3.473 \(\int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=142 \[ -\frac {(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {d^3 x}{a^3}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

d^3*x/a^3-1/15*(c-d)^2*(2*c+7*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(c-d)*(2*c^2+11*c*d+29*d^2)*cos(f*x+e)
/f/(a^3+a^3*sin(f*x+e))-1/5*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^3

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Rubi [A]  time = 0.33, antiderivative size = 142, 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, 3019, 2735, 2648} \[ -\frac {(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {d^3 x}{a^3}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*x)/a^3 - ((c - d)^2*(2*c + 7*d)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((c - d)*(2*c^2 + 11*c*d
+ 29*d^2)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(
a + a*Sin[e + f*x])^3)

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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && 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))^3} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {(c+d \sin (e+f x)) \left (-a \left (2 c^2+5 c d-2 d^2\right )-5 a d^2 \sin (e+f x)\right )}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a c \left (2 c^2+5 c d-2 d^2\right )+\left (-5 a c d^2-a d \left (2 c^2+5 c d-2 d^2\right )\right ) \sin (e+f x)-5 a d^3 \sin ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac {\int \frac {a^2 \left (2 c^3+9 c^2 d+18 c d^2-14 d^3\right )+15 a^2 d^3 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=\frac {d^3 x}{a^3}-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac {\left ((c-d) \left (2 c^2+11 c d+29 d^2\right )\right ) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=\frac {d^3 x}{a^3}-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}\\ \end {align*}

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Mathematica [B]  time = 5.61, size = 408, normalized size = 2.87 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (40 c^3 \sin \left (\frac {1}{2} (e+f x)\right )-4 c^3 \sin \left (\frac {5}{2} (e+f x)\right )+30 d \cos \left (\frac {1}{2} (e+f x)\right ) \left (3 c^2+6 c d+d^2 (5 e+5 f x-9)\right )+90 c^2 d \sin \left (\frac {1}{2} (e+f x)\right )-18 c^2 d \sin \left (\frac {5}{2} (e+f x)\right )-5 \cos \left (\frac {3}{2} (e+f x)\right ) \left (4 c^3+18 c^2 d+24 c d^2+d^3 (15 e+15 f x-46)\right )+240 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+90 c d^2 \sin \left (\frac {3}{2} (e+f x)\right )-42 c d^2 \sin \left (\frac {5}{2} (e+f x)\right )-370 d^3 \sin \left (\frac {1}{2} (e+f x)\right )+150 d^3 e \sin \left (\frac {1}{2} (e+f x)\right )+150 d^3 f x \sin \left (\frac {1}{2} (e+f x)\right )-90 d^3 \sin \left (\frac {3}{2} (e+f x)\right )+75 d^3 e \sin \left (\frac {3}{2} (e+f x)\right )+75 d^3 f x \sin \left (\frac {3}{2} (e+f x)\right )+64 d^3 \sin \left (\frac {5}{2} (e+f x)\right )-15 d^3 e \sin \left (\frac {5}{2} (e+f x)\right )-15 d^3 f x \sin \left (\frac {5}{2} (e+f x)\right )-15 d^3 e \cos \left (\frac {5}{2} (e+f x)\right )-15 d^3 f x \cos \left (\frac {5}{2} (e+f x)\right )\right )}{60 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(30*d*(3*c^2 + 6*c*d + d^2*(-9 + 5*e + 5*f*x))*Cos[(e + f*x)/2] - 5*(4*
c^3 + 18*c^2*d + 24*c*d^2 + d^3*(-46 + 15*e + 15*f*x))*Cos[(3*(e + f*x))/2] - 15*d^3*e*Cos[(5*(e + f*x))/2] -
15*d^3*f*x*Cos[(5*(e + f*x))/2] + 40*c^3*Sin[(e + f*x)/2] + 90*c^2*d*Sin[(e + f*x)/2] + 240*c*d^2*Sin[(e + f*x
)/2] - 370*d^3*Sin[(e + f*x)/2] + 150*d^3*e*Sin[(e + f*x)/2] + 150*d^3*f*x*Sin[(e + f*x)/2] + 90*c*d^2*Sin[(3*
(e + f*x))/2] - 90*d^3*Sin[(3*(e + f*x))/2] + 75*d^3*e*Sin[(3*(e + f*x))/2] + 75*d^3*f*x*Sin[(3*(e + f*x))/2]
- 4*c^3*Sin[(5*(e + f*x))/2] - 18*c^2*d*Sin[(5*(e + f*x))/2] - 42*c*d^2*Sin[(5*(e + f*x))/2] + 64*d^3*Sin[(5*(
e + f*x))/2] - 15*d^3*e*Sin[(5*(e + f*x))/2] - 15*d^3*f*x*Sin[(5*(e + f*x))/2]))/(60*a^3*f*(1 + Sin[e + f*x])^
3)

