3.3.53 \(\int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx\) [253]

Optimal. Leaf size=94 \[ -\frac {15 a^3 x}{2 c}+\frac {15 a^3 \cos (e+f x)}{2 c f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))} \]

[Out]

-15/2*a^3*x/c+15/2*a^3*cos(f*x+e)/c/f+2*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^3+5/2*a^3*cos(f*x+e)^3/f/(c-c*
sin(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2815, 2759, 2758, 2761, 8} \begin {gather*} \frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {15 a^3 \cos (e+f x)}{2 c f}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac {15 a^3 x}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^3*x)/(2*c) + (15*a^3*Cos[e + f*x])/(2*c*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^3) + (5
*a^3*Cos[e + f*x]^3)/(2*f*(c - c*Sin[e + f*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[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 2759

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 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 2815

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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac {1}{2} \left (15 a^3\right ) \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx\\ &=\frac {15 a^3 \cos (e+f x)}{2 c f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac {\left (15 a^3\right ) \int 1 \, dx}{2 c}\\ &=-\frac {15 a^3 x}{2 c}+\frac {15 a^3 \cos (e+f x)}{2 c f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.36, size = 153, normalized size = 1.63 \begin {gather*} \frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (\cos \left (\frac {1}{2} (e+f x)\right ) (30 (e+f x)-16 \cos (e+f x)-\sin (2 (e+f x)))+\sin \left (\frac {1}{2} (e+f x)\right ) (-64-30 e-30 f x+16 \cos (e+f x)+\sin (2 (e+f x)))\right )}{4 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (-1+\sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(Cos[(e + f*x)/2]*(30*(e + f*x) - 16*Cos[e + f
*x] - Sin[2*(e + f*x)]) + Sin[(e + f*x)/2]*(-64 - 30*e - 30*f*x + 16*Cos[e + f*x] + Sin[2*(e + f*x)])))/(4*c*f
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-1 + Sin[e + f*x]))

________________________________________________________________________________________

Maple [A]
time = 0.28, size = 96, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-4 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-4}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\)	\(96\)
default	\(\frac {2 a^{3} \left (-\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-4 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-4}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\)	\(96\)
risch	\(-\frac {15 a^{3} x}{2 c}+\frac {2 a^{3} {\mathrm e}^{i \left (f x +e \right )}}{c f}+\frac {2 a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{c f}+\frac {16 a^{3}}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {a^{3} \sin \left (2 f x +2 e \right )}{4 c f}\)	\(96\)
norman	\(\frac {\frac {15 a^{3} x}{2 c}-\frac {7 a^{3}}{c f}-\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c}+\frac {45 a^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}-\frac {45 a^{3} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}+\frac {45 a^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}-\frac {45 a^{3} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}+\frac {15 a^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}-\frac {15 a^{3} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c}-\frac {12 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {5 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {10 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {42 a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {17 a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {35 a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(319\)

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)),x,method=_RETURNVERBOSE)

[Out]

2/f*a^3/c*(-(1/2*tan(1/2*f*x+1/2*e)^3-4*tan(1/2*f*x+1/2*e)^2-1/2*tan(1/2*f*x+1/2*e)-4)/(1+tan(1/2*f*x+1/2*e)^2
)^2-15/2*arctan(tan(1/2*f*x+1/2*e))-8/(tan(1/2*f*x+1/2*e)-1))

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (95) = 190\).
time = 0.53, size = 471, normalized size = 5.01 \begin {gather*} -\frac {6 \, a^{3} {\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}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \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 )}{c}\right )} + a^{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}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c \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 )}{c}\right )} + 6 \, a^{3} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {2 \, a^{3}}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \end {gather*}

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)),x, algorithm="maxima")

[Out]

-(6*a^3*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(
f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*
x + e)/(cos(f*x + e) + 1))/c) + a^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)/(c - c*sin(f*x + e)/(cos(
f*x + e) + 1) + 2*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1
))/c) + 6*a^3*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e)/(cos(f*x + e) + 1))) - 2*a^3/
(c - c*sin(f*x + e)/(cos(f*x + e) + 1)))/f

________________________________________________________________________________________

Fricas [A]
time = 0.32, size = 137, normalized size = 1.46 \begin {gather*} \frac {a^{3} \cos \left (f x + e\right )^{3} - 15 \, a^{3} f x + 8 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} - {\left (15 \, a^{3} f x - 23 \, a^{3}\right )} \cos \left (f x + e\right ) + {\left (15 \, a^{3} f x + a^{3} \cos \left (f x + e\right )^{2} - 7 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3}\right )} \sin \left (f x + e\right )}{2 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

