3.594 \(\int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=251 \[ -\frac {d^{3/2} (5 c-3 d) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {(c+9 d) (c-d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac {d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a \sin (e+f x)+a}} \]
[Out]
-(5*c-3*d)*d^(3/2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/a^(3/2)/f-
1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)^(3/2)*(c+9*d)*arctanh(1/2*cos(f
*x+e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/a^(3/2)/f*2^(1/2)+1/2*(c-3*d)
*d*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.90, antiderivative size = 251, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used =
{2765, 2983, 2982, 2782, 208, 2775, 205} \[ -\frac {d^{3/2} (5 c-3 d) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {(c+9 d) (c-d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac {d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a \sin (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
-(((5*c - 3*d)*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]
])])/(a^(3/2)*f)) - ((c - d)^(3/2)*(c + 9*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Si
n[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*f) + ((c - 3*d)*d*Cos[e + f*x]*Sqrt[c + d*Sin[e + f
*x]])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(2*f*(a + a*Sin[e +
f*x])^(3/2))
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 2765
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Rule 2775
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(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]
Rule 2782
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(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]
Rule 2982
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], 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]
Rule 2983
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])
Rubi steps
\begin {align*} \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (-\frac {1}{2} a \left (c^2+6 c d-3 d^2\right )+a (c-3 d) d \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a^2 \left (c^3+7 c^2 d-7 c d^2+3 d^3\right )-a^2 (5 c-3 d) d^2 \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{2 a^3}\\ &=\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {\left ((5 c-3 d) d^2\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c+d \sin (e+f x)}} \, dx}{2 a^2}+\frac {\left ((c-d)^2 (c+9 d)\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{4 a}\\ &=\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left ((5 c-3 d) d^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{a f}-\frac {\left ((c-d)^2 (c+9 d)\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2-(a c-a d) x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{2 f}\\ &=-\frac {(5 c-3 d) d^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {(c-d)^{3/2} (c+9 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}
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Mathematica [C] time = 17.02, size = 1844, normalized size = 7.35 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-(d^2*Cos[(e + f*x)/2]) + d^2*Sin[(e + f*x)/2] - (c - d)^2/(2*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])) + (c^2*Sin[(e + f*x)/2] - 2*c*d*Sin[(e + f*x)/2] + d^2*Sin[(e + f*x)/2])/(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^2)*Sqrt[c + d*Sin[e + f*x]])/(f*(a*(1 + Sin[e + f*x]))^(3/2)) + ((((c - d)^(
3/2)*(c + 9*d)*Log[1 + Tan[(e + f*x)/2]])/Sqrt[2] + I*(5*c - 3*d)*d^(3/2)*Log[((-I)*((-I)*c + d + (1 - I)*Sqrt
[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c - I*d)*Tan[(e + f*x)/2]))/(d^(5/2)*(-5
*c + 3*d)*(-I + Tan[(e + f*x)/2]))] + I*d^(3/2)*(-5*c + 3*d)*Log[(I*(I*c + d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1
+ Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2]))/(d^(5/2)*(-5*c + 3*d)*(I + Tan[
(e + f*x)/2]))] - ((c - d)^(3/2)*(c + 9*d)*Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*
Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]])/Sqrt[2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c^3/(4*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (7*c^2*d)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sq
rt[c + d*Sin[e + f*x]]) - (7*c*d^2)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (3*d^
3)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (5*c*d^2*Sin[e + f*x])/(2*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) - (3*d^3*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(a*(1 + Sin[e + f*x]))^(3/2)*(((c - d)^(3/2)*(c + 9*d)*Sec[(e + f*x)/2]^2
)/(2*Sqrt[2]*(1 + Tan[(e + f*x)/2])) - ((c - d)^(3/2)*(c + 9*d)*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d
]*d*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqrt[c - d]*((1 + Cos[e + f*x])^(-1
))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]))/(Sqrt[2]*(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]
*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2])) - ((5*c - 3*d)*d^4*(-5*c + 3*d)*(-I + Tan[(e + f*x)/2]
)*(((-I)*(((c - I*d)*Sec[(e + f*x)/2]^2)/2 + ((1 - I)*d^(3/2)*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/(Sqr
t[2]*Sqrt[c + d*Sin[e + f*x]]) + ((1 - I)*Sqrt[d]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[
e + f*x]])/Sqrt[2]))/(d^(5/2)*(-5*c + 3*d)*(-I + Tan[(e + f*x)/2])) + ((I/2)*Sec[(e + f*x)/2]^2*((-I)*c + d +
(1 - I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c - I*d)*Tan[(e + f*x)/2]))/
(d^(5/2)*(-5*c + 3*d)*(-I + Tan[(e + f*x)/2])^2)))/((-I)*c + d + (1 - I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x
])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c - I*d)*Tan[(e + f*x)/2]) + (d^4*(-5*c + 3*d)^2*(I + Tan[(e + f*x)/2])*(
(I*(((c + I*d)*Sec[(e + f*x)/2]^2)/2 + ((1 + I)*d^(3/2)*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/(Sqrt[2]*S
qrt[c + d*Sin[e + f*x]]) + ((1 + I)*Sqrt[d]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*
x]])/Sqrt[2]))/(d^(5/2)*(-5*c + 3*d)*(I + Tan[(e + f*x)/2])) - ((I/2)*Sec[(e + f*x)/2]^2*(I*c + d + (1 + I)*Sq
rt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2]))/(d^(5/2)*(
-5*c + 3*d)*(I + Tan[(e + f*x)/2])^2)))/(I*c + d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[
