3.454 \(\int \frac{(c+d \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx\)
Optimal. Leaf size=121 \[ \frac{2 d \left (c^2-3 c d+d^2\right ) \cos (e+f x)}{a f}+\frac{3 d x \left (2 c^2-2 c d+d^2\right )}{2 a}+\frac{d^2 (2 c-3 d) \sin (e+f x) \cos (e+f x)}{2 a f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a \sin (e+f x)+a)} \]
[Out]
(3*d*(2*c^2 - 2*c*d + d^2)*x)/(2*a) + (2*d*(c^2 - 3*c*d + d^2)*Cos[e + f*x])/(a*f) + ((2*c - 3*d)*d^2*Cos[e +
f*x]*Sin[e + f*x])/(2*a*f) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]))
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Rubi [A] time = 0.125398, antiderivative size = 121, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used =
{2767, 2734} \[ \frac{2 d \left (c^2-3 c d+d^2\right ) \cos (e+f x)}{a f}+\frac{3 d x \left (2 c^2-2 c d+d^2\right )}{2 a}+\frac{d^2 (2 c-3 d) \sin (e+f x) \cos (e+f x)}{2 a f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a \sin (e+f x)+a)} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x]),x]
[Out]
(3*d*(2*c^2 - 2*c*d + d^2)*x)/(2*a) + (2*d*(c^2 - 3*c*d + d^2)*Cos[e + f*x])/(a*f) + ((2*c - 3*d)*d^2*Cos[e +
f*x]*Sin[e + f*x])/(2*a*f) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]))
Rule 2767
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(a + b*Sin[e + f*x])), x] - Dist[d/(a*b), Int[(c +
d*Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*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] && GtQ[n, 1] && (IntegerQ[2
*n] || EqQ[c, 0])
Rule 2734
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx &=-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a+a \sin (e+f x))}-\frac{d \int (-a (3 c-2 d)+a (2 c-3 d) \sin (e+f x)) (c+d \sin (e+f x)) \, dx}{a^2}\\ &=\frac{3 d \left (2 c^2-2 c d+d^2\right ) x}{2 a}+\frac{2 d \left (c^2-3 c d+d^2\right ) \cos (e+f x)}{a f}+\frac{(2 c-3 d) d^2 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a+a \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.561892, size = 192, normalized size = 1.59 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (d \cos \left (\frac{1}{2} (e+f x)\right ) \left (6 \left (2 c^2-2 c d+d^2\right ) (e+f x)-4 d (3 c-d) \cos (e+f x)+d^2 (-\sin (2 (e+f x)))\right )+\sin \left (\frac{1}{2} (e+f x)\right ) \left (2 \left (6 c^2 d (e+f x-2)+4 c^3-6 c d^2 (e+f x-2)+d^3 (3 e+3 f x-4)\right )-4 d^2 (3 c-d) \cos (e+f x)+d^3 (-\sin (2 (e+f x)))\right )\right )}{4 a f (\sin (e+f x)+1)} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x]),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(d*Cos[(e + f*x)/2]*(6*(2*c^2 - 2*c*d + d^2)*(e + f*x) - 4*(3*c - d)*d*
Cos[e + f*x] - d^2*Sin[2*(e + f*x)]) + Sin[(e + f*x)/2]*(2*(4*c^3 + 6*c^2*d*(-2 + e + f*x) - 6*c*d^2*(-2 + e +
f*x) + d^3*(-4 + 3*e + 3*f*x)) - 4*(3*c - d)*d^2*Cos[e + f*x] - d^3*Sin[2*(e + f*x)])))/(4*a*f*(1 + Sin[e + f
*x]))
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Maple [B] time = 0.068, size = 364, normalized size = 3. \begin{align*}{\frac{{d}^{3}}{af} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-6\,{\frac{c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}{d}^{2}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}{d}^{3}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{d}^{3}}{af}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-6\,{\frac{c{d}^{2}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{{d}^{3}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+6\,{\frac{d\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ){c}^{2}}{af}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c{d}^{2}}{af}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ){d}^{3}}{af}}-2\,{\frac{{c}^{3}}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+6\,{\frac{{c}^{2}d}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-6\,{\frac{c{d}^{2}}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{{d}^{3}}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \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)),x)
[Out]
1/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3*d^3-6/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^
2*c*d^2+2/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*d^3-1/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x
+1/2*e)*d^3-6/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*c*d^2+2/a/f/(1+tan(1/2*f*x+1/2*e)^2)^2*d^3+6/a/f*d*arctan(tan(1/2
*f*x+1/2*e))*c^2-6/a/f*arctan(tan(1/2*f*x+1/2*e))*c*d^2+3/a/f*arctan(tan(1/2*f*x+1/2*e))*d^3-2/f*c^3/a/(tan(1/
2*f*x+1/2*e)+1)+6/a/f/(tan(1/2*f*x+1/2*e)+1)*c^2*d-6/a/f/(tan(1/2*f*x+1/2*e)+1)*c*d^2+2/a/f/(tan(1/2*f*x+1/2*e
)+1)*d^3
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Maxima [B] time = 1.74515, size = 574, normalized size = 4.74 \begin{align*} \frac{d^{3}{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 4}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{2 \, a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac{3 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - 6 \, c d^{2}{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} + 6 \, c^{2} d{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac{2 \, c^{3}}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{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)),x, algorithm="maxima")
[Out]
(d^3*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e
) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 6*c*d^2*((sin(f*x + e)/
(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) +
1))/a) + 6*c^2*d*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) - 2*c
^3/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f
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Fricas [B] time = 1.6493, size = 537, normalized size = 4.44 \begin{align*} \frac{d^{3} \cos \left (f x + e\right )^{3} - 2 \, c^{3} + 6 \, c^{2} d - 6 \, c d^{2} + 2 \, d^{3} + 3 \,{\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} f x - 2 \,{\left (3 \, c d^{2} - d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (2 \, c^{3} - 6 \, c^{2} d + 12 \, c d^{2} - 3 \, d^{3} - 3 \,{\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} f x\right )} \cos \left (f x + e\right ) -{\left (d^{3} \cos \left (f x + e\right )^{2} - 2 \, c^{3} + 6 \, c^{2} d - 6 \, c d^{2} + 2 \, d^{3} - 3 \,{\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} f x +{\left (6 \, c d^{2} - d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\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)),x, algorithm="fricas")
