7+3940*cos(f*x+e)^6*d^7-4775*cos(f*x+e)^4*d^7-56
0*cos(f*x+e)^2*c^7+2800*cos(f*x+e)^2*d^7-896*sin(f*x+e)*c^7+640*sin(f*x+e)*d^7+230*cos(f*x+e)^10*d^7-35*cos(f*
x+e)^4*c^7-302*cos(f*x+e)^6*c^5*d^2+6*cos(f*x+e)^10*c^2*d^5+44*cos(f*x+e)^10*c*d^6+48*cos(f*x+e)^8*c^4*d^3+257
*cos(f*x+e)^8*c^3*d^4-21*sin(f*x+e)*cos(f*x+e)^4*c^7+3*sin(f*x+e)*cos(f*x+e)^8*c^3*d^4+37*sin(f*x+e)*cos(f*x+e
)^8*c^2*d^5+102*sin(f*x+e)*cos(f*x+e)^6*c^5*d^2+518*sin(f*x+e)*cos(f*x+e)^6*c^4*d^3+sin(f*x+e)*cos(f*x+e)^4*c^
6*d)/cos(f*x+e)^5/(cos(f*x+e)^2*d^2+c^2-d^2)^4
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Maxima [B] time = 2.25458, size = 949, normalized size = 3.74 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(9/2),x, algorithm="maxima")
[Out]
-2/105*((301*c^4 + 169*c^3*d + 75*c^2*d^2 + 15*c*d^3)*a^(5/2) - 3*(35*c^4 - 763*c^3*d - 297*c^2*d^2 - 85*c*d^3
- 10*d^4)*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 6*(182*c^4 - 127*c^3*d + 1059*c^2*d^2 + 251*c*d^3 + 35*d^
4)*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*(50*c^4 - 421*c^3*d + 201*c^2*d^2 - 535*c*d^3 - 55*d^4)*a^
(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*(11*c^4 - 36*c^3*d + 80*c^2*d^2 - 40*c*d^3 + 25*d^4)*a^(5/2)*s
in(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*(11*c^4 - 36*c^3*d + 80*c^2*d^2 - 40*c*d^3 + 25*d^4)*a^(5/2)*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5 + 14*(50*c^4 - 421*c^3*d + 201*c^2*d^2 - 535*c*d^3 - 55*d^4)*a^(5/2)*sin(f*x + e)^
6/(cos(f*x + e) + 1)^6 - 6*(182*c^4 - 127*c^3*d + 1059*c^2*d^2 + 251*c*d^3 + 35*d^4)*a^(5/2)*sin(f*x + e)^7/(c
os(f*x + e) + 1)^7 + 3*(35*c^4 - 763*c^3*d - 297*c^2*d^2 - 85*c*d^3 - 10*d^4)*a^(5/2)*sin(f*x + e)^8/(cos(f*x
+ e) + 1)^8 - (301*c^4 + 169*c^3*d + 75*c^2*d^2 + 15*c*d^3)*a^(5/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9)*(sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^2/((c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 + 2*(c^4 + 4*c^3*d + 6*c^2*
d^2 + 4*c*d^3 + d^4)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
)^(9/2)*f)
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Fricas [B] time = 2.37057, size = 2342, normalized size = 9.22 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(9/2),x, algorithm="fricas")
[Out]
-2/105*(224*a^2*c^3 - 608*a^2*c^2*d + 544*a^2*c*d^2 - 160*a^2*d^3 - 2*(3*a^2*c^2*d + 22*a^2*c*d^2 + 115*a^2*d^
3)*cos(f*x + e)^4 - (21*a^2*c^3 + 157*a^2*c^2*d + 827*a^2*c*d^2 + 115*a^2*d^3)*cos(f*x + e)^3 + (77*a^2*c^3 +
783*a^2*c^2*d - 425*a^2*c*d^2 + 405*a^2*d^3)*cos(f*x + e)^2 + 2*(161*a^2*c^3 + 163*a^2*c^2*d + 451*a^2*c*d^2 +
65*a^2*d^3)*cos(f*x + e) - (224*a^2*c^3 - 608*a^2*c^2*d + 544*a^2*c*d^2 - 160*a^2*d^3 + 2*(3*a^2*c^2*d + 22*a
^2*c*d^2 + 115*a^2*d^3)*cos(f*x + e)^3 - (21*a^2*c^3 + 151*a^2*c^2*d + 783*a^2*c*d^2 - 115*a^2*d^3)*cos(f*x +
e)^2 - 2*(49*a^2*c^3 + 467*a^2*c^2*d + 179*a^2*c*d^2 + 145*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x
+ e) + a)*sqrt(d*sin(f*x + e) + c)/((c^4*d^4 + 4*c^3*d^5 + 6*c^2*d^6 + 4*c*d^7 + d^8)*f*cos(f*x + e)^5 + (4*c
^5*d^3 + 17*c^4*d^4 + 28*c^3*d^5 + 22*c^2*d^6 + 8*c*d^7 + d^8)*f*cos(f*x + e)^4 - 2*(3*c^6*d^2 + 12*c^5*d^3 +
19*c^4*d^4 + 16*c^3*d^5 + 9*c^2*d^6 + 4*c*d^7 + d^8)*f*cos(f*x + e)^3 - 2*(2*c^7*d + 11*c^6*d^2 + 28*c^5*d^3 +
43*c^4*d^4 + 42*c^3*d^5 + 25*c^2*d^6 + 8*c*d^7 + d^8)*f*cos(f*x + e)^2 + (c^8 + 4*c^7*d + 12*c^6*d^2 + 28*c^5
*d^3 + 38*c^4*d^4 + 28*c^3*d^5 + 12*c^2*d^6 + 4*c*d^7 + d^8)*f*cos(f*x + e) + (c^8 + 8*c^7*d + 28*c^6*d^2 + 56
*c^5*d^3 + 70*c^4*d^4 + 56*c^3*d^5 + 28*c^2*d^6 + 8*c*d^7 + d^8)*f + ((c^4*d^4 + 4*c^3*d^5 + 6*c^2*d^6 + 4*c*d
^7 + d^8)*f*cos(f*x + e)^4 - 4*(c^5*d^3 + 4*c^4*d^4 + 6*c^3*d^5 + 4*c^2*d^6 + c*d^7)*f*cos(f*x + e)^3 - 2*(3*c
^6*d^2 + 14*c^5*d^3 + 27*c^4*d^4 + 28*c^3*d^5 + 17*c^2*d^6 + 6*c*d^7 + d^8)*f*cos(f*x + e)^2 + 4*(c^7*d + 4*c^
6*d^2 + 7*c^5*d^3 + 8*c^4*d^4 + 7*c^3*d^5 + 4*c^2*d^6 + c*d^7)*f*cos(f*x + e) + (c^8 + 8*c^7*d + 28*c^6*d^2 +
56*c^5*d^3 + 70*c^4*d^4 + 56*c^3*d^5 + 28*c^2*d^6 + 8*c*d^7 + d^8)*f)*sin(f*x + e))
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Sympy [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))**(5/2)/(c+d*sin(f*x+e))**(9/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(9/2),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(9/2), x)