3.503 \(\int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=419 \[ -\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^3 (c+d \sin (e+f x))^{3/2}}+\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 d^2 f (c-d) (c+d)^4 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f (c-d) (c+d)^4 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 f (c+d)^2 (c+d \sin (e+f x))^{5/2}}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{7 d f (c+d) (c+d \sin (e+f x))^{7/2}} \]

[Out]

2/7*(c-d)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/(c+d)/f/(c+d*sin(f*x+e))^(7/2)+8/35*a^3*(c-d)*(c+4*d)*cos(f*x+e)/d
^2/(c+d)^2/f/(c+d*sin(f*x+e))^(5/2)-4/105*a^3*(4*c^2+21*c*d+65*d^2)*cos(f*x+e)/d^2/(c+d)^3/f/(c+d*sin(f*x+e))^
(3/2)-4/105*a^3*(4*c^3+21*c^2*d+62*c*d^2-147*d^3)*cos(f*x+e)/(c-d)/d^2/(c+d)^4/f/(c+d*sin(f*x+e))^(1/2)+4/105*
a^3*(4*c^3+21*c^2*d+62*c*d^2-147*d^3)*(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)/(c-d)/d^3/(c+d)^4/f/((c+d*sin(f*x+e)
)/(c+d))^(1/2)-4/105*a^3*(4*c^2+21*c*d+65*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)*E
llipticF(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^3/(c+d)^3/f/(c+d*
sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.93, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2762, 2968, 3021, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a^3 \left (21 c^2 d+4 c^3+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 d^2 f (c-d) (c+d)^4 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^3 (c+d \sin (e+f x))^{3/2}}+\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (21 c^2 d+4 c^3+62 c d^2-147 d^3\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f (c-d) (c+d)^4 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 f (c+d)^2 (c+d \sin (e+f x))^{5/2}}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{7 d f (c+d) (c+d \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(9/2),x]

[Out]

(2*(c - d)*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(7*d*(c + d)*f*(c + d*Sin[e + f*x])^(7/2)) + (8*a^3*(c - d)*
(c + 4*d)*Cos[e + f*x])/(35*d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^(5/2)) - (4*a^3*(4*c^2 + 21*c*d + 65*d^2)*Cos
[e + f*x])/(105*d^2*(c + d)^3*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^3*(4*c^3 + 21*c^2*d + 62*c*d^2 - 147*d^3)*C
os[e + f*x])/(105*(c - d)*d^2*(c + d)^4*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^3*(4*c^3 + 21*c^2*d + 62*c*d^2 - 14
7*d^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(105*(c - d)*d^3*(c + d)^4*f*Sqr
t[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^3*(4*c^2 + 21*c*d + 65*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d
)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(105*d^3*(c + d)^3*f*Sqrt[c + d*Sin[e + f*x]])

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]

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 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 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 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 2754

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 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 2968

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

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{9/2}} \, dx &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}-\frac {(2 a) \int \frac {(a+a \sin (e+f x)) (a (c-8 d)-a (2 c+5 d) \sin (e+f x))}{(c+d \sin (e+f x))^{7/2}} \, dx}{7 d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}-\frac {(2 a) \int \frac {a^2 (c-8 d)+\left (a^2 (c-8 d)-a^2 (2 c+5 d)\right ) \sin (e+f x)-a^2 (2 c+5 d) \sin ^2(e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx}{7 d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}+\frac {(4 a) \int \frac {\frac {5}{2} a^2 (c-d) d (c+13 d)+\frac {1}{2} a^2 (c-d) \left (4 c^2+17 c d+49 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{35 (c-d) d^2 (c+d)^2}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f (c+d \sin (e+f x))^{3/2}}-\frac {(8 a) \int \frac {-\frac {3}{4} a^2 (c-d)^2 d (c+49 d)-\frac {1}{4} a^2 (c-d)^2 \left (4 c^2+21 c d+65 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{105 (c-d)^2 d^2 (c+d)^3}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 (c-d) d^2 (c+d)^4 f \sqrt {c+d \sin (e+f x)}}+\frac {(16 a) \int \frac {-\frac {1}{8} a^2 (c-d)^2 d \left (c^2-126 c d+65 d^2\right )-\frac {1}{8} a^2 (c-d)^2 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{105 (c-d)^3 d^2 (c+d)^4}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 (c-d) d^2 (c+d)^4 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 a^3 \left (4 c^2+21 c d+65 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{105 d^3 (c+d)^3}-\frac {\left (2 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{105 (c-d) d^3 (c+d)^4}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 (c-d) d^2 (c+d)^4 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{105 (c-d) d^3 (c+d)^4 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^3 \left (4 c^2+21 c d+65 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}{105 d^3 (c+d)^3 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{7 d (c+d) f (c+d \sin (e+f x))^{7/2}}+\frac {8 a^3 (c-d) (c+4 d) \cos (e+f x)}{35 d^2 (c+d)^2 f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^3 \left (4 c^2+21 c d+65 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \cos (e+f x)}{105 (c-d) d^2 (c+d)^4 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\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)}}{105 (c-d) d^3 (c+d)^4 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^3 \left (4 c^2+21 c d+65 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}}}{105 d^3 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 3.70, size = 351, normalized size = 0.84 \[ -\frac {2 a^3 (\sin (e+f x)+1)^3 \left (d \cos (e+f x) \left (2 (c-d) \left (4 c^2+21 c d+65 d^2\right ) (c+d) (c+d \sin (e+f x))^2+2 \left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) (c+d \sin (e+f x))^3-9 (c-d)^2 (3 c+7 d) (c+d)^2 (c+d \sin (e+f x))+15 (c-d)^3 (c+d)^3\right )-2 (c+d \sin (e+f x))^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (d^2 \left (c^2-126 c d+65 d^2\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )+\left (4 c^3+21 c^2 d+62 c d^2-147 d^3\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )\right )\right )}{105 d^3 f (c-d) (c+d)^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6 (c+d \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(9/2),x]

