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

Optimal. Leaf size=161 \[ -\frac {2 a^2 c (A+B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\frac {4 \sqrt {2} a^2 (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

-2/5*a^2*B*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)-2/3*a^2*(A+B)*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+4*a
^2*(A+B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/2)/f/c^(1/2)-4*a^2*(A+B)*cos(f*x+
e)/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2967, 2860, 2679, 2649, 206} \[ -\frac {2 a^2 c (A+B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\frac {4 \sqrt {2} a^2 (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^2*(A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^
2*B*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) - (2*a^2*(A + B)*c*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e
+ f*x])^(3/2)) - (4*a^2*(A + B)*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

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])

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2860

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}+\left (a^2 (A+B) c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (2 a^2 (A+B) c\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\left (4 a^2 (A+B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {\left (8 a^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}\\ &=\frac {4 \sqrt {2} a^2 (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.18, size = 175, normalized size = 1.09 \[ -\frac {a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (2 (5 A+11 B) \sin (e+f x)+70 A-3 B \cos (2 (e+f x))+79 B)+(120+120 i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right )\right )}{15 f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*((120 + 120*I)*(-1)^(1/4)*(A + B)*ArcTan
[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])] + (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(70*A + 79*B - 3*B*Cos
[2*(e + f*x)] + 2*(5*A + 11*B)*Sin[e + f*x])))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c - c*Sin[e + f
*x]])

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fricas [B]  time = 0.45, size = 310, normalized size = 1.93 \[ \frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} c \cos \left (f x + e\right ) - {\left (A + B\right )} a^{2} c \sin \left (f x + e\right ) + {\left (A + B\right )} a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{3} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - {\left (35 \, A + 41 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - {\left (5 \, A + 11 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{15 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(2)*((A + B)*a^2*c*cos(f*x + e) - (A + B)*a^2*c*sin(f*x + e) + (A + B)*a^2*c)*log(-(cos(f*x + e)^
2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sq
rt(c) + 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(c) + (
3*B*a^2*cos(f*x + e)^3 + (5*A + 14*B)*a^2*cos(f*x + e)^2 - (35*A + 41*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)*a^
2 + (3*B*a^2*cos(f*x + e)^2 - (5*A + 11*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)*a^2)*sin(f*x + e))*sqrt(-c*sin(f
*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

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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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),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)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f*(2/sqrt(c*tan((f*x+exp(1))/2)^2+c)/(c*tan((f*x+exp(1))/2)^2+c)^2*(tan((f*x+exp(1
))/2)*(tan((f*x+exp(1))/2)*(tan((f*x+exp(1))/2)*(tan((f*x+exp(1))/2)*(1/3600*tan((f*x+exp(1))/2)*(4200*A*a^2*c
^2*sign(tan((f*x+exp(1))/2)-1)+4560*B*a^2*c^2*sign(tan((f*x+exp(1))/2)-1))+1/3600*(5400*A*a^2*c^2*sign(tan((f*
x+exp(1))/2)-1)+7200*B*a^2*c^2*sign(tan((f*x+exp(1))/2)-1)))+1/3600*(9600*A*a^2*c^2*sign(tan((f*x+exp(1))/2)-1
)+13200*B*a^2*c^2*sign(tan((f*x+exp(1))/2)-1)))+1/3600*(9600*A*a^2*c^2*sign(tan((f*x+exp(1))/2)-1)+13200*B*a^2
*c^2*sign(tan((f*x+exp(1))/2)-1)))+1/3600*(5400*A*a^2*c^2*sign(tan((f*x+exp(1))/2)-1)+7200*B*a^2*c^2*sign(tan(
(f*x+exp(1))/2)-1)))+1/3600*(4200*A*a^2*c^2*sign(tan((f*x+exp(1))/2)-1)+4560*B*a^2*c^2*sign(tan((f*x+exp(1))/2
)-1)))+sqrt(2)*(4*A*a^2+4*B*a^2)*atan((-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))/s
qrt(2)/sqrt(-c))/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1)+(-60*A*a^2*c*sqrt(2)*atan(sqrt(c)/sqrt(-c))-40*A*a^2*sqr
t(-c)*sqrt(2)*sqrt(c)-60*B*a^2*c*sqrt(2)*atan(sqrt(c)/sqrt(-c))-52*B*a^2*sqrt(-c)*sqrt(2)*sqrt(c))/15/c/sqrt(-
c)*sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 1.49, size = 197, normalized size = 1.22 \[ \frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (-30 c^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) A -30 c^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) B +3 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}+5 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c +5 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c +30 A \,c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}+30 B \,c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(sin(f*x+e)-1)*(c*(1+sin(f*x+e)))^(1/2)*a^2*(-30*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(
1/2)/c^(1/2))*A-30*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*B+3*B*(c*(1+sin(f*x+e
)))^(5/2)+5*A*(c*(1+sin(f*x+e)))^(3/2)*c+5*B*(c*(1+sin(f*x+e)))^(3/2)*c+30*A*c^2*(c*(1+sin(f*x+e)))^(1/2)+30*B
*c^2*(c*(1+sin(f*x+e)))^(1/2))/c^3/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {A}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 A \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {A \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 B \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin ^{3}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A/sqrt(-c*sin(e + f*x) + c), x) + Integral(2*A*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x) + Int
egral(A*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x)
+ Integral(2*B*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*sin(e + f*x)**3/sqrt(-c*sin(e + f*x)
+ c), x))

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