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

Optimal. Leaf size=174 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f} \]

[Out]

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

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Rubi [A]  time = 0.437845, antiderivative size = 174, normalized size of antiderivative = 1., 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, 2855, 2675, 2649, 206} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2675

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

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))^3 \sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{2 a^3 c^2}\\ &=-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \sec ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{4 a^3 c}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 a^3}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(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^3 f}\\ &=\frac{(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^3 \sqrt{c} f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}\\ \end{align*}

Mathematica [C]  time = 0.785356, size = 204, normalized size = 1.17 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-15 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4-10 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(-15-15 i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+12 (B-A)\right )}{60 a^3 f (\sin (e+f x)+1)^3 \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(12*(-A + B) - 10*(A + B)*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^2 - 15*(A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (15 + 15*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])^5))/(60*a
^3*f*(1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 1.209, size = 200, normalized size = 1.2 \begin{align*}{\frac{-1+\sin \left ( fx+e \right ) }{120\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 74\,{c}^{9/2}A+80\,A{c}^{9/2}\sin \left ( fx+e \right ) +30\,A{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{2}A+26\,{c}^{9/2}B+80\,B{c}^{9/2}\sin \left ( fx+e \right ) +30\,B{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{2}B \right ){c}^{-{\frac{9}{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))^3/(c-c*sin(f*x+e))^(1/2),x)

[Out]

1/120/a^3*(-1+sin(f*x+e))/c^(9/2)/(1+sin(f*x+e))^2*(74*c^(9/2)*A+80*A*c^(9/2)*sin(f*x+e)+30*A*c^(9/2)*sin(f*x+
e)^2-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*c^2*A+26*c^(9/2
)*B+80*B*c^(9/2)*sin(f*x+e)+30*B*c^(9/2)*sin(f*x+e)^2-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/
c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*c^2*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [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))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.76415, size = 729, normalized size = 4.19 \begin{align*} \frac{15 \, \sqrt{2}{\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \,{\left (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 (15 \,{\left (A + B\right )} \cos \left (f x + e\right )^{2} - 40 \,{\left (A + B\right )} \sin \left (f x + e\right ) - 52 \, A - 28 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{240 \,{\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c 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))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/240*(15*sqrt(2)*((A + B)*cos(f*x + e)^3 - 2*(A + B)*cos(f*x + e)*sin(f*x + e) - 2*(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*(15*(A + B)*cos(f*x + e)^2 - 40*(A + B)*sin(f*x + e) - 52*A - 28*B)*sqrt(-c*sin(f*x
+ e) + c))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(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))**3/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.14438, size = 1511, normalized size = 8.68 \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))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/60*((870*sqrt(2)*A*c*arctan(sqrt(c)/sqrt(-c)) + 870*sqrt(2)*B*c*arctan(sqrt(c)/sqrt(-c)) - 1230*A*c*arctan(s
qrt(c)/sqrt(-c)) - 1230*B*c*arctan(sqrt(c)/sqrt(-c)) - 850*sqrt(2)*A*sqrt(-c)*sqrt(c) - 40*sqrt(2)*B*sqrt(-c)*
sqrt(c) + 1203*A*sqrt(-c)*sqrt(c) + 57*B*sqrt(-c)*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(41*sqrt(2)*a^3*sqrt(
-c)*c - 58*a^3*sqrt(-c)*c) + 15*sqrt(2)*(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^3*sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 2*(105*(sqrt(c)*t
an(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*A - 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(
1/2*f*x + 1/2*e)^2 + c))^9*B + 435*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*A*sqr
t(c) + 75*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*B*sqrt(c) + 580*(sqrt(c)*tan(1
/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*A*c + 100*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1
/2*f*x + 1/2*e)^2 + c))^7*B*c - 620*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*A*c^
(3/2) - 140*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*B*c^(3/2) - 1258*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A*c^2 - 442*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))^5*B*c^2 + 490*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))
^4*A*c^(5/2) + 250*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*B*c^(5/2) + 900*(sqrt
(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c^3 + 420*(sqrt(c)*tan(1/2*f*x + 1/2*e) - s
qrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^3 - 860*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^2*A*c^(7/2) - 380*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*c^(7/2) + 265
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^4 + 145*(sqrt(c)*tan(1/2*f*x + 1/2*e)
- sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^4 - 37*A*c^(9/2) - 13*B*c^(9/2))/(((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)^5*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f