3.5.51 \(\int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^4} \, dx\) [451]

Optimal. Leaf size=207 \[ \frac {5 a^3 \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d)^3 \sqrt {c^2-d^2} f}+\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))} \]

[Out]

1/3*(c-d)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/(c+d)/f/(c+d*sin(f*x+e))^3+1/6*a^3*(c-d)*(2*c+7*d)*cos(f*x+e)/d^2/
(c+d)^2/f/(c+d*sin(f*x+e))^2-1/6*a^3*(2*c^2+9*c*d+22*d^2)*cos(f*x+e)/d^2/(c+d)^3/f/(c+d*sin(f*x+e))+5*a^3*arct
an((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/(c+d)^3/f/(c^2-d^2)^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2841, 3047, 3100, 2833, 12, 2739, 632, 210} \begin {gather*} \frac {5 a^3 \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d)^3 \sqrt {c^2-d^2}}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 f (c+d)^3 (c+d \sin (e+f x))}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 f (c+d)^2 (c+d \sin (e+f x))^2}+\frac {(c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{3 d f (c+d) (c+d \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^4,x]

[Out]

(5*a^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^3*Sqrt[c^2 - d^2]*f) + ((c - d)*Cos[e + f*x]
*(a^3 + a^3*Sin[e + f*x]))/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^3) + (a^3*(c - d)*(2*c + 7*d)*Cos[e + f*x])/(6*
d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^2) - (a^3*(2*c^2 + 9*c*d + 22*d^2)*Cos[e + f*x])/(6*d^2*(c + d)^3*f*(c +
d*Sin[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*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[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 3047

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 3100

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^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^4} \, dx &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a \int \frac {(a+a \sin (e+f x)) (a (c-7 d)-2 a (c+2 d) \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a \int \frac {a^2 (c-7 d)+\left (a^2 (c-7 d)-2 a^2 (c+2 d)\right ) \sin (e+f x)-2 a^2 (c+2 d) \sin ^2(e+f x)}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}+\frac {a \int \frac {2 a^2 (c-d) d (c+11 d)+a^2 (c-d) \left (2 c^2+7 c d+15 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{6 (c-d) d^2 (c+d)^2}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}-\frac {a \int -\frac {15 a^2 (c-d)^2 d^2}{c+d \sin (e+f x)} \, dx}{6 (c-d)^2 d^2 (c+d)^3}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}+\frac {\left (5 a^3\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 (c+d)^3}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}+\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^3 f}\\ &=\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}-\frac {\left (10 a^3\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^3 f}\\ &=\frac {5 a^3 \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d)^3 \sqrt {c^2-d^2} f}+\frac {(c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{3 d (c+d) f (c+d \sin (e+f x))^3}+\frac {a^3 (c-d) (2 c+7 d) \cos (e+f x)}{6 d^2 (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^3 \left (2 c^2+9 c d+22 d^2\right ) \cos (e+f x)}{6 d^2 (c+d)^3 f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.50, size = 178, normalized size = 0.86 \begin {gather*} \frac {a^3 \cos (e+f x) \left (\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right )}{(-c-d)^{5/2} \sqrt {c-d} \sqrt {\cos ^2(e+f x)}}-\frac {(1+\sin (e+f x))^2}{(c+d \sin (e+f x))^3}-\frac {5 (1+\sin (e+f x))}{2 (c+d) (c+d \sin (e+f x))^2}-\frac {15}{2 (c+d)^2 (c+d \sin (e+f x))}\right )}{3 (c+d) f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^4,x]

[Out]

(a^3*Cos[e + f*x]*((15*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/((
-c - d)^(5/2)*Sqrt[c - d]*Sqrt[Cos[e + f*x]^2]) - (1 + Sin[e + f*x])^2/(c + d*Sin[e + f*x])^3 - (5*(1 + Sin[e
+ f*x]))/(2*(c + d)*(c + d*Sin[e + f*x])^2) - 15/(2*(c + d)^2*(c + d*Sin[e + f*x]))))/(3*(c + d)*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(465\) vs. \(2(196)=392\).
time = 0.89, size = 466, normalized size = 2.25

