3.113 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=136 \[ \frac{(3 A-B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}+\frac{(3 A-B) \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B) \sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.330427, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {2967, 2855, 2650, 2649, 206} \[ \frac{(3 A-B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}+\frac{(3 A-B) \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B) \sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

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

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]))

Rule 2855

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

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

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

Mathematica [C]  time = 0.561778, size = 284, normalized size = 2.09 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 (B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+(A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+2 (A+B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-(1+i) \sqrt [4]{-1} (3 A-B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2\right )}{4 a f (\sin (e+f x)+1) (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(-A + B)*(Cos[(e + f*x)/2] - S
in[(e + f*x)/2])^2 + (A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (1
+ I)*(-1)^(1/4)*(3*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])^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2])))/(4*a*f*(1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.98, size = 225, normalized size = 1.7 \begin{align*} -{\frac{1}{8\,af\cos \left ( fx+e \right ) } \left ( \sin \left ( fx+e \right ) \left ( 3\,A\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c-B\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c+c\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ) c-6\,A{c}^{3/2}+2\,B{c}^{3/2} \right ) -3\,A\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c+B\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c+c\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ) c+2\,A{c}^{3/2}-6\,B{c}^{3/2} \right ){c}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{c-c\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))/(c-c*sin(f*x+e))^(3/2),x)

[Out]

-1/8/c^(5/2)/a*(sin(f*x+e)*(3*A*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1
/2))*c-B*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c-6*A*c^(3/2)+2*B*
c^(3/2))-3*A*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c+B*(c+c*sin(f
*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c+2*A*c^(3/2)-6*B*c^(3/2))/cos(f*x+e)
/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{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))/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(3/2)), x)

________________________________________________________________________________________

Fricas [A]  time = 1.55292, size = 608, normalized size = 4.47 \begin{align*} -\frac{\sqrt{2}{\left ({\left (3 \, A - B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (3 \, A - B\right )} \cos \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\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 ) + 4 \,{\left ({\left (3 \, A - B\right )} \sin \left (f x + e\right ) - A + 3 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{16 \,{\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}} \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))/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/16*(sqrt(2)*((3*A - B)*cos(f*x + e)*sin(f*x + e) - (3*A - B)*cos(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 -
2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x
+ e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*((
3*A - B)*sin(f*x + e) - A + 3*B)*sqrt(-c*sin(f*x + e) + c))/(a*c^2*f*cos(f*x + e)*sin(f*x + e) - a*c^2*f*cos(f
*x + e))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 2.40715, size = 878, normalized size = 6.46 \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))/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

1/4*(sqrt(2)*(3*A - B)*arctan(-1/2*sqrt(2)*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
- sqrt(c))/sqrt(-c))/(a*sqrt(-c)*c*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 4*((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))*A - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B - A*sqr
t(c) + B*sqrt(c))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2
*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 2*(3*(sq
rt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A + 3*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(
c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*s
qrt(c) - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*sqrt(c) - (sqrt(c)*tan(1/2*f*
x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/
2*e)^2 + c))*B*c - A*c^(3/2) - B*c^(3/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)^2 - 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^2*a*c*sgn(tan(1/2*f*x
+ 1/2*e) - 1)))/f