[Out]

(-2*a^3*(1 + Sin[e + f*x])^3*(-2*(d^2*(c^2 - 126*c*d + 65*d^2)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]
+ (4*c^3 + 21*c^2*d + 62*c*d^2 - 147*d^3)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*Ellipt
icF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]))*(c + d*Sin[e + f*x])^3*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + d*Cos[
e + f*x]*(15*(c - d)^3*(c + d)^3 - 9*(c - d)^2*(c + d)^2*(3*c + 7*d)*(c + d*Sin[e + f*x]) + 2*(c - d)*(c + d)*
(4*c^2 + 21*c*d + 65*d^2)*(c + d*Sin[e + f*x])^2 + 2*(4*c^3 + 21*c^2*d + 62*c*d^2 - 147*d^3)*(c + d*Sin[e + f*
x])^3)))/(105*(c - d)*d^3*(c + d)^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c + d*Sin[e + f*x])^(7/2))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{5 \, c d^{4} \cos \left (f x + e\right )^{4} + c^{5} + 10 \, c^{3} d^{2} + 5 \, c d^{4} - 10 \, {\left (c^{3} d^{2} + c d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (d^{5} \cos \left (f x + e\right )^{4} + 5 \, c^{4} d + 10 \, c^{2} d^{3} + d^{5} - 2 \, {\left (5 \, c^{2} d^{3} + d^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

integral(-(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c)/
(5*c*d^4*cos(f*x + e)^4 + c^5 + 10*c^3*d^2 + 5*c*d^4 - 10*(c^3*d^2 + c*d^4)*cos(f*x + e)^2 + (d^5*cos(f*x + e)
^4 + 5*c^4*d + 10*c^2*d^3 + d^5 - 2*(5*c^2*d^3 + d^5)*cos(f*x + e)^2)*sin(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(9/2), x)

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maple [B]  time = 9.86, size = 2079, normalized size = 4.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(9/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a^3*((-c^3+3*c^2*d-3*c*d^2+d^3)/d^3*(2/7/(c^2-d^2)/d^3*(-(-d*sin(f*x+e
)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^4+24/35/(c^2-d^2)^2/d^2*c*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(s
in(f*x+e)+c/d)^3+2/105*(71*c^2+25*d^2)/d/(c^2-d^2)^3*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^
2+32/105*d*cos(f*x+e)^2/(c^2-d^2)^4*c*(11*c^2+13*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(105*c^4+254*c
^2*d^2+25*d^4)/(105*c^8-420*c^6*d^2+630*c^4*d^4-420*c^2*d^6+105*d^8)*(c/d-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))+32/105*c*d*(11*c^2+13*d^2)/(c^2-d^2)^4*(c/d-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)*((-c/d-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))))+1/d^3*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1
/2)+2*c/(c^2-d^2)*(c/d-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/(c^2-d^2)*d*(c/d-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)*((-c/d-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))))+3*(-c+d)/d^3*(2/3/(c^2-d^2)/d*(-(-d*
sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)*cos(
f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(c/d-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))+8/3*c*d/(c^2-d^2)^2*(c/d-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)*((-c/d-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))))+3*(c^2-2*c*d+d^2)/d^3*(2/5/(c^2-d^2)/d^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^3+16/1
5*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+2/15*d*cos(f*x+e)^2/(c^2-d^2)^3*(
23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(15*c^3+17*c*d^2)/(15*c^6-45*c^4*d^2+45*c^2*d^4-15*d^6
)*(c/d-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))+2/15*d*(23*c^2
+9*d^2)/(c^2-d^2)^3*(c/d-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)*((-c/d-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)))))/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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c + d*sin(e + f*x))^(9/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c + d*sin(e + f*x))^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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