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\frac {\left (3 c^{3}-6 c^{2} d -6 c \,d^{2}-2 d^{3}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (6 c^{4}-3 c^{3} d +30 c^{2} d^{2}+18 d^{3} c +4 d^{4}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c^{2} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {d \left (66 c^{4}+27 c^{3} d +50 c^{2} d^{2}+18 d^{3} c +4 d^{4}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c^{3} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (8 c^{4}+6 c^{3} d +30 c^{2} d^{2}+9 d^{3} c +2 d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{2} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (3 c^{3}+38 c^{2} d +12 c \,d^{2}+2 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {22 c^{2}+9 c d +2 d^{2}}{6 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}}{\left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{3}}+\frac {5 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(466\)
default	\(\frac {2 a^{3} \left (\frac {\frac {\left (3 c^{3}-6 c^{2} d -6 c \,d^{2}-2 d^{3}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (6 c^{4}-3 c^{3} d +30 c^{2} d^{2}+18 d^{3} c +4 d^{4}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c^{2} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {d \left (66 c^{4}+27 c^{3} d +50 c^{2} d^{2}+18 d^{3} c +4 d^{4}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c^{3} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (8 c^{4}+6 c^{3} d +30 c^{2} d^{2}+9 d^{3} c +2 d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{2} \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {\left (3 c^{3}+38 c^{2} d +12 c \,d^{2}+2 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {22 c^{2}+9 c d +2 d^{2}}{6 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}}{\left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{3}}+\frac {5 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(466\)
risch	\(\frac {i a^{3} \left (8 c^{5} {\mathrm e}^{3 i \left (f x +e \right )}-22 i d^{5}+9 d^{5} {\mathrm e}^{5 i \left (f x +e \right )}-9 d^{5} {\mathrm e}^{i \left (f x +e \right )}-54 i c^{3} d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-90 i c^{2} d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+9 i c \,d^{4} {\mathrm e}^{4 i \left (f x +e \right )}+12 i c^{4} d \,{\mathrm e}^{2 i \left (f x +e \right )}+180 i c^{2} d^{3} {\mathrm e}^{2 i \left (f x +e \right )}+54 i c^{3} d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+36 i {\mathrm e}^{2 i \left (f x +e \right )} c \,d^{4}-12 i c^{4} d \,{\mathrm e}^{4 i \left (f x +e \right )}-2 i c^{2} d^{3}-9 i c \,d^{4}-6 c^{3} d^{2} {\mathrm e}^{i \left (f x +e \right )}+36 c^{4} d \,{\mathrm e}^{3 i \left (f x +e \right )}+100 c^{3} d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+54 c^{2} d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+132 c \,d^{4} {\mathrm e}^{3 i \left (f x +e \right )}-36 c^{2} d^{3} {\mathrm e}^{i \left (f x +e \right )}-114 c \,d^{4} {\mathrm e}^{i \left (f x +e \right )}+48 i d^{5} {\mathrm e}^{2 i \left (f x +e \right )}-18 i d^{5} {\mathrm e}^{4 i \left (f x +e \right )}-18 c^{2} d^{3} {\mathrm e}^{5 i \left (f x +e \right )}-18 c \,d^{4} {\mathrm e}^{5 i \left (f x +e \right )}-6 d^{2} c^{3} {\mathrm e}^{5 i \left (f x +e \right )}\right )}{3 \left (c +d \right )^{3} \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )^{3} f \,d^{3}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} f}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} f}\)	\(592\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

