\(\int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx\) [528]

Optimal result

Integrand size = 27, antiderivative size = 152 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {3+3 \sin (e+f x)}}\right )}{4 \sqrt {d} (c+d)^{5/2} f}-\frac {3 \cos (e+f x)}{2 (c+d) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {9 \cos (e+f x)}{4 (c+d)^2 f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2851, 2852, 214} \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=-\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{4 \sqrt {d} f (c+d)^{5/2}}-\frac {3 a \cos (e+f x)}{4 f (c+d)^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \]

[In]

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

[Out]

(-3*Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(4*Sqrt[d]*(c + d)
^(5/2)*f) - (a*Cos[e + f*x])/(2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2) - (3*a*Cos[e + f*x]
)/(4*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2851

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 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*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*} \text {integral}& = -\frac {a \cos (e+f x)}{2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {3 \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx}{4 (c+d)} \\ & = -\frac {a \cos (e+f x)}{2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {3 a \cos (e+f x)}{4 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {3 \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 (c+d)^2} \\ & = -\frac {a \cos (e+f x)}{2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {3 a \cos (e+f x)}{4 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {(3 a) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{4 (c+d)^2 f} \\ & = -\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {d} (c+d)^{5/2} f}-\frac {a \cos (e+f x)}{2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {3 a \cos (e+f x)}{4 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 5.78 (sec) , antiderivative size = 923, normalized size of antiderivative = 6.07 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {3} \sqrt {1+\sin (e+f x)} \left (\frac {3 \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right ) \left ((-1+i) x \cos (e)+\frac {\text {RootSum}\left [-d+2 i c e^{i e} \text {$\#$1}^2+d e^{2 i e} \text {$\#$1}^4\&,\frac {(1+i) d \sqrt {e^{-i e}} f x-(2-2 i) d \sqrt {e^{-i e}} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right )-i \sqrt {d} \sqrt {c+d} f x \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}+\frac {(1-i) c f x \text {$\#$1}^2}{\sqrt {e^{-i e}}}+\frac {(2+2 i) c \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-i e}}}-\sqrt {d} \sqrt {c+d} e^{i e} f x \text {$\#$1}^3-2 i \sqrt {d} \sqrt {c+d} e^{i e} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^3}{d-i c e^{i e} \text {$\#$1}^2}\&\right ] (\cos (e)+i (-1+\sin (e))) \sqrt {\cos (e)-i \sin (e)}}{4 f}+(1+i) x \sin (e)\right )}{\sqrt {d} (c+d)^{5/2} (\cos (e)+i (-1+\sin (e))) \sqrt {\cos (e)-i \sin (e)}}+\frac {3 \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right ) \left ((1-i) x \cos (e)-(1+i) x \sin (e)+\frac {\text {RootSum}\left [-d+2 i c e^{i e} \text {$\#$1}^2+d e^{2 i e} \text {$\#$1}^4\&,\frac {(1-i) d \sqrt {e^{-i e}} f x+(2+2 i) d \sqrt {e^{-i e}} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right )+\sqrt {d} \sqrt {c+d} f x \text {$\#$1}+2 i \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}-\frac {(1+i) c f x \text {$\#$1}^2}{\sqrt {e^{-i e}}}+\frac {(2-2 i) c \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-i e}}}-i \sqrt {d} \sqrt {c+d} e^{i e} f x \text {$\#$1}^3+2 \sqrt {d} \sqrt {c+d} e^{i e} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^3}{d-i c e^{i e} \text {$\#$1}^2}\&\right ] \sqrt {\cos (e)-i \sin (e)} (-1-i \cos (e)+\sin (e))}{4 f}\right )}{\sqrt {d} (c+d)^{5/2} (\cos (e)+i (-1+\sin (e))) \sqrt {\cos (e)-i \sin (e)}}-\frac {(4-4 i) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) f (c+d \sin (e+f x))^2}-\frac {(6-6 i) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^2 f (c+d \sin (e+f x))}\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )} \]

[In]

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

[Out]