1/2*(a^3*cos(f*x + e)^3 - 15*a^3*f*x + 8*a^3*cos(f*x + e)^2 + 16*a^3 - (15*a^3*f*x - 23*a^3)*cos(f*x + e) + (1
5*a^3*f*x + a^3*cos(f*x + e)^2 - 7*a^3*cos(f*x + e) + 16*a^3)*sin(f*x + e))/(c*f*cos(f*x + e) - c*f*sin(f*x +
e) + c*f)

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1168 vs. \(2 (83) = 166\).
time = 2.47, size = 1168, normalized size = 12.43 \begin {gather*} \begin {cases} - \frac {15 a^{3} f x \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} + \frac {15 a^{3} f x \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} - \frac {30 a^{3} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} + \frac {30 a^{3} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} - \frac {15 a^{3} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} + \frac {15 a^{3} f x}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} - \frac {34 a^{3} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} + \frac {18 a^{3} \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} - \frac {78 a^{3} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} + \frac {14 a^{3} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} - \frac {48 a^{3}}{2 c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 4 c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 4 c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 2 c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 2 c f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )^{3}}{- c \sin {\left (e \right )} + c} & \text {otherwise} \end {cases} \end {gather*}

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)),x)

[Out]

Piecewise((-15*a**3*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan
(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 15*a**3*f*x*tan(e/2 + f*x/2)*
*4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)
**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 30*a**3*f*x*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan
(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 3
0*a**3*f*x*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)
**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 15*a**3*f*x*tan(e/2 + f*x/2)/(2*c*f*tan(e/
2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(
e/2 + f*x/2) - 2*c*f) + 15*a**3*f*x/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f
*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 34*a**3*tan(e/2 + f*x/2)**4/(2*c*f*ta
n(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*
tan(e/2 + f*x/2) - 2*c*f) + 18*a**3*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4
+ 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 78*a**3*tan(e/2 +
f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2
+ f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 14*a**3*tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*ta
n(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) -
48*a**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f
*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f), Ne(f, 0)), (x*(a*sin(e) + a)**3/(-c*sin(e) + c), True))

________________________________________________________________________________________

Giac [A]
time = 0.48, size = 117, normalized size = 1.24 \begin {gather*} -\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c} + \frac {32 \, a^{3}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

-1/2*(15*(f*x + e)*a^3/c + 32*a^3/(c*(tan(1/2*f*x + 1/2*e) - 1)) + 2*(a^3*tan(1/2*f*x + 1/2*e)^3 - 8*a^3*tan(1
/2*f*x + 1/2*e)^2 - a^3*tan(1/2*f*x + 1/2*e) - 8*a^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*c))/f

________________________________________________________________________________________

Mupad [B]
time = 9.12, size = 219, normalized size = 2.33 \begin {gather*} -\frac {15\,a^3\,x}{2\,c}-\frac {\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\frac {a^3\,\left (15\,e+15\,f\,x-14\right )}{2}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-48\right )}{2}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\frac {a^3\,\left (15\,e+15\,f\,x-34\right )}{2}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (30\,e+30\,f\,x-18\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (30\,e+30\,f\,x-78\right )}{2}\right )}{c\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x)),x)

[Out]

- (15*a^3*x)/(2*c) - ((15*a^3*(e + f*x))/2 - tan(e/2 + (f*x)/2)*((15*a^3*(e + f*x))/2 - (a^3*(15*e + 15*f*x -
14))/2) - (a^3*(15*e + 15*f*x - 48))/2 + tan(e/2 + (f*x)/2)^4*((15*a^3*(e + f*x))/2 - (a^3*(15*e + 15*f*x - 34
))/2) - tan(e/2 + (f*x)/2)^3*(15*a^3*(e + f*x) - (a^3*(30*e + 30*f*x - 18))/2) + tan(e/2 + (f*x)/2)^2*(15*a^3*
(e + f*x) - (a^3*(30*e + 30*f*x - 78))/2))/(c*f*(tan(e/2 + (f*x)/2) - 1)*(tan(e/2 + (f*x)/2)^2 + 1)^2)

________________________________________________________________________________________