2/f*a^3*((1/2*(3*c^3-6*c^2*d-6*c*d^2-2*d^3)/c/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5-1/2*(6*c^4-3*c^3*
d+30*c^2*d^2+18*c*d^3+4*d^4)/c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^4-1/3/c^3*d*(66*c^4+27*c^3*d+50*
c^2*d^2+18*c*d^3+4*d^4)/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^3-(8*c^4+6*c^3*d+30*c^2*d^2+9*c*d^3+2*d^4
)/c^2/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^2-1/2*(3*c^3+38*c^2*d+12*c*d^2+2*d^3)/c/(c^3+3*c^2*d+3*c*d^
2+d^3)*tan(1/2*f*x+1/2*e)-1/6*(22*c^2+9*c*d+2*d^2)/(c^3+3*c^2*d+3*c*d^2+d^3))/(c*tan(1/2*f*x+1/2*e)^2+2*d*tan(
1/2*f*x+1/2*e)+c)^3+5/2/(c^3+3*c^2*d+3*c*d^2+d^3)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2
-d^2)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (204) = 408\).
time = 0.41, size = 1135, normalized size = 5.48 \begin {gather*} \left [-\frac {2 \, {\left (2 \, a^{3} c^{4} + 9 \, a^{3} c^{3} d + 20 \, a^{3} c^{2} d^{2} - 9 \, a^{3} c d^{3} - 22 \, a^{3} d^{4}\right )} \cos \left (f x + e\right )^{3} - 6 \, {\left (3 \, a^{3} c^{4} + 16 \, a^{3} c^{3} d - 16 \, a^{3} c d^{3} - 3 \, a^{3} d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 15 \, {\left (3 \, a^{3} c d^{2} \cos \left (f x + e\right )^{2} - a^{3} c^{3} - 3 \, a^{3} c d^{2} + {\left (a^{3} d^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{3} c^{2} d - a^{3} d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 12 \, {\left (4 \, a^{3} c^{4} + 3 \, a^{3} c^{3} d - 3 \, a^{3} c d^{3} - 4 \, a^{3} d^{4}\right )} \cos \left (f x + e\right )}{12 \, {\left (3 \, {\left (c^{6} d^{2} + 3 \, c^{5} d^{3} + 2 \, c^{4} d^{4} - 2 \, c^{3} d^{5} - 3 \, c^{2} d^{6} - c d^{7}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{8} + 3 \, c^{7} d + 5 \, c^{6} d^{2} + 7 \, c^{5} d^{3} + 3 \, c^{4} d^{4} - 7 \, c^{3} d^{5} - 9 \, c^{2} d^{6} - 3 \, c d^{7}\right )} f + {\left ({\left (c^{5} d^{3} + 3 \, c^{4} d^{4} + 2 \, c^{3} d^{5} - 2 \, c^{2} d^{6} - 3 \, c d^{7} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - {\left (3 \, c^{7} d + 9 \, c^{6} d^{2} + 7 \, c^{5} d^{3} - 3 \, c^{4} d^{4} - 7 \, c^{3} d^{5} - 5 \, c^{2} d^{6} - 3 \, c d^{7} - d^{8}\right )} f\right )} \sin \left (f x + e\right )\right )}}, -\frac {{\left (2 \, a^{3} c^{4} + 9 \, a^{3} c^{3} d + 20 \, a^{3} c^{2} d^{2} - 9 \, a^{3} c d^{3} - 22 \, a^{3} d^{4}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{3} c^{4} + 16 \, a^{3} c^{3} d - 16 \, a^{3} c d^{3} - 3 \, a^{3} d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 15 \, {\left (3 \, a^{3} c d^{2} \cos \left (f x + e\right )^{2} - a^{3} c^{3} - 3 \, a^{3} c d^{2} + {\left (a^{3} d^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{3} c^{2} d - a^{3} d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) - 6 \, {\left (4 \, a^{3} c^{4} + 3 \, a^{3} c^{3} d - 3 \, a^{3} c d^{3} - 4 \, a^{3} d^{4}\right )} \cos \left (f x + e\right )}{6 \, {\left (3 \, {\left (c^{6} d^{2} + 3 \, c^{5} d^{3} + 2 \, c^{4} d^{4} - 2 \, c^{3} d^{5} - 3 \, c^{2} d^{6} - c d^{7}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{8} + 3 \, c^{7} d + 5 \, c^{6} d^{2} + 7 \, c^{5} d^{3} + 3 \, c^{4} d^{4} - 7 \, c^{3} d^{5} - 9 \, c^{2} d^{6} - 3 \, c d^{7}\right )} f + {\left ({\left (c^{5} d^{3} + 3 \, c^{4} d^{4} + 2 \, c^{3} d^{5} - 2 \, c^{2} d^{6} - 3 \, c d^{7} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - {\left (3 \, c^{7} d + 9 \, c^{6} d^{2} + 7 \, c^{5} d^{3} - 3 \, c^{4} d^{4} - 7 \, c^{3} d^{5} - 5 \, c^{2} d^{6} - 3 \, c d^{7} - d^{8}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