((1/16 + I/16)*Sqrt[3]*Sqrt[1 + Sin[e + f*x]]*((3*(Cos[e/2] + I*Sin[e/2])*((-1 + I)*x*Cos[e] + (RootSum[-d + (
2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 + I)*d*Sqrt[E^((-I)*e)]*f*x - (2 - 2*I)*d*Sqrt[E^((-I)*e)]*Lo
g[E^((I/2)*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 + ((1
- I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 + 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - Sqrt[d]*Sqrt
[c + d]*E^(I*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1
^2) & ]*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]])/(4*f) + (1 + I)*x*Sin[e]))/(Sqrt[d]*(c + d)^(5/2)*
(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + (3*(Cos[e/2] + I*Sin[e/2])*((1 - I)*x*Cos[e] - (1 + I)*x
*Sin[e] + (RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 - I)*d*Sqrt[E^((-I)*e)]*f*x + (2 + 2
*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] + Sqrt[d]*Sqrt[c + d]*f*x*#1 + (2*I)*Sqrt[d]*Sqrt[c + d]*Log[E^
((I/2)*f*x) - #1]*#1 - ((1 + I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 - 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt
[E^((-I)*e)] - I*Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*
#1^3)/(d - I*c*E^(I*e)*#1^2) & ]*Sqrt[Cos[e] - I*Sin[e]]*(-1 - I*Cos[e] + Sin[e]))/(4*f)))/(Sqrt[d]*(c + d)^(5
/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((4 - 4*I)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((
c + d)*f*(c + d*Sin[e + f*x])^2) - ((6 - 6*I)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)^2*f*(c + d*Sin[e
+ f*x]))))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )+1\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) a \,d^{2}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a c d +3 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, \sin \left (f x +e \right ) d +3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a \,c^{2}+5 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, c +2 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, d \right )}{4 \left (c +d \right )^{2} \left (c +d \sin \left (f x +e \right )\right )^{2} \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(254\)

[In]

int((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*(sin(f*x+e)+1)*(-a*(sin(f*x+e)-1))^(1/2)*(3*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*
x+e)^2*a*d^2+6*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a*c*d+3*(-a*(sin(f*x+e)-1))^(
1/2)*(a*(c+d)*d)^(1/2)*sin(f*x+e)*d+3*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a*c^2+5*(-a*(sin(
f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*c+2*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*d)/(c+d)^2/(c+d*sin(f*x+e))
^2/(a*(c+d)*d)^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (130) = 260\).

Time = 0.39 (sec) , antiderivative size = 1250, normalized size of antiderivative = 8.22 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/16*(3*(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (
d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(
f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^
2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*co
s(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x +
e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2
*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x +
e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(3*d*cos(f*x + e)^2 + (5*c + 2*d)*cos(f*x +
e) + (3*d*cos(f*x + e) - 5*c + d)*sin(f*x + e) + 5*c - d)*sqrt(a*sin(f*x + e) + a))/((c^2*d^2 + 2*c*d^3 + d^4
)*f*cos(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c
*d^3 + d^4)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(
f*x + e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin
(f*x + e)), -1/8*(3*(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f
*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arc
tan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(3*d*co
s(f*x + e)^2 + (5*c + 2*d)*cos(f*x + e) + (3*d*cos(f*x + e) - 5*c + d)*sin(f*x + e) + 5*c - d)*sqrt(a*sin(f*x
+ e) + a))/((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*f*cos(f*x + e)^
2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f
+ ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^4 + 4*c^3*d
+ 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\int { \frac {\sqrt {a \sin \left (f x + e\right ) + a}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt {-c d - d^{2}}} + \frac {2 \, {\left (6 \, d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2} {\left (c^{2} + 2 \, c d + d^{2}\right )}}\right )}}{8 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/8*sqrt(2)*sqrt(a)*(3*sqrt(2)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))*sgn(cos(-1/4
*pi + 1/2*f*x + 1/2*e))/((c^2 + 2*c*d + d^2)*sqrt(-c*d - d^2)) + 2*(6*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*si
n(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 5*d*
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c
- d)^2*(c^2 + 2*c*d + d^2)))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\int \frac {\sqrt {a+a\,\sin \left (e+f\,x\right )}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

[In]

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

[Out]

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