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

Optimal. Leaf size=126 \[ \frac{a (B c-A d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{\sqrt{a} (A d+B (c+2 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.261733, antiderivative size = 126, 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 = {2980, 2773, 208} \[ \frac{a (B c-A d) \cos (e+f x)}{d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{\sqrt{a} (A d+B (c+2 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}} \]

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])^2,x]

[Out]

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

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), 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, A
, B}, 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))^2} \, dx &=\frac{a (B c-A d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+-\frac{(-a A d-B (a c+2 a d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d (a c+a d)}\\ &=\frac{a (B c-A d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{(a (A d+B (c+2 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 (c+d) f}\\ &=-\frac{\sqrt{a} (A d+B (c+2 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} (c+d)^{3/2} f}+\frac{a (B c-A d) \cos (e+f x)}{d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [C]  time = 8.71981, size = 901, normalized size = 7.15 \[ \frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{a (\sin (e+f x)+1)} \left (\frac{(A d+B (c+2 d)) \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 )}{(c+d)^{3/2} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}+\frac{(A d+B (c+2 d)) \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 )}{(c+d)^{3/2} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}-\frac{(2-2 i) \sqrt{d} (A d-B c) \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 )}{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])^2,x]

[Out]

((1/4 + I/4)*Sqrt[a*(1 + Sin[e + f*x])]*(((A*d + B*(c + 2*d))*(Cos[e/2] + I*Sin[e/2])*((-1 + I)*x*Cos[e] + (Ro
otSum[-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]))/((c + d)^(3
/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((A*d + B*(c + 2*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)*Lo
g[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)
))/((c + d)^(3/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((2 - 2*I)*Sqrt[d]*(-(B*c) + A*d)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*f*(c + d*Sin[e + f*x]))))/(d^(3/2)*(Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]))

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Maple [B]  time = 1.881, size = 274, normalized size = 2.2 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{d \left ( c+d \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \sin \left ( fx+e \right ){\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) ad \left ( Ad+Bc+2\,Bd \right ) +A{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) acd+B{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) a{c}^{2}+2\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) acd+A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}d-B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}c \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+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x)

[Out]

-(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a
*d*(A*d+B*c+2*B*d)+A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d+B*arctanh((a-a*sin(f*x+e))^(1
/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2+2*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d+A*(a-a*sin(f*
x+e))^(1/2)*(a*(c+d)*d)^(1/2)*d-B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c)/d/(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., 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}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 10.5287, size = 2354, normalized size = 18.68 \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))^2,x, algorithm="fricas")

[Out]

[-1/4*((B*c^2 + (A + 3*B)*c*d + (A + 2*B)*d^2 - (B*c*d + (A + 2*B)*d^2)*cos(f*x + e)^2 + (B*c^2 + (A + 2*B)*c*
d)*cos(f*x + e) + (B*c^2 + (A + 3*B)*c*d + (A + 2*B)*d^2 + (B*c*d + (A + 2*B)*d^2)*cos(f*x + e))*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*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*(B*c -
A*d + (B*c - A*d)*cos(f*x + e) - (B*c - A*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^2 + d^3)*f*cos(f*x
+ e)^2 - (c^2*d + c*d^2)*f*cos(f*x + e) - (c^2*d + 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*cos(f*x + e) + (c^2*d +
2*c*d^2 + d^3)*f)*sin(f*x + e)), 1/2*((B*c^2 + (A + 3*B)*c*d + (A + 2*B)*d^2 - (B*c*d + (A + 2*B)*d^2)*cos(f*
x + e)^2 + (B*c^2 + (A + 2*B)*c*d)*cos(f*x + e) + (B*c^2 + (A + 3*B)*c*d + (A + 2*B)*d^2 + (B*c*d + (A + 2*B)*
d^2)*cos(f*x + e))*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*c - A*d + (B*c - A*d)*cos(f*x + e) - (B*c - A*d)*sin(f*x
+ e))*sqrt(a*sin(f*x + e) + a))/((c*d^2 + d^3)*f*cos(f*x + e)^2 - (c^2*d + c*d^2)*f*cos(f*x + e) - (c^2*d + 2*
c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*cos(f*x + e) + (c^2*d + 2*c*d^2 + d^3)*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+B*sin(f*x+e))*(a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**2,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+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError