3.331 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=192 \[ -\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

(35*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) + (35*Cos[e
+ f*x])/(64*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) + (7*Sec[e + f*x])/(24*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) -
(35*Sec[e + f*x])/(48*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - Sec[e + f*x]^3/(3*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x
]])

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Rubi [A]  time = 0.330746, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2736, 2687, 2681, 2650, 2649, 206} \[ -\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

(35*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) + (35*Cos[e
+ f*x])/(64*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) + (7*Sec[e + f*x])/(24*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) -
(35*Sec[e + f*x])/(48*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - Sec[e + f*x]^3/(3*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x
]])

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 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 2681

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*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*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{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx &=\frac{\int \frac{\sec ^4(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{a^2 c^2}\\ &=-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{7 \int \frac{\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{6 a^2 c}\\ &=\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{35 \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{48 a^2 c^2}\\ &=\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{35 \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{32 a^2 c}\\ &=\frac{35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{35 \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{128 a^2 c^2}\\ &=\frac{35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{35 \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 )}{64 a^2 c^2 f}\\ &=\frac{35 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{35 \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.1363, size = 156, normalized size = 0.81 \[ -\frac{\left (\frac{1}{1536}+\frac{i}{1536}\right ) \sec ^3(e+f x) \left ((1-i) (-329 \sin (e+f x)-105 \sin (3 (e+f x))+70 \cos (2 (e+f x))+102)+840 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{a^2 c^2 f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

((-1/1536 - I/1536)*Sec[e + f*x]^3*(840*(-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])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 - I)*(102 + 70*Cos[2*(e + f*x)
] - 329*Sin[e + f*x] - 105*Sin[3*(e + f*x)])))/(a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.728, size = 233, normalized size = 1.2 \begin{align*} -{\frac{1}{384\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 105\, \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 ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}+70\,{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-210\,{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}-210\, \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}^{2}+322\,{c}^{7/2}\sin \left ( fx+e \right ) +105\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2} \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}-86\,{c}^{7/2} \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(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/384/c^(11/2)/a^2*(105*(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)
)*sin(f*x+e)^2*c^2+70*c^(7/2)*sin(f*x+e)^2-210*c^(7/2)*sin(f*x+e)^3-210*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arcta
nh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2+322*c^(7/2)*sin(f*x+e)+105*2^(1/2)*arctanh(1/2
*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2*(c*(1+sin(f*x+e)))^(3/2)-86*c^(7/2))/(1+sin(f*x+e))/(-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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.24951, size = 643, normalized size = 3.35 \begin{align*} \frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - \cos \left (f x + e\right )^{3}\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 (35 \, \cos \left (f x + e\right )^{2} - 7 \,{\left (15 \, \cos \left (f x + e\right )^{2} + 8\right )} \sin \left (f x + e\right ) + 8\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{768 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) - cos(f*x + e)^3)*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*s
qrt(-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*(35*cos(f*x +
e)^2 - 7*(15*cos(f*x + e)^2 + 8)*sin(f*x + e) + 8)*sqrt(-c*sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*
x + e) - a^2*c^3*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))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 3.08634, size = 1083, normalized size = 5.64 \begin{align*} \text{result too large to display} \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))^(5/2),x, algorithm="giac")

[Out]

1/192*(105*sqrt(2)*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) - sq
rt(c))/sqrt(-c))/(a^2*sqrt(-c)*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 16*(15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^5 + 33*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*
sqrt(c) - 22*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c - 66*(sqrt(c)*tan(1/2*f*x
+ 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(3/2) + 51*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*
f*x + 1/2*e)^2 + c))*c^2 - 11*c^(5/2))/(((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)^3*a^2*c^2*sgn(tan(1/2*f*
x + 1/2*e) - 1)) + 6*(53*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7 - 179*(sqrt(c)*
tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*sqrt(c) + 127*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c + 195*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))
^4*c^(3/2) + 7*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^2 - 121*(sqrt(c)*tan(1/
2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(5/2) - 67*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan
(1/2*f*x + 1/2*e)^2 + c))*c^3 - 15*c^(7/2))/(((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)^4*a^2*c^2*sgn(tan(1
/2*f*x + 1/2*e) - 1)))/f