[-1/12*(2*(2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^2*d^2 - 9*a^3*c*d^3 - 22*a^3*d^4)*cos(f*x + e)^3 - 6*(3*a^3*c^4
+ 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x + e)*sin(f*x + e) + 15*(3*a^3*c*d^2*cos(f*x + e)^2 - a^3*c^
3 - 3*a^3*c*d^2 + (a^3*d^3*cos(f*x + e)^2 - 3*a^3*c^2*d - a^3*d^3)*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2
- d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt
(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) - 12*(4*a^3*c^4 + 3*a^3*c^3*d - 3*a^3*c*d
^3 - 4*a^3*d^4)*cos(f*x + e))/(3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^3*d^5 - 3*c^2*d^6 - c*d^7)*f*cos(f*x +
e)^2 - (c^8 + 3*c^7*d + 5*c^6*d^2 + 7*c^5*d^3 + 3*c^4*d^4 - 7*c^3*d^5 - 9*c^2*d^6 - 3*c*d^7)*f + ((c^5*d^3 +
3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f*cos(f*x + e)^2 - (3*c^7*d + 9*c^6*d^2 + 7*c^5*d^3 - 3*c^4
*d^4 - 7*c^3*d^5 - 5*c^2*d^6 - 3*c*d^7 - d^8)*f)*sin(f*x + e)), -1/6*((2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^2*d^
2 - 9*a^3*c*d^3 - 22*a^3*d^4)*cos(f*x + e)^3 - 3*(3*a^3*c^4 + 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x
+ e)*sin(f*x + e) + 15*(3*a^3*c*d^2*cos(f*x + e)^2 - a^3*c^3 - 3*a^3*c*d^2 + (a^3*d^3*cos(f*x + e)^2 - 3*a^3*
c^2*d - a^3*d^3)*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) -
6*(4*a^3*c^4 + 3*a^3*c^3*d - 3*a^3*c*d^3 - 4*a^3*d^4)*cos(f*x + e))/(3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^
3*d^5 - 3*c^2*d^6 - c*d^7)*f*cos(f*x + e)^2 - (c^8 + 3*c^7*d + 5*c^6*d^2 + 7*c^5*d^3 + 3*c^4*d^4 - 7*c^3*d^5 -
9*c^2*d^6 - 3*c*d^7)*f + ((c^5*d^3 + 3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f*cos(f*x + e)^2 - (3
*c^7*d + 9*c^6*d^2 + 7*c^5*d^3 - 3*c^4*d^4 - 7*c^3*d^5 - 5*c^2*d^6 - 3*c*d^7 - d^8)*f)*sin(f*x + e))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**4,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 667 vs. \(2 (204) = 408\).
time = 0.54, size = 667, normalized size = 3.22 \begin {gather*} \frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} a^{3}}{{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {9 \, a^{3} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18 \, a^{3} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18 \, a^{3} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 6 \, a^{3} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18 \, a^{3} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 9 \, a^{3} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 90 \, a^{3} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 54 \, a^{3} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 12 \, a^{3} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 132 \, a^{3} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 54 \, a^{3} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 100 \, a^{3} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 36 \, a^{3} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, a^{3} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 48 \, a^{3} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 36 \, a^{3} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 180 \, a^{3} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 54 \, a^{3} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, a^{3} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, a^{3} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 114 \, a^{3} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 36 \, a^{3} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a^{3} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 22 \, a^{3} c^{5} - 9 \, a^{3} c^{4} d - 2 \, a^{3} c^{3} d^{2}}{{\left (c^{6} + 3 \, c^{5} d + 3 \, c^{4} d^{2} + c^{3} d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{3}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^4,x, algorithm="giac")

[Out]

1/3*(15*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))*a^3/(
(c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sqrt(c^2 - d^2)) + (9*a^3*c^5*tan(1/2*f*x + 1/2*e)^5 - 18*a^3*c^4*d*tan(1/2*f*
x + 1/2*e)^5 - 18*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 6*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 - 18*a^3*c^5*tan(1
/2*f*x + 1/2*e)^4 + 9*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^4 - 90*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^4 - 54*a^3*c^2*d^
3*tan(1/2*f*x + 1/2*e)^4 - 12*a^3*c*d^4*tan(1/2*f*x + 1/2*e)^4 - 132*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 54*a^3
*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 100*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 36*a^3*c*d^4*tan(1/2*f*x + 1/2*e)^3
- 8*a^3*d^5*tan(1/2*f*x + 1/2*e)^3 - 48*a^3*c^5*tan(1/2*f*x + 1/2*e)^2 - 36*a^3*c^4*d*tan(1/2*f*x + 1/2*e)^2
- 180*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 54*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 - 12*a^3*c*d^4*tan(1/2*f*x +
1/2*e)^2 - 9*a^3*c^5*tan(1/2*f*x + 1/2*e) - 114*a^3*c^4*d*tan(1/2*f*x + 1/2*e) - 36*a^3*c^3*d^2*tan(1/2*f*x +
1/2*e) - 6*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e) - 22*a^3*c^5 - 9*a^3*c^4*d - 2*a^3*c^3*d^2)/((c^6 + 3*c^5*d + 3*c^
4*d^2 + c^3*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^3))/f

