3.313 \(\int \frac{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=157 \[ \frac{a^3 c^2 \cos ^5(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac{15 a^3 \cos (e+f x)}{4 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{15 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} c^{5/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

(15*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f) + (a^3*c^2*C
os[e + f*x]^5)/(2*f*(c - c*Sin[e + f*x])^(9/2)) - (5*a^3*Cos[e + f*x]^3)/(4*f*(c - c*Sin[e + f*x])^(5/2)) - (1
5*a^3*Cos[e + f*x])/(4*c^2*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.325292, antiderivative size = 157, 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} \[ \frac{a^3 c^2 \cos ^5(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac{15 a^3 \cos (e+f x)}{4 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{15 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} c^{5/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.998543, size = 187, normalized size = 1.19 \[ \frac{a^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-5 \sin \left (\frac{1}{2} (e+f x)\right )+15 \sin \left (\frac{3}{2} (e+f x)\right )+2 \sin \left (\frac{5}{2} (e+f x)\right )-5 \cos \left (\frac{1}{2} (e+f x)\right )-15 \cos \left (\frac{3}{2} (e+f x)\right )+2 \cos \left (\frac{5}{2} (e+f x)\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) (4 \sin (e+f x)+\cos (2 (e+f x))-3)\right )}{4 c^2 f (\sin (e+f x)-1)^2 \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-5*Cos[(e + f*x)/2] - 15*Cos[(3*(e + f*x))/2] + 2*Cos[(5*(e + f*x)
)/2] - 5*Sin[(e + f*x)/2] + (15 + 15*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-3 +
Cos[2*(e + f*x)] + 4*Sin[e + f*x]) + 15*Sin[(3*(e + f*x))/2] + 2*Sin[(5*(e + f*x))/2]))/(4*c^2*f*(-1 + Sin[e
+ f*x])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.744, size = 239, normalized size = 1.5 \begin{align*} -{\frac{{a}^{3}}{ \left ( -4+4\,\sin \left ( fx+e \right ) \right ) \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 ) ^{2}{c}^{2}-8\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-30\,\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}+18\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{c}+16\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2}\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}-36\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{9}{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))^(5/2),x)

[Out]

-1/4/c^(9/2)*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)^2*c^2-8*(c*(1+si
n(f*x+e)))^(1/2)*c^(3/2)*sin(f*x+e)^2-30*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+18*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+16*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*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^2-36*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2
)/(-1+sin(f*x+e))/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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(5/2), x)

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Fricas [B]  time = 1.19308, size = 984, normalized size = 6.27 \begin{align*} \frac{15 \, \sqrt{2}{\left (a^{3} \cos \left (f x + e\right )^{3} + 3 \, a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) - 4 \, a^{3} -{\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) - 4 \, 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 (4 \, a^{3} \cos \left (f x + e\right )^{3} - 13 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3} +{\left (4 \, a^{3} \cos \left (f x + e\right )^{2} + 17 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{8 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/8*(15*sqrt(2)*(a^3*cos(f*x + e)^3 + 3*a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) - 4*a^3 - (a^3*cos(f*x + e)^2
- 2*a^3*cos(f*x + e) - 4*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*(4*a^3*cos(f*x + e)^3 - 13*a^3*cos(f
*x + e)^2 - 13*a^3*cos(f*x + e) + 4*a^3 + (4*a^3*cos(f*x + e)^2 + 17*a^3*cos(f*x + e) + 4*a^3)*sin(f*x + e))*s
qrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c
^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

sage2