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

Optimal. Leaf size=150 \[ \frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}-\frac{10 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac{10 a^3 \cos (e+f x)}{c f \sqrt{c-c \sin (e+f x)}} \]

[Out]

(-10*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) + (a^3*c^2*Co
s[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^(7/2)) + (5*a^3*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) + (10*a
^3*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.320458, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2736, 2680, 2679, 2649, 206} \[ \frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}-\frac{10 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac{10 a^3 \cos (e+f x)}{c f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) + (a^3*c^2*Co
s[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^(7/2)) + (5*a^3*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) + (10*a
^3*Cos[e + f*x])/(c*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 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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), 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]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 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+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{3/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}-\frac{1}{2} \left (5 a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\left (5 a^3\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac{10 a^3 \cos (e+f x)}{c f \sqrt{c-c \sin (e+f x)}}-\frac{\left (10 a^3\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac{10 a^3 \cos (e+f x)}{c f \sqrt{c-c \sin (e+f x)}}+\frac{\left (20 a^3\right ) \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 )}{c f}\\ &=-\frac{10 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac{a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac{10 a^3 \cos (e+f x)}{c f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.798989, size = 173, normalized size = 1.15 \[ -\frac{a^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (50 \sin \left (\frac{1}{2} (e+f x)\right )-25 \sin \left (\frac{3}{2} (e+f x)\right )+\sin \left (\frac{5}{2} (e+f x)\right )+50 \cos \left (\frac{1}{2} (e+f x)\right )+25 \cos \left (\frac{3}{2} (e+f x)\right )+\cos \left (\frac{5}{2} (e+f x)\right )-(120+120 i) \sqrt [4]{-1} (\sin (e+f x)-1) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right )\right )}{6 c f (\sin (e+f x)-1) \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(50*Cos[(e + f*x)/2] + 25*Cos[(3*(e + f*x))/2] + Cos[(5*(e + f*x))
/2] + 50*Sin[(e + f*x)/2] - (120 + 120*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-1
+ Sin[e + f*x]) - 25*Sin[(3*(e + f*x))/2] + Sin[(5*(e + f*x))/2]))/(6*c*f*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[
e + f*x]])

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Maple [A]  time = 0.599, size = 189, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{3}}{3\,f\cos \left ( fx+e \right ) } \left ( 15\,\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}-12\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2}\sin \left ( fx+e \right ) - \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}\sqrt{c}\sin \left ( fx+e \right ) -15\,\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}+18\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2}+ \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}\sqrt{c} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \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((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x)

[Out]

2/3*a^3*(15*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^2-12*(c*(1+sin(f*x+e)))
^(1/2)*c^(3/2)*sin(f*x+e)-(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)*sin(f*x+e)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e))
)^(1/2)*2^(1/2)/c^(1/2))*c^2+18*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)+(c*(1+sin(f*x+e)))^(3/2)*c^(1/2))*(c*(1+sin(f
*x+e)))^(1/2)/c^(7/2)/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{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [B]  time = 1.12119, size = 844, normalized size = 5.63 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left (a^{3} c \cos \left (f x + e\right )^{2} - a^{3} c \cos \left (f x + e\right ) - 2 \, a^{3} c +{\left (a^{3} c \cos \left (f x + e\right ) + 2 \, a^{3} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac{2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt{c}} + 3 \, \cos \left (f x + e\right ) + 2}{\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 )}{\sqrt{c}} - 2 \,{\left (a^{3} \cos \left (f x + e\right )^{3} + 13 \, a^{3} \cos \left (f x + e\right )^{2} + 18 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 12 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} 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))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/3*(15*sqrt(2)*(a^3*c*cos(f*x + e)^2 - a^3*c*cos(f*x + e) - 2*a^3*c + (a^3*c*cos(f*x + e) + 2*a^3*c)*sin(f*x
+ e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x +
e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f
*x + e) - 2))/sqrt(c) - 2*(a^3*cos(f*x + e)^3 + 13*a^3*cos(f*x + e)^2 + 18*a^3*cos(f*x + e) + 6*a^3 + (a^3*cos
(f*x + e)^2 - 12*a^3*cos(f*x + e) + 6*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^
2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*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))**3/(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((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

sage2