3.114 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{3 (5 A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}+\frac{3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

(3*(5*A - 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) +
(3*(5*A - 3*B)*Cos[e + f*x])/(32*a*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + B)*Sec[e + f*x])/(4*a*c*f*(c - c*Si
n[e + f*x])^(3/2)) - ((5*A - 3*B)*Sec[e + f*x])/(8*a*c^2*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.424661, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2967, 2859, 2687, 2650, 2649, 206} \[ -\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{3 (5 A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}+\frac{3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

(3*(5*A - 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) +
(3*(5*A - 3*B)*Cos[e + f*x])/(32*a*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + B)*Sec[e + f*x])/(4*a*c*f*(c - c*Si
n[e + f*x])^(3/2)) - ((5*A - 3*B)*Sec[e + f*x])/(8*a*c^2*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 2859

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*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

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))^{5/2}} \, dx &=\frac{\int \frac{\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(5 A-3 B) \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(3 (5 A-3 B)) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 a c}\\ &=\frac{3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(3 (5 A-3 B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{64 a c^2}\\ &=\frac{3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(3 (5 A-3 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 )}{32 a c^2 f}\\ &=\frac{3 (5 A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}+\frac{3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.884677, size = 404, normalized size = 2.24 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4+(7 A-B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (7 A-B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+8 (A+B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-(3+3 i) \sqrt [4]{-1} (5 A-3 B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{32 a f (\sin (e+f x)+1) (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(-A + B)*(Cos[(e + f*x)/2] - S
in[(e + f*x)/2])^4 + 4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (
7*A - B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (3 + 3*I)*(-1)^(1/4)*
(5*A - 3*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])^4*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]) + 8*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(7*A
- B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(32*a*f*
(1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))

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Maple [B]  time = 1.285, size = 350, normalized size = 1.9 \begin{align*} -{\frac{1}{64\,a \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( \sin \left ( fx+e \right ) \left ( -30\,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}^{2}+40\,A{c}^{5/2}+18\,B\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}^{2}-24\,B{c}^{5/2} \right ) + \left ( -15\,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}^{2}+30\,A{c}^{5/2}+9\,B\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}^{2}-18\,B{c}^{5/2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+30\,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}^{2}-24\,A{c}^{5/2}-18\,B\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}^{2}+40\,B{c}^{5/2} \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))/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/64/c^(9/2)/a*(sin(f*x+e)*(-30*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^2+40*A*c^(5/2)+18*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^2-24*B*c^(5/2))+(-15*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^2+30*A*c^(5/2)+9*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^2-1
8*B*c^(5/2))*cos(f*x+e)^2+30*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^2-24*A*c^(5/2)-18*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+40*B*c^(5/2))/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(5/2)), x)

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Fricas [A]  time = 1.51291, size = 753, normalized size = 4.18 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \,{\left (5 \, A - 3 \, 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 (3 \,{\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left (5 \, A - 3 \, B\right )} \sin \left (f x + e\right ) - 12 \, A + 20 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{128 \,{\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

-1/128*(3*sqrt(2)*((5*A - 3*B)*cos(f*x + e)^3 + 2*(5*A - 3*B)*cos(f*x + e)*sin(f*x + e) - 2*(5*A - 3*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*(5*A - 3*B)*cos(f*x + e)^2 + 4*(5*A - 3*B)*sin(f*x + e) - 12*A + 20*B
)*sqrt(-c*sin(f*x + e) + c))/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*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))/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(5/2),x, algorithm="giac")

[Out]

sage2