3.315 \(\int \frac{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=191 \[ \frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} c^{9/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}} \]

[Out]

(-5*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*c^(9/2)*f) + (a^3*c^2
*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(13/2)) - (5*a^3*Cos[e + f*x]^3)/(24*f*(c - c*Sin[e + f*x])^(9/2))
+ (5*a^3*Cos[e + f*x])/(32*c^2*f*(c - c*Sin[e + f*x])^(5/2)) - (5*a^3*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e +
f*x])^(3/2))

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Rubi [A]  time = 0.353228, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2736, 2680, 2650, 2649, 206} \[ \frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} c^{9/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(-5*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*c^(9/2)*f) + (a^3*c^2
*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(13/2)) - (5*a^3*Cos[e + f*x]^3)/(24*f*(c - c*Sin[e + f*x])^(9/2))
+ (5*a^3*Cos[e + f*x])/(32*c^2*f*(c - c*Sin[e + f*x])^(5/2)) - (5*a^3*Cos[e + f*x])/(128*c^3*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))^3}{(c-c \sin (e+f x))^{9/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{15/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{1}{8} \left (5 a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{\left (5 a^3\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx}{16 c}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{\left (5 a^3\right ) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{64 c^3}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (5 a^3\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{256 c^4}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (5 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 )}{128 c^4 f}\\ &=-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} c^{9/2} f}+\frac{a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac{5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac{5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 2.57538, size = 371, normalized size = 1.94 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (768 \sin \left (\frac{1}{2} (e+f x)\right )-15 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7-30 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+236 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+472 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-544 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-1088 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+384 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8\right )}{384 f (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(384*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 544*(Cos[(e + f*x)/2]
- Sin[(e + f*x)/2])^3 + 236*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - 15*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
^7 + (15 + 15*I)*(-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])^8 + 768*Sin[(e + f*x)/2] - 1088*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 472*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 30*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)
/2])*(1 + Sin[e + f*x])^3)/(384*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(9/2))

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Maple [A]  time = 0.991, size = 299, normalized size = 1.6 \begin{align*} -{\frac{{a}^{3}}{768\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) f} \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 ) \left ( \sin \left ( fx+e \right ) \right ) ^{4}{c}^{5}+30\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{7/2}{c}^{3/2}+60\,\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 ) ^{3}{c}^{5}+292\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{5/2}-90\,\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}^{5}-440\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{7/2}+60\,\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}^{5}+240\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{9/2}-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}^{5} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{19}{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))^(9/2),x)

[Out]

-1/768/c^(19/2)*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)^4*c^5+30*(c*
(1+sin(f*x+e)))^(7/2)*c^(3/2)+60*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^
5+292*(c*(1+sin(f*x+e)))^(5/2)*c^(5/2)-90*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^5-440*(c*(1+sin(f*x+e)))^(3/2)*c^(7/2)+60*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^5+240*(c*(1+sin(f*x+e)))^(1/2)*c^(9/2)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)
/c^(1/2))*c^5)*(c*(1+sin(f*x+e)))^(1/2)/(-1+sin(f*x+e))^3/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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(9/2), x)

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Fricas [B]  time = 1.23502, size = 1338, normalized size = 7.01 \begin{align*} \frac{15 \, \sqrt{2}{\left (a^{3} \cos \left (f x + e\right )^{5} + 5 \, a^{3} \cos \left (f x + e\right )^{4} - 8 \, a^{3} \cos \left (f x + e\right )^{3} - 20 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3} -{\left (a^{3} \cos \left (f x + e\right )^{4} - 4 \, a^{3} \cos \left (f x + e\right )^{3} - 12 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3}\right )} \sin \left (f x + e\right )\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 (15 \, a^{3} \cos \left (f x + e\right )^{4} - 191 \, a^{3} \cos \left (f x + e\right )^{3} - 338 \, a^{3} \cos \left (f x + e\right )^{2} + 252 \, a^{3} \cos \left (f x + e\right ) + 384 \, a^{3} -{\left (15 \, a^{3} \cos \left (f x + e\right )^{3} + 206 \, a^{3} \cos \left (f x + e\right )^{2} - 132 \, a^{3} \cos \left (f x + e\right ) - 384 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{1536 \,{\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f -{\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} 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))^(9/2),x, algorithm="fricas")

[Out]

1/1536*(15*sqrt(2)*(a^3*cos(f*x + e)^5 + 5*a^3*cos(f*x + e)^4 - 8*a^3*cos(f*x + e)^3 - 20*a^3*cos(f*x + e)^2 +
8*a^3*cos(f*x + e) + 16*a^3 - (a^3*cos(f*x + e)^4 - 4*a^3*cos(f*x + e)^3 - 12*a^3*cos(f*x + e)^2 + 8*a^3*cos(
f*x + e) + 16*a^3)*sin(f*x + e))*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*(15*a^3*cos(f*x + e)^4 - 191*a^3*cos(f*x + e)^3
- 338*a^3*cos(f*x + e)^2 + 252*a^3*cos(f*x + e) + 384*a^3 - (15*a^3*cos(f*x + e)^3 + 206*a^3*cos(f*x + e)^2 -
132*a^3*cos(f*x + e) - 384*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(
f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f - (c^5*f*cos(f
*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*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))**(9/2),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))^(9/2),x, algorithm="giac")

[Out]

sage2