3.290 \(\int \frac{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=100 \[ \frac{2 \sqrt{a} (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt{c+d}}-\frac{2 a B \cos (e+f x)}{d f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.245729, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {2981, 2773, 208} \[ \frac{2 \sqrt{a} (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt{c+d}}-\frac{2 a B \cos (e+f x)}{d f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

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

Mathematica [C]  time = 8.78459, size = 903, normalized size = 9.03 \[ \frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \left (-\frac{(2-2 i) B \sqrt{d} \cos \left (\frac{f x}{2}\right ) \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right )}{f}+\frac{(A d-B c) \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{c+d} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}+\frac{(A d-B c) \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{c+d} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}+\frac{(2-2 i) B \sqrt{d} \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) \sin \left (\frac{f x}{2}\right )}{f}\right ) \sqrt{a (\sin (e+f x)+1)}}{d^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/2 + I/2)*(((-2 + 2*I)*B*Sqrt[d]*Cos[(f*x)/2]*(Cos[e/2] - Sin[e/2]))/f + ((-(B*c) + A*d)*(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]*Sqr
t[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[c + d]*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((-(B*c) + A*d)*(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[c + d]*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((2 - 2*I)*B*Sqrt
[d]*(Cos[e/2] + Sin[e/2])*Sin[(f*x)/2])/f)*Sqrt[a*(1 + Sin[e + f*x])])/(d^(3/2)*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2]))

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Maple [A]  time = 1.545, size = 139, normalized size = 1.4 \begin{align*} -2\,{\frac{ \left ( 1+\sin \left ( fx+e \right ) \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{d\sqrt{a \left ( c+d \right ) d}\cos \left ( fx+e \right ) \sqrt{a+a\sin \left ( fx+e \right ) }f} \left ( A{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) ad-B{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) ac+B\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( c+d \right ) d} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(A*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a*d-B*
arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a*c+B*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2))/d/
(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{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)/(d*sin(f*x + e) + c), x)

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Fricas [A]  time = 9.3318, size = 1539, normalized size = 15.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((B*c - A*d + (B*c - A*d)*cos(f*x + e) + (B*c - A*d)*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)*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*c
os(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*(B*cos(f*x + e) - B*sin(f*x + e) + B)*sqrt(a*
sin(f*x + e) + a))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), ((B*c - A*d + (B*c - A*d)*cos(f*x + e) + (B*c
- A*d)*sin(f*x + e))*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*(B*cos(f*x + e) - B*sin(f*x + e) + B)*sqrt(a*sin(f*x + e) + a))/(d*f*cos
(f*x + e) + d*f*sin(f*x + e) + d*f)]

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

[Out]

Timed out

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

[Out]

Timed out