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fricas [B]  time = 0.44, size = 352, normalized size = 2.48 \[ -\frac {60 \, d^{3} f x - {\left (15 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 21 \, c d^{2} + 32 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, c^{3} + 9 \, c^{2} d - 9 \, c d^{2} + 3 \, d^{3} - {\left (45 \, d^{3} f x + 4 \, c^{3} + 18 \, c^{2} d - 3 \, c d^{2} - 19 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, d^{3} f x - 3 \, c^{3} - 6 \, c^{2} d - 9 \, c d^{2} + 18 \, d^{3}\right )} \cos \left (f x + e\right ) + {\left (60 \, d^{3} f x + 3 \, c^{3} - 9 \, c^{2} d + 9 \, c d^{2} - 3 \, d^{3} - {\left (15 \, d^{3} f x + 2 \, c^{3} + 9 \, c^{2} d + 21 \, c d^{2} - 32 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 6 \, c d^{2} + 17 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

-1/15*(60*d^3*f*x - (15*d^3*f*x - 2*c^3 - 9*c^2*d - 21*c*d^2 + 32*d^3)*cos(f*x + e)^3 - 3*c^3 + 9*c^2*d - 9*c*
d^2 + 3*d^3 - (45*d^3*f*x + 4*c^3 + 18*c^2*d - 3*c*d^2 - 19*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 3*c^3 - 6*c^
2*d - 9*c*d^2 + 18*d^3)*cos(f*x + e) + (60*d^3*f*x + 3*c^3 - 9*c^2*d + 9*c*d^2 - 3*d^3 - (15*d^3*f*x + 2*c^3 +
9*c^2*d + 21*c*d^2 - 32*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 2*c^3 - 9*c^2*d - 6*c*d^2 + 17*d^3)*cos(f*x + e
))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(
f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

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giac [B]  time = 0.23, size = 280, normalized size = 1.97 \[ \frac {\frac {15 \, {\left (f x + e\right )} d^{3}}{a^{3}} - \frac {2 \, {\left (15 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 75 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 60 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 145 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 95 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 22 \, d^{3}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]

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))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*d^3/a^3 - 2*(15*c^3*tan(1/2*f*x + 1/2*e)^4 - 15*d^3*tan(1/2*f*x + 1/2*e)^4 + 30*c^3*tan(1/2
*f*x + 1/2*e)^3 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 75*d^3*tan(1/2*f*x + 1/2*e)^3 + 40*c^3*tan(1/2*f*x + 1/2*e
)^2 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^2 + 60*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 145*d^3*tan(1/2*f*x + 1/2*e)^2 + 20*
c^3*tan(1/2*f*x + 1/2*e) + 45*c^2*d*tan(1/2*f*x + 1/2*e) + 30*c*d^2*tan(1/2*f*x + 1/2*e) - 95*d^3*tan(1/2*f*x
+ 1/2*e) + 7*c^3 + 9*c^2*d + 6*c*d^2 - 22*d^3)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

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maple [B]  time = 0.25, size = 438, normalized size = 3.08 \[ \frac {2 d^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{3} f}-\frac {2 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 d^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {4 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {6 c^{2} d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {2 d^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {12 c^{2} d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {12 c \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {4 d^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8 c^{3}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {24 c^{2} d}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {24 c \,d^{2}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {8 d^{3}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {16 c^{3}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {12 c^{2} d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8 c \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 d^{3}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}} \]

