3.311 \(\int \frac{(a+a \sin (e+f x))^3}{\sqrt{c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=151 \[ -\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{8 a^3 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\frac{8 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f} \]

[Out]

(8*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^3*c^2*Co
s[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) - (4*a^3*c*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (
8*a^3*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.315871, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2736, 2679, 2649, 206} \[ -\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{8 a^3 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\frac{8 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^3*c^2*Co
s[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) - (4*a^3*c*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (
8*a^3*Cos[e + f*x])/(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 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}{\sqrt{c-c \sin (e+f x)}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=-\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}+\left (2 a^3 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=-\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (4 a^3 c\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{8 a^3 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\left (8 a^3\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{8 a^3 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}-\frac{\left (16 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 )}{f}\\ &=\frac{8 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac{4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{8 a^3 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.723938, size = 156, normalized size = 1.03 \[ -\frac{a^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (330 \sin \left (\frac{1}{2} (e+f x)\right )+35 \sin \left (\frac{3}{2} (e+f x)\right )-3 \sin \left (\frac{5}{2} (e+f x)\right )+330 \cos \left (\frac{1}{2} (e+f x)\right )-35 \cos \left (\frac{3}{2} (e+f x)\right )-3 \cos \left (\frac{5}{2} (e+f x)\right )+(480+480 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right )\right )}{30 f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*((480 + 480*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(
e + f*x)/4])] + 330*Cos[(e + f*x)/2] - 35*Cos[(3*(e + f*x))/2] - 3*Cos[(5*(e + f*x))/2] + 330*Sin[(e + f*x)/2]
+ 35*Sin[(3*(e + f*x))/2] - 3*Sin[(5*(e + f*x))/2]))/(30*f*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.767, size = 129, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){a}^{3}}{15\,{c}^{3}\cos \left ( fx+e \right ) f}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \left ( 60\,{c}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) -3\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}-10\,c \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}-60\,{c}^{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \right ){\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))^(1/2),x)

[Out]

-2/15*(-1+sin(f*x+e))*(c*(1+sin(f*x+e)))^(1/2)*a^3*(60*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^
(1/2)/c^(1/2))-3*(c*(1+sin(f*x+e)))^(5/2)-10*c*(c*(1+sin(f*x+e)))^(3/2)-60*c^2*(c*(1+sin(f*x+e)))^(1/2))/c^3/c
os(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}}{\sqrt{-c \sin \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.15413, size = 707, normalized size = 4.68 \begin{align*} \frac{2 \,{\left (\frac{30 \, \sqrt{2}{\left (a^{3} c \cos \left (f x + e\right ) - a^{3} c \sin \left (f x + e\right ) + a^{3} c\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}} +{\left (3 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 76 \, a^{3} \cos \left (f x + e\right ) - 92 \, a^{3} +{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 16 \, a^{3} \cos \left (f x + e\right ) - 92 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}\right )}}{15 \,{\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\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))^(1/2),x, algorithm="fricas")

[Out]

2/15*(30*sqrt(2)*(a^3*c*cos(f*x + e) - a^3*c*sin(f*x + e) + a^3*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*s
in(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) + (3*a^3*cos(f*x + e)^3 + 1
9*a^3*cos(f*x + e)^2 - 76*a^3*cos(f*x + e) - 92*a^3 + (3*a^3*cos(f*x + e)^2 - 16*a^3*cos(f*x + e) - 92*a^3)*si
n(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

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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))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.52584, size = 466, normalized size = 3.09 \begin{align*} \frac{\frac{960 \, \sqrt{2} a^{3} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{{\left ({\left ({\left ({\left (\frac{73 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{c^{7}} + \frac{105 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{190 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{190 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{105 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{73 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{7}}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c\right )}^{\frac{5}{2}}} - \frac{4 \, \sqrt{2}{\left (240 \, a^{3} c^{\frac{21}{2}} \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + 23 \, a^{3} \sqrt{-c} c\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{-c} c^{\frac{21}{2}}}}{60 \, f} \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))^(1/2),x, algorithm="giac")

[Out]

1/60*(960*sqrt(2)*a^3*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) -
sqrt(c))/sqrt(-c))/(sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + (((((73*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1
/2*f*x + 1/2*e)/c^7 + 105*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^7)*tan(1/2*f*x + 1/2*e) + 190*a^3*sgn(tan(1/2*f*
x + 1/2*e) - 1)/c^7)*tan(1/2*f*x + 1/2*e) + 190*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^7)*tan(1/2*f*x + 1/2*e) +
105*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^7)*tan(1/2*f*x + 1/2*e) + 73*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^7)/(c
*tan(1/2*f*x + 1/2*e)^2 + c)^(5/2) - 4*sqrt(2)*(240*a^3*c^(21/2)*arctan(sqrt(c)/sqrt(-c)) + 23*a^3*sqrt(-c)*c)
*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(-c)*c^(21/2)))/f