[Out]
1/2*(d^3*cos(f*x + e)^3 - 2*c^3 + 6*c^2*d - 6*c*d^2 + 2*d^3 + 3*(2*c^2*d - 2*c*d^2 + d^3)*f*x - 2*(3*c*d^2 - d
^3)*cos(f*x + e)^2 - (2*c^3 - 6*c^2*d + 12*c*d^2 - 3*d^3 - 3*(2*c^2*d - 2*c*d^2 + d^3)*f*x)*cos(f*x + e) - (d^
3*cos(f*x + e)^2 - 2*c^3 + 6*c^2*d - 6*c*d^2 + 2*d^3 - 3*(2*c^2*d - 2*c*d^2 + d^3)*f*x + (6*c*d^2 - d^3)*cos(f
*x + e))*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
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Sympy [A] time = 9.64225, size = 3499, 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)),x)
[Out]
Piecewise((-4*c**3*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2
+ f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 8*c**3*tan(e/2 + f*x/2)**2/(2*a*f*
tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*
f*tan(e/2 + f*x/2) + 2*a*f) - 4*c**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 +
f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 6*c**2*d*f*x*tan(e/2 + f*x/2)**5/(2*
a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 +
2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 6*c**2*d*f*x*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2
+ f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 12*c**
2*d*f*x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3
+ 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 12*c**2*d*f*x*tan(e/2 + f*x/2)**2/(2*a*f*tan(
e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*ta
n(e/2 + f*x/2) + 2*a*f) + 6*c**2*d*f*x*tan(e/2 + f*x/2)/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4
+ 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 6*c**2*d*f*x/(2*a
*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2
*a*f*tan(e/2 + f*x/2) + 2*a*f) + 12*c**2*d*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*
x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 24*c**2*d*
tan(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f
*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 12*c**2*d/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2
+ f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 6*c*d*
*2*f*x*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3
+ 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 6*c*d**2*f*x*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/
2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(
e/2 + f*x/2) + 2*a*f) - 12*c*d**2*f*x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)*
*4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 12*c*d**2*f*x*t
an(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*
tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 6*c*d**2*f*x*tan(e/2 + f*x/2)/(2*a*f*tan(e/2 + f*x/2)*
*5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2
) + 2*a*f) - 6*c*d**2*f*x/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 +
4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 12*c*d**2*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 +
f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2
+ f*x/2) + 2*a*f) + 12*c*d**2*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a
*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 12*c*d**2*tan(e/2 + f*x
/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*
x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 12*c*d**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 +
4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 3*d**3*f*x*tan(e/2
+ f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2
+ f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 3*d**3*f*x*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2
*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*
a*f) + 6*d**3*f*x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 +
f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 6*d**3*f*x*tan(e/2 + f*x/2)**2/(2*a
*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2
*a*f*tan(e/2 + f*x/2) + 2*a*f) + 3*d**3*f*x*tan(e/2 + f*x/2)/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/
2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 3*d**3*f*x/(
2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2
+ 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 2*d**3*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*
x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 4*d**3*tan
(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*ta
n(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 2*d**3*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 +
2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2
*a*f) + 6*d**3*tan(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*
x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 6*d**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*
a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a
*f), Ne(f, 0)), (x*(c + d*sin(e))**3/(a*sin(e) + a), True))
________________________________________________________________________________________
Giac [A] time = 1.36096, size = 232, normalized size = 1.92 \begin{align*} \frac{\frac{3 \,{\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )}{\left (f x + e\right )}}{a} - \frac{4 \,{\left (c^{3} - 3 \, c^{2} d + 3 \, c d^{2} - d^{3}\right )}}{a{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{2 \,{\left (d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, c d^{2} + 2 \, d^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} a}}{2 \, 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)),x, algorithm="giac")
[Out]
1/2*(3*(2*c^2*d - 2*c*d^2 + d^3)*(f*x + e)/a - 4*(c^3 - 3*c^2*d + 3*c*d^2 - d^3)/(a*(tan(1/2*f*x + 1/2*e) + 1)
) + 2*(d^3*tan(1/2*f*x + 1/2*e)^3 - 6*c*d^2*tan(1/2*f*x + 1/2*e)^2 + 2*d^3*tan(1/2*f*x + 1/2*e)^2 - d^3*tan(1/
2*f*x + 1/2*e) - 6*c*d^2 + 2*d^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a))/f