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Mupad [B]
time = 10.36, size = 649, normalized size = 3.14 \begin {gather*} \frac {5\,a^3\,\mathrm {atan}\left (\frac {\left (\frac {5\,a^3\,\left (2\,c^3\,d+6\,c^2\,d^2+6\,c\,d^3+2\,d^4\right )}{2\,{\left (c+d\right )}^{7/2}\,\sqrt {c-d}\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {5\,a^3\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (c+d\right )}^{7/2}\,\sqrt {c-d}}\right )\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}{5\,a^3}\right )}{f\,{\left (c+d\right )}^{7/2}\,\sqrt {c-d}}-\frac {\frac {22\,a^3\,c^2+9\,a^3\,c\,d+2\,a^3\,d^2}{3\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (-3\,c^3+6\,c^2\,d+6\,c\,d^2+2\,d^3\right )}{c\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,c^3+38\,c^2\,d+12\,c\,d^2+2\,d^3\right )}{c\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,c^4+6\,c^3\,d+30\,c^2\,d^2+9\,c\,d^3+2\,d^4\right )}{c^2\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (6\,c^4-3\,c^3\,d+30\,c^2\,d^2+18\,c\,d^3+4\,d^4\right )}{c^2\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}+\frac {2\,a^3\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,c^2+2\,d^2\right )\,\left (22\,c^2+9\,c\,d+2\,d^2\right )}{3\,c^3\,\left (c^3+3\,c^2\,d+3\,c\,d^2+d^3\right )}}{f\,\left (c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (3\,c^3+12\,c\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (3\,c^3+12\,c\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (12\,c^2\,d+8\,d^3\right )+c^3+6\,c^2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+6\,c^2\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c + d*sin(e + f*x))^4,x)

[Out]

(5*a^3*atan((((5*a^3*(6*c*d^3 + 2*c^3*d + 2*d^4 + 6*c^2*d^2))/(2*(c + d)^(7/2)*(c - d)^(1/2)*(3*c*d^2 + 3*c^2*
d + c^3 + d^3)) + (5*a^3*c*tan(e/2 + (f*x)/2))/((c + d)^(7/2)*(c - d)^(1/2)))*(3*c*d^2 + 3*c^2*d + c^3 + d^3))
/(5*a^3)))/(f*(c + d)^(7/2)*(c - d)^(1/2)) - ((22*a^3*c^2 + 2*a^3*d^2 + 9*a^3*c*d)/(3*(3*c*d^2 + 3*c^2*d + c^3
+ d^3)) + (a^3*tan(e/2 + (f*x)/2)^5*(6*c*d^2 + 6*c^2*d - 3*c^3 + 2*d^3))/(c*(3*c*d^2 + 3*c^2*d + c^3 + d^3))
+ (a^3*tan(e/2 + (f*x)/2)*(12*c*d^2 + 38*c^2*d + 3*c^3 + 2*d^3))/(c*(3*c*d^2 + 3*c^2*d + c^3 + d^3)) + (2*a^3*
tan(e/2 + (f*x)/2)^2*(9*c*d^3 + 6*c^3*d + 8*c^4 + 2*d^4 + 30*c^2*d^2))/(c^2*(3*c*d^2 + 3*c^2*d + c^3 + d^3)) +
(a^3*tan(e/2 + (f*x)/2)^4*(18*c*d^3 - 3*c^3*d + 6*c^4 + 4*d^4 + 30*c^2*d^2))/(c^2*(3*c*d^2 + 3*c^2*d + c^3 +
d^3)) + (2*a^3*d*tan(e/2 + (f*x)/2)^3*(3*c^2 + 2*d^2)*(9*c*d + 22*c^2 + 2*d^2))/(3*c^3*(3*c*d^2 + 3*c^2*d + c^
3 + d^3)))/(f*(c^3*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^2*(12*c*d^2 + 3*c^3) + tan(e/2 + (f*x)/2)^4*(12*c
*d^2 + 3*c^3) + tan(e/2 + (f*x)/2)^3*(12*c^2*d + 8*d^3) + c^3 + 6*c^2*d*tan(e/2 + (f*x)/2) + 6*c^2*d*tan(e/2 +
(f*x)/2)^5))

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