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))^3,x)

[Out]

2/a^3/f*d^3*arctan(tan(1/2*f*x+1/2*e))-2*c^3/a^3/f/(tan(1/2*f*x+1/2*e)+1)+2/a^3/f/(tan(1/2*f*x+1/2*e)+1)*d^3+4
*c^3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2-6/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2*c^2*d+2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2*d^
3+4*c^3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4-12/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*c^2*d+12/a^3/f/(tan(1/2*f*x+1/2*e)+1)
^4*c*d^2-4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*d^3-8/5*c^3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5+24/5/a^3/f/(tan(1/2*f*x+1
/2*e)+1)^5*c^2*d-24/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*c*d^2+8/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*d^3-16/3*c^3/a^3
/f/(tan(1/2*f*x+1/2*e)+1)^3+12/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*c^2*d-8/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*c*d^2+4/3
/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*d^3

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maxima [B]  time = 0.44, size = 784, normalized size = 5.52 \[ \text {result too large to display} \]

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))^3,x, algorithm="maxima")

[Out]

2/15*(d^3*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1
) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) +
1))/a^3) - c^3*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^
3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) +
1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*c*d^2*(5*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 9*c^2*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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mupad [B]  time = 9.90, size = 240, normalized size = 1.69 \[ \frac {d^3\,x}{a^3}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,c^3-2\,d^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {16\,c^3}{3}+6\,c^2\,d+8\,c\,d^2-\frac {58\,d^3}{3}\right )+\frac {4\,c\,d^2}{5}+\frac {6\,c^2\,d}{5}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,c^3+6\,c^2\,d-10\,d^3\right )+\frac {14\,c^3}{15}-\frac {44\,d^3}{15}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {8\,c^3}{3}+6\,c^2\,d+4\,c\,d^2-\frac {38\,d^3}{3}\right )}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]

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))^3,x)

[Out]

(d^3*x)/a^3 - (tan(e/2 + (f*x)/2)^4*(2*c^3 - 2*d^3) + tan(e/2 + (f*x)/2)^2*(8*c*d^2 + 6*c^2*d + (16*c^3)/3 - (
58*d^3)/3) + (4*c*d^2)/5 + (6*c^2*d)/5 + tan(e/2 + (f*x)/2)^3*(6*c^2*d + 4*c^3 - 10*d^3) + (14*c^3)/15 - (44*d
^3)/15 + tan(e/2 + (f*x)/2)*(4*c*d^2 + 6*c^2*d + (8*c^3)/3 - (38*d^3)/3))/(f*(10*a^3*tan(e/2 + (f*x)/2)^2 + 10
*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x
)/2)))

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sympy [A]  time = 25.97, size = 2640, normalized size = 18.59 \[ \text {result too large to display} \]

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))**3,x)

[Out]

Piecewise((-30*c**3*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a
**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c**3
*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x
/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*c**3*tan(e/2 + f*x/2)**
2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f
*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*c**3*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 +
f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 +
75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 14*c**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)
**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f
) - 90*c**2*d*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*
tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 90*c**2*d*tan
(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)*
*3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 90*c**2*d*tan(e/2 + f*x/2)/(15
*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(
e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 18*c**2*d/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3
*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 +
f*x/2) + 15*a**3*f) - 120*c*d**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x
/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**
3*f) - 60*c*d**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*
tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 12*c*d**2/(15
*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(
e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15*d**3*f*x*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2
+ f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2
+ 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 75*d**3*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 +
75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*ta
n(e/2 + f*x/2) + 15*a**3*f) + 150*d**3*f*x*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(
e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2)
+ 15*a**3*f) + 150*d**3*f*x*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**
4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f)
+ 75*d**3*f*x*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan
(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15*d**3*f*x/(15*
a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e
/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 30*d**3*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*
x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75
*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 150*d**3*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*
f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 +
f*x/2) + 15*a**3*f) + 290*d**3*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)
**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f
) + 190*d**3*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(
e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 44*d**3/(15*a**3*
f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 +
f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*(c + d*sin(e))**3/(a*sin(e) + a)**3, True))

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