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

Optimal. Leaf size=130 \[ -\frac{2 (3 A d+3 B c-2 B d) \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (A-B) (c-d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 B d \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.269964, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2968, 3023, 2751, 2649, 206} \[ -\frac{2 (3 A d+3 B c-2 B d) \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (A-B) (c-d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 B d \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.465693, size = 135, normalized size = 1.04 \[ -\frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (3 A d+3 B c+B d \sin (e+f x)-B d)-(6+6 i) (-1)^{3/4} (A-B) (c-d) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{3 f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((-6 - 6*I)*(-1)^(3/4)*(A - B)*(c - d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*
(-1 + Tan[(e + f*x)/4])] + 2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*B*c + 3*A*d - B*d + B*d*Sin[e + f*x])))/
(3*f*Sqrt[a*(1 + Sin[e + f*x])])

________________________________________________________________________________________

Maple [B]  time = 1.293, size = 232, normalized size = 1.8 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{3\,{a}^{2}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( 3\,A{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) c-3\,A{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) d-3\,B{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) c+3\,B{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) d-2\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}d+6\,Aad\sqrt{a-a\sin \left ( fx+e \right ) }+6\,Bac\sqrt{a-a\sin \left ( fx+e \right ) } \right ){\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))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x)

[Out]

-1/3*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(3*A*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)
/a^(1/2))*c-3*A*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d-3*B*a^(3/2)*2^(1/2)*arct
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c+3*B*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/
2)/a^(1/2))*d-2*B*(a-a*sin(f*x+e))^(3/2)*d+6*A*a*d*(a-a*sin(f*x+e))^(1/2)+6*B*a*c*(a-a*sin(f*x+e))^(1/2))/a^2/
cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.83923, size = 782, normalized size = 6.02 \begin{align*} \frac{\frac{3 \, \sqrt{2}{\left ({\left (A - B\right )} a c -{\left (A - B\right )} a d +{\left ({\left (A - B\right )} a c -{\left (A - B\right )} a d\right )} \cos \left (f x + e\right ) +{\left ({\left (A - B\right )} a c -{\left (A - B\right )} a d\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac{2 \, \sqrt{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 )}}{\sqrt{a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt{a}} - 4 \,{\left (B d \cos \left (f x + e\right )^{2} + 3 \, B c +{\left (3 \, A - 2 \, B\right )} d +{\left (3 \, B c +{\left (3 \, A - B\right )} d\right )} \cos \left (f x + e\right ) +{\left (B d \cos \left (f x + e\right ) - 3 \, B c -{\left (3 \, A - 2 \, B\right )} d\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{6 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(2)*((A - B)*a*c - (A - B)*a*d + ((A - B)*a*c - (A - B)*a*d)*cos(f*x + e) + ((A - B)*a*c - (A - B)*
a*d)*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)
*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x
+ e) - cos(f*x + e) - 2))/sqrt(a) - 4*(B*d*cos(f*x + e)^2 + 3*B*c + (3*A - 2*B)*d + (3*B*c + (3*A - B)*d)*cos(
f*x + e) + (B*d*cos(f*x + e) - 3*B*c - (3*A - 2*B)*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e
) + a*f*sin(f*x + e) + a*f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sin{\left (e + f x \right )}\right ) \left (c + d \sin{\left (e + f x \right )}\right )}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)

________________________________________________________________________________________

Giac [B]  time = 1.58862, size = 720, normalized size = 5.54 \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))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/3*(6*sqrt(2)*(A*c - B*c - A*d + B*d)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x
+ 1/2*e)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) + ((((3*B*a*c*sgn(tan(1/2*f*x +
1/2*e) + 1) + 3*A*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - B*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1))*tan(1/2*f*x + 1/2*e
)/a^6 - 3*(B*a*c*sgn(tan(1/2*f*x + 1/2*e) + 1) + A*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - B*a*d*sgn(tan(1/2*f*x +
1/2*e) + 1))/a^6)*tan(1/2*f*x + 1/2*e) + 3*(B*a*c*sgn(tan(1/2*f*x + 1/2*e) + 1) + A*a*d*sgn(tan(1/2*f*x + 1/2
*e) + 1) - B*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^6)*tan(1/2*f*x + 1/2*e) - (3*B*a*c*sgn(tan(1/2*f*x + 1/2*e)
+ 1) + 3*A*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - B*a*d*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^6)/(a*tan(1/2*f*x + 1/2*
e)^2 + a)^(3/2) - (6*sqrt(2)*A*a^7*c*arctan(sqrt(a)/sqrt(-a)) - 6*sqrt(2)*B*a^7*c*arctan(sqrt(a)/sqrt(-a)) - 6
*sqrt(2)*A*a^7*d*arctan(sqrt(a)/sqrt(-a)) + 6*sqrt(2)*B*a^7*d*arctan(sqrt(a)/sqrt(-a)) - 3*sqrt(2)*B*sqrt(-a)*
sqrt(a)*c - 3*sqrt(2)*A*sqrt(-a)*sqrt(a)*d + 2*sqrt(2)*B*sqrt(-a)*sqrt(a)*d)*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sq
rt(-a)*a^7))/f