c + d*Sin[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2])))
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fricas [B] time = 1.22, size = 3420, normalized size = 13.63 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
[1/8*(2*sqrt(1/2)*(2*a*c^2 + 16*a*c*d - 18*a*d^2 - (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 8*a*c
*d - 9*a*d^2)*cos(f*x + e) + (2*a*c^2 + 16*a*c*d - 18*a*d^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e))*sin(f*
x + e))*sqrt((c - d)/a)*log(-(4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt((c - d)/a)*(c
os(f*x + e) - sin(f*x + e) + 1) + (c - 3*d)*cos(f*x + e)^2 + (3*c - d)*cos(f*x + e) - ((c - 3*d)*cos(f*x + e)
- 2*c - 2*d)*sin(f*x + e) + 2*c + 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))
+ (10*a*c*d - 6*a*d^2 - (5*a*c*d - 3*a*d^2)*cos(f*x + e)^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e) + (10*a*c*d - 6*
a*d^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/a)*log((128*d^4*cos(f*x + e)^5 + 128*(2*c*d^3
- d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x
+ e)^3 - 32*(c^3*d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 + 8*(16*d^3*cos(f*x + e)^4 + 24*(c*d^2 - d^3
)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3
- 7*c^2*d + 31*c*d^2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c*d^2 - 51*d^3 - 8*
(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x
+ e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d^3 + 289*d^4)*cos(f*x +
e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 -
32*(5*c^2*d^2 - 6*c*d^3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - 9*d^4)*cos(f*x + e))*sin
(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) + 4*(2*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 3
*d^2)*cos(f*x + e) + (2*d^2*cos(f*x + e) - c^2 + 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*si
n(f*x + e) + c))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x
+ e)), 1/4*(sqrt(1/2)*(2*a*c^2 + 16*a*c*d - 18*a*d^2 - (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e)^2 + (a*c^2 +
8*a*c*d - 9*a*d^2)*cos(f*x + e) + (2*a*c^2 + 16*a*c*d - 18*a*d^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e))*s
in(f*x + e))*sqrt((c - d)/a)*log(-(4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt((c - d)/
a)*(cos(f*x + e) - sin(f*x + e) + 1) + (c - 3*d)*cos(f*x + e)^2 + (3*c - d)*cos(f*x + e) - ((c - 3*d)*cos(f*x
+ e) - 2*c - 2*d)*sin(f*x + e) + 2*c + 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) -
2)) - (10*a*c*d - 6*a*d^2 - (5*a*c*d - 3*a*d^2)*cos(f*x + e)^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e) + (10*a*c*d
- 6*a*d^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/a)*arctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2
+ 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(d/a)/(2*d
^3*cos(f*x + e)^3 - (3*c*d^2 - d^3)*cos(f*x + e)*sin(f*x + e) - (c^2*d - c*d^2 + 2*d^3)*cos(f*x + e))) + 2*(2*
d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 3*d^2)*cos(f*x + e) + (2*d^2*cos(f*x + e) - c^2 + 2*c*
d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*
x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e)), 1/8*(4*sqrt(1/2)*(2*a*c^2 + 16*a*c*d - 18*a*d
^2 - (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e) + (2*a*c^2 + 16*a*c
*d - 18*a*d^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-(c - d)/a)*arctan(-2*sqrt(1/2)*s
qrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-(c - d)/a)/((c - d)*cos(f*x + e))) + (10*a*c*d - 6*a*d^
2 - (5*a*c*d - 3*a*d^2)*cos(f*x + e)^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e) + (10*a*c*d - 6*a*d^2 + (5*a*c*d - 3
*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/a)*log((128*d^4*cos(f*x + e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^
4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3*d
- 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 + 8*(16*d^3*cos(f*x + e)^4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c
^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^2
- 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c*d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*co
s(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin
(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d^3 + 289*d^4)*cos(f*x + e) + (128*d^4*cos(f
*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 - 6*c
*d^3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - 9*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f*x
+ e) + sin(f*x + e) + 1)) + 4*(2*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 3*d^2)*cos(f*x + e) +
(2*d^2*cos(f*x + e) - c^2 + 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(a^
2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e)), 1/4*(2*sqrt(
1/2)*(2*a*c^2 + 16*a*c*d - 18*a*d^2 - (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)
*cos(f*x + e) + (2*a*c^2 + 16*a*c*d - 18*a*d^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(
-(c - d)/a)*arctan(-2*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-(c - d)/a)/((c - d)*co
s(f*x + e))) - (10*a*c*d - 6*a*d^2 - (5*a*c*d - 3*a*d^2)*cos(f*x + e)^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e) + (
10*a*c*d - 6*a*d^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/a)*arctan(1/4*(8*d^2*cos(f*x + e)^
2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(d
/a)/(2*d^3*cos(f*x + e)^3 - (3*c*d^2 - d^3)*cos(f*x + e)*sin(f*x + e) - (c^2*d - c*d^2 + 2*d^3)*cos(f*x + e)))
+ 2*(2*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 3*d^2)*cos(f*x + e) + (2*d^2*cos(f*x + e) - c^
2 + 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(a^2*f*cos(f*x + e)^2 - a^2*
f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^(3/2), x)
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maple [F] time = 0.55, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x)
[Out]
int((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^(3/2),x)
[Out]
int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^(3/2), 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((c+d*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**(3/2),x)
[Out]
Timed out
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