3.528 \(\int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=154 \[ -\frac{3 a \cos (e+f x)}{4 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a \cos (e+f x)}{2 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^2}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 \sqrt{d} f (c+d)^{5/2}} \]

[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]))

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Rubi [A]  time = 0.269156, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2772, 2773, 208} \[ -\frac{3 a \cos (e+f x)}{4 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a \cos (e+f x)}{2 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^2}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 \sqrt{d} f (c+d)^{5/2}} \]

Antiderivative was successfully verified.

[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 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]

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx &=-\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) \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 )}{4 (c+d)^2 f}\\ &=-\frac{3 \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 )}{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]  time = 7.24634, size = 920, normalized size = 5.97 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{a (\sin (e+f x)+1)} \left (\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 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)^{5/2} (\cos (e)+i (\sin (e)-1)) \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 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)^{5/2} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}-\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))}-\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}\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])^3,x]

[Out]

((1/16 + I/16)*Sqrt[a*(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)]*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)^(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])

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Maple [A]  time = 1., size = 254, normalized size = 1.7 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{4\, \left ( c+d \right ) ^{2} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( 3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}a{d}^{2}+6\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) \sin \left ( fx+e \right ) acd+3\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d}\sin \left ( fx+e \right ) d+3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) a{c}^{2}+5\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d}c+2\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d}d \right ){\frac{1}{\sqrt{a \left ( c+d \right ) d}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

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))^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sin \left (f x + e\right ) + a}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}

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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.13945, size = 2890, normalized size = 18.77 \begin{align*} \text{result too large to display} \end{align*}

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))^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))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**3,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError