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

Optimal. Leaf size=194 \[ -\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {\sqrt {c-d} (c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-2*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-1/4*(c+5
*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))*(c-d)^(1
/2)/a^(3/2)/f*2^(1/2)-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(3/2)

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Rubi [A]  time = 0.56, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2765, 2982, 2782, 208, 2775, 205} \[ -\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {\sqrt {c-d} (c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*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) - (Sqrt[c - d]*(c + 5*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sq
rt[c + d*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*f) - ((c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(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]

Rubi steps

\begin {align*} \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a \left (c^2+4 c d-d^2\right )-2 a d^2 \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {d^2 \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c+d \sin (e+f x)}} \, dx}{a^2}+\frac {((c-d) (c+5 d)) \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{4 a}\\ &=-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (2 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 {((c-d) (c+5 d)) \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 {2 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 {\sqrt {c-d} (c+5 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-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 17.12, size = 1625, normalized size = 8.38 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*((-c + d)/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (c*Sin[(e + f*x
)/2] - d*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)) + ((Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*L
og[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]] -
(4*I)*Sqrt[c - d]*d^(3/2)*(Log[(c - I*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*S
in[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2])/(2*d^(5/2)*(I + Tan[(e + f*x)/2]))] - Log[(c + I*d + (1 + I)*Sq
rt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2])/(2*d^(5/2)*
(-I + Tan[(e + f*x)/2]))]))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c^2/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2])*Sqrt[c + d*Sin[e + f*x]]) + (c*d)/((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) - d^2/(
4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (d^2*Sin[e + f*x])/((Cos[(e + f*x)/2] + Si
n[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(a*(1 + Sin[e + f*x]))^(3/2)*(((c^2 + 4*c*d - 5*d^2)*Sec[(e + f
*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*(((-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]]))/(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]) - (4*I)*Sqrt[c - d]*d^(3/2)*((2*d^(5/2)*(I + Tan[(e
+ f*x)/2])*(((((-I)*c + d)*Sec[(e + f*x)/2]^2)/2 - I*(((1 + I)*d^(3/2)*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-
1)])/(Sqrt[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]))/(2*d^(5/2)*(I + Tan[(e + f*x)/2])) - (Sec[(e + f*x)/2]^2*(c - I*(d + (1 + I)*Sqr
t[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2]))/(4*d^(5
/2)*(I + Tan[(e + f*x)/2])^2)))/(c - I*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*S
in[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2]) - (2*d^(5/2)*(-I + Tan[(e + f*x)/2])*((((I*c + 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]*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])/(2*d^(5/2)*(-I
+ Tan[(e + f*x)/2])) - (Sec[(e + f*x)/2]^2*(c + I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*S
qrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2]))/(4*d^(5/2)*(-I + Tan[(e + f*x)/2])^2)))/(c + I*d + (1 +
I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2]))))

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fricas [B]  time = 1.08, size = 2883, normalized size = 14.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(1/2)*((a*c + 5*a*d)*cos(f*x + e)^2 - 2*a*c - 10*a*d - (a*c + 5*a*d)*cos(f*x + e) - (2*a*c + 10*a*d
+ (a*c + 5*a*d)*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*s
in(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)) + (a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*
d)*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)) + 2*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*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
*(sqrt(1/2)*((a*c + 5*a*d)*cos(f*x + e)^2 - 2*a*c - 10*a*d - (a*c + 5*a*d)*cos(f*x + e) - (2*a*c + 10*a*d + (a
*c + 5*a*d)*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)*(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)) + 2*(a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*d)
*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*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c
- d)*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)*((a*c + 5*a*d)*cos(f*x + e)^2 - 2*a*c - 10*a*
d - (a*c + 5*a*d)*cos(f*x + e) - (2*a*c + 10*a*d + (a*c + 5*a*d)*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)*cos(f*x + e)))
- (a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*d)*sin(f*x + e))*sqrt(-d/a)*log((1
28*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)) - 2*((c - d)*cos(f*x + e) -
(c - d)*sin(f*x + e) + c - d)*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)), -1/2*(sqrt(1/2)*((a*c + 5*a*d)*cos(f*
x + e)^2 - 2*a*c - 10*a*d - (a*c + 5*a*d)*cos(f*x + e) - (2*a*c + 10*a*d + (a*c + 5*a*d)*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)*cos(f*x + e))) - (a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*d)*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*s
in(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))) - ((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*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*co
s(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 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.47, size = 6681, normalized size = 34.44 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*sin(e + f*x))**(3/2)/(a*(sin(e + f*x) + 1))**(3/2), x)

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