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

Optimal. Leaf size=113 \[ -\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}-\frac{a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

-(a*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) + (a*Cos[e + f*x
])/(2*f*(c - c*Sin[e + f*x])^(5/2)) - (a*Cos[e + f*x])/(8*c*f*(c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.163481, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2680, 2650, 2649, 206} \[ -\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}-\frac{a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) + (a*Cos[e + f*x
])/(2*f*(c - c*Sin[e + f*x])^(5/2)) - (a*Cos[e + f*x])/(8*c*f*(c - c*Sin[e + f*x])^(3/2))

Rule 2736

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 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+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c}\\ &=\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}-\frac{a \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 c^2}\\ &=\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{a \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 )}{8 c^2 f}\\ &=-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}+\frac{a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.911654, size = 176, normalized size = 1.56 \[ \frac{a \left (2 \sqrt{2} \sqrt{-c (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )-2 \sqrt{c} (-8 \sin (e+f x)+\cos (2 (e+f x))-7)\right )}{32 c^{5/2} f \sqrt{c-c \sin (e+f x)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-2*Sqrt[c]*(-7 + Cos[2*(e + f*x)] - 8*Sin[e + f*x]) + 2*Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt
[2]*Sqrt[c])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(32*c^(5/2)*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.714, size = 189, normalized size = 1.7 \begin{align*} -{\frac{a}{ \left ( -16+16\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{3}+2\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{3/2}+2\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{3}+4\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{5/2}-\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{3} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{11}{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+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/16/c^(11/2)*a*(-2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^3+2*(c*(1+sin(
f*x+e)))^(3/2)*c^(3/2)+2*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^3+4*(c*(1+
sin(f*x+e)))^(1/2)*c^(5/2)-2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3)*(c*(1+sin(f*x+e)
))^(1/2)/(-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{a \sin \left (f x + e\right ) + a}{{\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+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.14269, size = 878, normalized size = 7.77 \begin{align*} \frac{\sqrt{2}{\left (a \cos \left (f x + e\right )^{3} + 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) - 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\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 (a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) + 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{32 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*(a*cos(f*x + e)^3 + 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e) - (a*cos(f*x + e)^2 - 2*a*cos(f*x + e)
- 4*a)*sin(f*x + e) - 4*a)*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*(a*cos(f*x + e)^2 - 3*a*cos(f*x + e) - (a*cos(f*x +
e) + 4*a)*sin(f*x + e) - 4*a)*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^
3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(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+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+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

sage2