3.316 \(\int \frac{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=225 \[ \frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}} \]

[Out]

(-3*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(512*Sqrt[2]*c^(11/2)*f) + (a^3*c^
2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(15/2)) - (a^3*Cos[e + f*x]^3)/(8*f*(c - c*Sin[e + f*x])^(11/2)) +
(a^3*Cos[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(7/2)) - (a^3*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e + f*x])
^(5/2)) - (3*a^3*Cos[e + f*x])/(512*c^4*f*(c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.387612, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(-3*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(512*Sqrt[2]*c^(11/2)*f) + (a^3*c^
2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(15/2)) - (a^3*Cos[e + f*x]^3)/(8*f*(c - c*Sin[e + f*x])^(11/2)) +
(a^3*Cos[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(7/2)) - (a^3*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e + f*x])
^(5/2)) - (3*a^3*Cos[e + f*x])/(512*c^4*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))^{11/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{17/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{1}{2} \left (a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{\left (3 a^3\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx}{16 c}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{a^3 \int \frac{1}{(c-c \sin (e+f x))^{5/2}} \, dx}{32 c^3}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{\left (3 a^3\right ) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{256 c^4}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (3 a^3\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{1024 c^5}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (3 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 )}{512 c^5 f}\\ &=-\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}+\frac{a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac{a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac{a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 4.14668, size = 435, normalized size = 1.93 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (4096 \sin \left (\frac{1}{2} (e+f x)\right )-15 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9-30 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8-20 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7-40 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+992 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+1984 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-2688 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-5376 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+2048 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{10}\right )}{2560 f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(2048*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 2688*(Cos[(e + f*x)/2
] - Sin[(e + f*x)/2])^3 + 992*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - 20*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2
])^7 - 15*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9 + (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])^10 + 4096*Sin[(e + f*x)/2] - 5376*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 1984*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 40*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2] - 30*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*Sin[(e + f*
x)/2])*(1 + Sin[e + f*x])^3)/(2560*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(11/2))

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Maple [A]  time = 0.837, size = 353, normalized size = 1.6 \begin{align*}{\frac{{a}^{3}}{5120\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{4}\cos \left ( fx+e \right ) f} \left ( 480\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{13/2}-1120\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{11/2}+1024\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{9/2}+280\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{7/2}{c}^{7/2}-30\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{9/2}{c}^{5/2}+15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{5}{c}^{7}-75\,\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}^{7}+150\,\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}^{7}-150\,\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}^{7}+75\,\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}^{7}-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}^{7} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{25}{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))^(11/2),x)

[Out]

1/5120*a^3*(480*(c*(1+sin(f*x+e)))^(1/2)*c^(13/2)-1120*(c*(1+sin(f*x+e)))^(3/2)*c^(11/2)+1024*(c*(1+sin(f*x+e)
))^(5/2)*c^(9/2)+280*(c*(1+sin(f*x+e)))^(7/2)*c^(7/2)-30*(c*(1+sin(f*x+e)))^(9/2)*c^(5/2)+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)^5*c^7-75*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^7+150*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^7-150*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^7+75*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^7-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^
(1/2)/c^(1/2))*c^7)*(c*(1+sin(f*x+e)))^(1/2)/c^(25/2)/(-1+sin(f*x+e))^4/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{11}{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))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(11/2), x)

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Fricas [B]  time = 1.25892, size = 1553, normalized size = 6.9 \begin{align*} \text{result too large to display} \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))^(11/2),x, algorithm="fricas")

[Out]

1/10240*(15*sqrt(2)*(a^3*cos(f*x + e)^6 - 5*a^3*cos(f*x + e)^5 - 18*a^3*cos(f*x + e)^4 + 20*a^3*cos(f*x + e)^3
+ 48*a^3*cos(f*x + e)^2 - 16*a^3*cos(f*x + e) - 32*a^3 + (a^3*cos(f*x + e)^5 + 6*a^3*cos(f*x + e)^4 - 12*a^3*
cos(f*x + e)^3 - 32*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 32*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)^5 - 65*a^3*cos(f*x + e)^4 + 812*a^3*cos(f*x + e)^3 + 1796*a^3*cos(f*x + e)^2 - 1144*
a^3*cos(f*x + e) - 2048*a^3 + (15*a^3*cos(f*x + e)^4 + 80*a^3*cos(f*x + e)^3 + 892*a^3*cos(f*x + e)^2 - 904*a^
3*cos(f*x + e) - 2048*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^6*f*cos(f*x + e)^6 - 5*c^6*f*cos(f*x +
e)^5 - 18*c^6*f*cos(f*x + e)^4 + 20*c^6*f*cos(f*x + e)^3 + 48*c^6*f*cos(f*x + e)^2 - 16*c^6*f*cos(f*x + e) - 3
2*c^6*f + (c^6*f*cos(f*x + e)^5 + 6*c^6*f*cos(f*x + e)^4 - 12*c^6*f*cos(f*x + e)^3 - 32*c^6*f*cos(f*x + e)^2 +
16*c^6*f*cos(f*x + e) + 32*c^6*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))**(11/2),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))^(11/2),x, algorithm="giac")

[Out]

sage2