3.463 \(\int \frac{(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=120 \[ \frac{d^2 (c-4 d) \cos (e+f x)}{3 a^2 f}+\frac{d^2 x (3 c-2 d)}{a^2}-\frac{(c+6 d) (c-d)^2 \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2} \]
[Out]
((3*c - 2*d)*d^2*x)/a^2 + ((c - 4*d)*d^2*Cos[e + f*x])/(3*a^2*f) - ((c - d)^2*(c + 6*d)*Cos[e + f*x])/(3*a^2*f
*(1 + Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*f*(a + a*Sin[e + f*x])^2)
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Rubi [A] time = 0.367196, antiderivative size = 120, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{2765, 2968, 3023, 2735, 2648} \[ \frac{d^2 (c-4 d) \cos (e+f x)}{3 a^2 f}+\frac{d^2 x (3 c-2 d)}{a^2}-\frac{(c+6 d) (c-d)^2 \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
[Out]
((3*c - 2*d)*d^2*x)/a^2 + ((c - 4*d)*d^2*Cos[e + f*x])/(3*a^2*f) - ((c - d)^2*(c + 6*d)*Cos[e + f*x])/(3*a^2*f
*(1 + Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*f*(a + a*Sin[e + f*x])^2)
Rule 2765
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 2735
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]
Rule 2648
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx &=-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a+a \sin (e+f x))^2}-\frac{\int \frac{(c+d \sin (e+f x)) \left (-a \left (c^2+4 c d-2 d^2\right )+a (c-4 d) d \sin (e+f x)\right )}{a+a \sin (e+f x)} \, dx}{3 a^2}\\ &=-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a+a \sin (e+f x))^2}-\frac{\int \frac{-a c \left (c^2+4 c d-2 d^2\right )+\left (a c (c-4 d) d-a d \left (c^2+4 c d-2 d^2\right )\right ) \sin (e+f x)+a (c-4 d) d^2 \sin ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{3 a^2}\\ &=\frac{(c-4 d) d^2 \cos (e+f x)}{3 a^2 f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a+a \sin (e+f x))^2}-\frac{\int \frac{-a^2 c \left (c^2+4 c d-2 d^2\right )-3 a^2 (3 c-2 d) d^2 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{3 a^3}\\ &=\frac{(3 c-2 d) d^2 x}{a^2}+\frac{(c-4 d) d^2 \cos (e+f x)}{3 a^2 f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a+a \sin (e+f x))^2}+\frac{\left ((c-d)^2 (c+6 d)\right ) \int \frac{1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=\frac{(3 c-2 d) d^2 x}{a^2}+\frac{(c-4 d) d^2 \cos (e+f x)}{3 a^2 f}-\frac{(c-d)^2 (c+6 d) \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a+a \sin (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 0.368427, size = 212, normalized size = 1.77 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 d^2 (3 c-2 d) (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (c-d)^3 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+2 (c+8 d) (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-3 d^3 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{3 a^2 f (\sin (e+f x)+1)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)^3*Sin[(e + f*x)/2] - (c - d)^3*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) + 2*(c - d)^2*(c + 8*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 3*(3*c - 2*d)*d^2*
(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 3*d^3*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3
))/(3*a^2*f*(1 + Sin[e + f*x])^2)
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Maple [B] time = 0.076, size = 340, normalized size = 2.8 \begin{align*} -2\,{\frac{{d}^{3}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+6\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{{a}^{2}f}}-4\,{\frac{{d}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}-2\,{\frac{{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+6\,{\frac{c{d}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-4\,{\frac{{d}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-6\,{\frac{{c}^{2}d}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+6\,{\frac{c{d}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-2\,{\frac{{d}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{4\,{c}^{3}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+4\,{\frac{{c}^{2}d}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-4\,{\frac{c{d}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}+{\frac{4\,{d}^{3}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x)
[Out]
-2/f/a^2*d^3/(1+tan(1/2*f*x+1/2*e)^2)+6/f/a^2*d^2*arctan(tan(1/2*f*x+1/2*e))*c-4/f/a^2*d^3*arctan(tan(1/2*f*x+
1/2*e))-2/f/a^2/(tan(1/2*f*x+1/2*e)+1)*c^3+6/f/a^2/(tan(1/2*f*x+1/2*e)+1)*c*d^2-4/f/a^2/(tan(1/2*f*x+1/2*e)+1)
*d^3+2/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*c^3-6/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*c^2*d+6/f/a^2/(tan(1/2*f*x+1/2*e)+1
)^2*c*d^2-2/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*d^3-4/3/f/a^2/(tan(1/2*f*x+1/2*e)+1)^3*c^3+4/f/a^2/(tan(1/2*f*x+1/2
*e)+1)^3*c^2*d-4/f/a^2/(tan(1/2*f*x+1/2*e)+1)^3*c*d^2+4/3/f/a^2/(tan(1/2*f*x+1/2*e)+1)^3*d^3
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Maxima [B] time = 1.7316, size = 798, normalized size = 6.65 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
-2/3*(2*d^3*((12*sin(f*x + e)/(cos(f*x + e) + 1) + 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1)
+ 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a
^2) - 3*c*d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*si
n(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) +
1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x +
e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*c^2*d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 +
3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3))/f
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Fricas [B] time = 1.62919, size = 699, normalized size = 5.82 \begin{align*} -\frac{3 \, d^{3} \cos \left (f x + e\right )^{3} - c^{3} + 3 \, c^{2} d - 3 \, c d^{2} + d^{3} + 6 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x -{\left (c^{3} + 6 \, c^{2} d - 15 \, c d^{2} + 11 \, d^{3} + 3 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right )^{2} -{\left (2 \, c^{3} + 3 \, c^{2} d - 12 \, c d^{2} + 13 \, d^{3} - 3 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right ) -{\left (3 \, d^{3} \cos \left (f x + e\right )^{2} - c^{3} + 3 \, c^{2} d - 3 \, c d^{2} + d^{3} - 6 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x +{\left (c^{3} + 6 \, c^{2} d - 15 \, c d^{2} + 14 \, d^{3} - 3 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
