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+1/3*(c-4*d)*d^2*cos(f*x+e)/a^2/f-1/3*(c-d)^2*(c+6*d)*cos(f*x+e)/a^2/f/(1+sin(f*x+e))-1/3*(
c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^2
________________________________________________________________________________________
Rubi [A] time = 0.37, antiderivative size = 120, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]
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 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]
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 = 0.35, 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)
________________________________________________________________________________________
fricas [B] time = 0.45, size = 308, normalized size = 2.57 \[ -\frac {3 \, d^{3} \cos \left (f x + e\right )^{3} - c^{3} + 3 \, c^{2} d - 3 \, c d^{2} + d^{3} + 6 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x - {\left (c^{3} + 6 \, c^{2} d - 15 \, c d^{2} + 11 \, d^{3} + 3 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, c^{3} + 3 \, c^{2} d - 12 \, c d^{2} + 13 \, d^{3} - 3 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right ) - {\left (3 \, d^{3} \cos \left (f x + e\right )^{2} - c^{3} + 3 \, c^{2} d - 3 \, c d^{2} + d^{3} - 6 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x + {\left (c^{3} + 6 \, c^{2} d - 15 \, c d^{2} + 14 \, d^{3} - 3 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} f x\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]
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))
________________________________________________________________________________________
giac [A] time = 0.95, size = 209, normalized size = 1.74 \[ \frac {\frac {3 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} {\left (f x + e\right )}}{a^{2}} - \frac {6 \, d^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} a^{2}} - \frac {2 \, {\left (3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 27 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, c^{3} + 3 \, c^{2} d - 12 \, c d^{2} + 7 \, d^{3}\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \]
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
________________________________________________________________________________________
maple [B] time = 0.27, size = 340, normalized size = 2.83 \[ -\frac {2 d^{3}}{a^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}+\frac {6 d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c}{a^{2} f}-\frac {4 d^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{2} f}-\frac {2 c^{3}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {6 c \,d^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4 d^{3}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 c^{3}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {6 c^{2} d}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {6 c \,d^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 d^{3}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 c^{3}}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 c^{2} d}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {4 c \,d^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 d^{3}}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}} \]
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/a^2/f*d^3/(1+tan(1/2*f*x+1/2*e)^2)+6/a^2/f*d^2*arctan(tan(1/2*f*x+1/2*e))*c-4/a^2/f*d^3*arctan(tan(1/2*f*x+
1/2*e))-2/a^2/f/(tan(1/2*f*x+1/2*e)+1)*c^3+6/a^2/f/(tan(1/2*f*x+1/2*e)+1)*c*d^2-4/a^2/f/(tan(1/2*f*x+1/2*e)+1)
*d^3+2/a^2/f/(tan(1/2*f*x+1/2*e)+1)^2*c^3-6/a^2/f/(tan(1/2*f*x+1/2*e)+1)^2*c^2*d+6/a^2/f/(tan(1/2*f*x+1/2*e)+1
)^2*c*d^2-2/a^2/f/(tan(1/2*f*x+1/2*e)+1)^2*d^3-4/3/a^2/f/(tan(1/2*f*x+1/2*e)+1)^3*c^3+4/a^2/f/(tan(1/2*f*x+1/2
*e)+1)^3*c^2*d-4/a^2/f/(tan(1/2*f*x+1/2*e)+1)^3*c*d^2+4/3/a^2/f/(tan(1/2*f*x+1/2*e)+1)^3*d^3
________________________________________________________________________________________
maxima [B] time = 0.44, size = 591, normalized size = 4.92 \[ -\frac {2 \, {\left (2 \, d^{3} {\left (\frac {\frac {12 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {11 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, \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}} + 5}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {4 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{2} \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^{2}}\right )} - 3 \, c d^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 4}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac {c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, c^{2} d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]
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
________________________________________________________________________________________
mupad [B] time = 8.39, size = 298, normalized size = 2.48 \[ \frac {2\,d^2\,\mathrm {atan}\left (\frac {2\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,c-2\,d\right )}{6\,c\,d^2-4\,d^3}\right )\,\left (3\,c-2\,d\right )}{a^2\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^3+6\,c^2\,d-18\,c\,d^2+12\,d^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {10\,c^3}{3}+2\,c^2\,d-14\,c\,d^2+\frac {44\,d^3}{3}\right )-8\,c\,d^2+2\,c^2\,d+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,c^3-6\,c\,d^2+4\,d^3\right )+\frac {4\,c^3}{3}+\frac {20\,d^3}{3}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c^3+6\,c^2\,d-18\,c\,d^2+16\,d^3\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*sin(e + f*x))^3/(a + a*sin(e + f*x))^2,x)
[Out]
(2*d^2*atan((2*d^2*tan(e/2 + (f*x)/2)*(3*c - 2*d))/(6*c*d^2 - 4*d^3))*(3*c - 2*d))/(a^2*f) - (tan(e/2 + (f*x)/
2)^3*(6*c^2*d - 18*c*d^2 + 2*c^3 + 12*d^3) + tan(e/2 + (f*x)/2)^2*(2*c^2*d - 14*c*d^2 + (10*c^3)/3 + (44*d^3)/
3) - 8*c*d^2 + 2*c^2*d + tan(e/2 + (f*x)/2)^4*(2*c^3 - 6*c*d^2 + 4*d^3) + (4*c^3)/3 + (20*d^3)/3 + tan(e/2 + (
f*x)/2)*(6*c^2*d - 18*c*d^2 + 2*c^3 + 16*d^3))/(f*(4*a^2*tan(e/2 + (f*x)/2)^2 + 4*a^2*tan(e/2 + (f*x)/2)^3 + 3
*a^2*tan(e/2 + (f*x)/2)^4 + a^2*tan(e/2 + (f*x)/2)^5 + a^2 + 3*a^2*tan(e/2 + (f*x)/2)))
________________________________________________________________________________________
sympy [A] time = 14.92, size = 3585, normalized size = 29.88 \[ \text {result too large to display} \]
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((-6*c**3*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) - 6*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) - 10*c**3*tan(e/2 + f*x/2)**2/(3*a**2*f*ta
n(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**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) - 4*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*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*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*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)**3/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*t
an(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 + 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)**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) + 54*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) + 42
*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) + 54*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) + 24*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) - 24*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) - 12*d**3*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*ta
n(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/(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) - 44*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) - 48*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*ta
n(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/(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))
________________________________________________________________________________________