∫ ∘ _________ a+a sin(e+fx)

3.527 $\int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx$

Optimal. Leaf size=105 \[ -\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f (c+d)^{3/2}} \]

[Out]

-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)^(3/2)/f/d^(1/2)-a*cos(f*

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

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Rubi [A] time = 0.19, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.111, Rules used = {2772, 2773, 208} \[ -\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f (c+d)^{3/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

/2)*f)) - (a*Cos[e + f*x])/((c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

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 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 2773

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

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Mathematica [C] time = 6.02, size = 871, normalized size = 8.30 \[ \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {a (\sin (e+f x)+1)} \left (\frac {\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 e^{2 i e} \text {$\#$1}^4+2 i c e^{i e} \text {$\#$1}^2-d\& ,\frac {-\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+\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} f x \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}+(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 )}{d-i c e^{i e} \text {$\#$1}^2}\& \right ] (\cos (e)+i (\sin (e)-1)) \sqrt {\cos (e)-i \sin (e)}}{4 f}\right )}{\sqrt {d} (c+d)^{3/2} (\cos (e)+i (\sin (e)-1)) \sqrt {\cos (e)-i \sin (e)}}+\frac {\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 e^{2 i e} \text {$\#$1}^4+2 i c e^{i e} \text {$\#$1}^2-d\& ,\frac {-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-\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} f x \text {$\#$1}+2 i \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}+(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 )}{d-i c e^{i e} \text {$\#$1}^2}\& \right ] \sqrt {\cos (e)-i \sin (e)} (-i \cos (e)+\sin (e)-1)}{4 f}\right )}{\sqrt {d} (c+d)^{3/2} (\cos (e)+i (\sin (e)-1)) \sqrt {\cos (e)-i \sin (e)}}-\frac {(2-2 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))}\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1/4 + I/4)*Sqrt[a*(1 + Sin[e + f*x])]*(((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)]*Log[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)^(3/2)*(Cos[e]

+ I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((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)^(3/2)*(Cos[e

] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((2 - 2*I)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*f*(

c + d*Sin[e + f*x]))))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])

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fricas [B] time = 0.56, size = 786, normalized size = 7.49 \[ \left [\frac {{\left (d \cos \left (f x + e\right )^{2} - c \cos \left (f x + e\right ) - {\left (d \cos \left (f x + e\right ) + c + d\right )} \sin \left (f x + e\right ) - c - d\right )} \sqrt {\frac {a}{c d + d^{2}}} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} - {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} + {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {a}{c d + d^{2}}} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{4 \, {\left ({\left (c d + d^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{2} + c d\right )} f \cos \left (f x + e\right ) - {\left (c^{2} + 2 \, c d + d^{2}\right )} f - {\left ({\left (c d + d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{2} + 2 \, c d + d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}, -\frac {{\left (d \cos \left (f x + e\right )^{2} - c \cos \left (f x + e\right ) - {\left (d \cos \left (f x + e\right ) + c + d\right )} \sin \left (f x + e\right ) - c - d\right )} \sqrt {-\frac {a}{c d + d^{2}}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {a}{c d + d^{2}}}}{2 \, a \cos \left (f x + e\right )}\right ) - 2 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{2 \, {\left ({\left (c d + d^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{2} + c d\right )} f \cos \left (f x + e\right ) - {\left (c^{2} + 2 \, c d + d^{2}\right )} f - {\left ({\left (c d + d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{2} + 2 \, c d + d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((d*cos(f*x + e)^2 - c*cos(f*x + e) - (d*cos(f*x + e) + c + d)*sin(f*x + e) - c - d)*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)*cos(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*sqrt(a*sin(f*x + e) + a)*(co

s(f*x + e) - sin(f*x + e) + 1))/((c*d + d^2)*f*cos(f*x + e)^2 - (c^2 + c*d)*f*cos(f*x + e) - (c^2 + 2*c*d + d^

2)*f - ((c*d + d^2)*f*cos(f*x + e) + (c^2 + 2*c*d + d^2)*f)*sin(f*x + e)), -1/2*((d*cos(f*x + e)^2 - c*cos(f*x

+ e) - (d*cos(f*x + e) + c + d)*sin(f*x + e) - c - d)*sqrt(-a/(c*d + d^2))*arctan(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*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e)

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

*d + d^2)*f*cos(f*x + e) + (c^2 + 2*c*d + d^2)*f)*sin(f*x + e))]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/

x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)4*sqrt(2*a)*(-sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(2*f*

x+2*exp(1)-pi))/(-4*c-4*d)/(-2*d*sin(1/4*(2*f*x+2*exp(1)-pi))^2+c+d)-1/2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*at

an(sqrt(2)*d*sin(1/4*(2*f*x+2*exp(1)-pi))/sqrt(-d^2-c*d))/sqrt(2)/sqrt(-d^2-c*d)/(2*c+2*d))/f

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maple [A] time = 1.27, size = 155, normalized size = 1.48 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a d +\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a c +\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\right )}{\left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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3.527 \(\int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx\)