-1/3*(3*d^3*cos(f*x + e)^3 - c^3 + 3*c^2*d - 3*c*d^2 + d^3 + 6*(3*c*d^2 - 2*d^3)*f*x - (c^3 + 6*c^2*d - 15*c*d
^2 + 11*d^3 + 3*(3*c*d^2 - 2*d^3)*f*x)*cos(f*x + e)^2 - (2*c^3 + 3*c^2*d - 12*c*d^2 + 13*d^3 - 3*(3*c*d^2 - 2*
d^3)*f*x)*cos(f*x + e) - (3*d^3*cos(f*x + e)^2 - c^3 + 3*c^2*d - 3*c*d^2 + d^3 - 6*(3*c*d^2 - 2*d^3)*f*x + (c^
3 + 6*c^2*d - 15*c*d^2 + 14*d^3 - 3*(3*c*d^2 - 2*d^3)*f*x)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 -
a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
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Sympy [A] time = 29.065, size = 3471, normalized size = 28.92 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))**3/(a+a*sin(f*x+e))**2,x)
[Out]
Piecewise((2*c**3*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f
*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 2*c**3*tan(e/2
+ f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*
a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 2*c**3*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(
e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 2*c**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4
+ 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 18*
c**2*d*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 +
f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*c**2*d*tan(e/2 + f*x/2)*
*2/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*ta
n(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 18*c**2*d*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x
/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2
*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*c**2*d/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a
**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*c*d**2*f
*x*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/
2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 27*c*d**2*f*x*tan(e/2 + f*x/2)
**4/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*t
an(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 36*c*d**2*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e
/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 36*c*d**2*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*
a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2
+ f*x/2) + 3*a**2*f) + 27*c*d**2*f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2
)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) +
9*c*d**2*f*x/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 1
2*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*c*d**2*tan(e/2 + f*x/2)**5/(3*a**2*f*
tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2
)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 30*c*d**2*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9
*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2
+ f*x/2) + 3*a**2*f) + 18*c*d**2*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2
)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) +
36*c*d**2*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 +
f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 18*c*d**2/(3*a**2*f*tan(e
/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*d**3*f*x*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**
2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f
*x/2) + 3*a**2*f) - 18*d**3*f*x*tan(e/2 + f*x/2)**4/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)*
*4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 2
4*d**3*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/
2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 24*d**3*f*x*tan(e/2 +
f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a*
*2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 18*d**3*f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(
e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*d**3*f*x/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2
)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) +
4*d**3*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 +
f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 20*d**3*tan(e/2 + f*x/2)*
*3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*ta
n(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 28*d**3*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*
x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**
2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 36*d**3*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 +
f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**
2*f) - 16*d**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 +
12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(c + d*sin(e))**3/(a*sin(
e) + a)**2, True))
________________________________________________________________________________________
Giac [A] time = 1.28716, size = 282, normalized size = 2.35 \begin{align*} \frac{\frac{3 \,{\left (3 \, c d^{2} - 2 \, d^{3}\right )}{\left (f x + e\right )}}{a^{2}} - \frac{6 \, d^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} a^{2}} - \frac{2 \,{\left (3 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 6 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 \, c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 27 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, c^{3} + 3 \, c^{2} d - 12 \, c d^{2} + 7 \, d^{3}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/3*(3*(3*c*d^2 - 2*d^3)*(f*x + e)/a^2 - 6*d^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^2) - 2*(3*c^3*tan(1/2*f*x + 1/2
*e)^2 - 9*c*d^2*tan(1/2*f*x + 1/2*e)^2 + 6*d^3*tan(1/2*f*x + 1/2*e)^2 + 3*c^3*tan(1/2*f*x + 1/2*e) + 9*c^2*d*t
an(1/2*f*x + 1/2*e) - 27*c*d^2*tan(1/2*f*x + 1/2*e) + 15*d^3*tan(1/2*f*x + 1/2*e) + 2*c^3 + 3*c^2*d - 12*c*d^2
+ 7*d^3)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f