3.330 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}} \]

[Out]

(5*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) + (5*Cos[e +
f*x])/(8*a^2*f*(c - c*Sin[e + f*x])^(3/2)) - (5*Sec[e + f*x])/(6*a^2*c*f*Sqrt[c - c*Sin[e + f*x]]) - (Sec[e +
f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*c^2*f)

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Rubi [A]  time = 0.249886, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2736, 2675, 2687, 2650, 2649, 206} \[ -\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

(5*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) + (5*Cos[e +
f*x])/(8*a^2*f*(c - c*Sin[e + f*x])^(3/2)) - (5*Sec[e + f*x])/(6*a^2*c*f*Sqrt[c - c*Sin[e + f*x]]) - (Sec[e +
f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*c^2*f)

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 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 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{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac{\int \sec ^4(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=-\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{5 \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{6 a^2 c}\\ &=-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{5 \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{5 \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 a^2 c}\\ &=\frac{5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}-\frac{5 \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 a^2 c f}\\ &=\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}\\ \end{align*}

Mathematica [C]  time = 0.785651, size = 164, normalized size = 1.06 \[ \frac{\left (\frac{1}{96}+\frac{i}{96}\right ) \cos (e+f x) \left ((1-i) (-20 \sin (e+f x)+15 \cos (2 (e+f x))+11)+60 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{a^2 c f (\sin (e+f x)-1) (\sin (e+f x)+1)^2 \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

((1/96 + I/96)*Cos[e + f*x]*(60*(-1)^(1/4)*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])^3 + (1 - I)*(11 + 15*Cos[2*(e + f*x)] - 20*Si
n[e + f*x])))/(a^2*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.569, size = 157, normalized size = 1. \begin{align*} -{\frac{1}{48\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 15\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\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-15\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c-20\,{c}^{5/2}\sin \left ( fx+e \right ) -30\,{c}^{5/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+26\,{c}^{5/2} \right ){c}^{-{\frac{7}{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(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x)

[Out]

-1/48/c^(7/2)/a^2*(15*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*s
in(f*x+e)*c-15*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c-20*c^(
5/2)*sin(f*x+e)-30*c^(5/2)*sin(f*x+e)^2+26*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(-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(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.1494, size = 512, normalized size = 3.3 \begin{align*} \frac{15 \, \sqrt{2} \sqrt{c} \cos \left (f x + e\right )^{3} \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 \, \cos \left (f x + e\right )^{2} - 10 \, \sin \left (f x + e\right ) - 2\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/96*(15*sqrt(2)*sqrt(c)*cos(f*x + e)^3*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*cos(f*x + e)^2 - 10*sin(f*x + e) - 2)*sqrt(-
c*sin(f*x + e) + c))/(a^2*c^2*f*cos(f*x + e)^3)

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

[Out]

sage2