\(\int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx\) [448]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\frac {27 \left (2 c^2-6 c d+7 d^2\right ) x}{2 d^3}-\frac {54 (c-d)^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \sqrt {c^2-d^2} f}+\frac {27 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) (27+27 \sin (e+f x))}{2 d f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2842, 3047, 3102, 2814, 2739, 632, 210} \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=-\frac {2 a^3 (c-d)^3 \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d^3 f \sqrt {c^2-d^2}}+\frac {a^3 x \left (2 c^2-6 c d+7 d^2\right )}{2 d^3}+\frac {a^3 (2 c-5 d) \cos (e+f x)}{2 d^2 f}-\frac {\cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f} \]

[In]

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

[Out]

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

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 2814

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 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/(
d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d
*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 3102

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.85 \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\frac {27 \left (2 \left (2 c^2-6 c d+7 d^2\right ) (e+f x)-\frac {8 (c-d)^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+4 (c-3 d) d \cos (e+f x)-d^2 \sin (2 (e+f x))\right )}{4 d^3 f} \]

[In]

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

[Out]

(27*(2*(2*c^2 - 6*c*d + 7*d^2)*(e + f*x) - (8*(c - d)^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt
[c^2 - d^2] + 4*(c - 3*d)*d*Cos[e + f*x] - d^2*Sin[2*(e + f*x)]))/(4*d^3*f)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\left (-c^{3}+3 c^{2} d -3 c \,d^{2}+d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}+\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}}{2}+\left (c d -3 d^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+c d -3 d^{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (2 c^{2}-6 c d +7 d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{3}}\right )}{f}\)	\(184\)
default	\(\frac {2 a^{3} \left (\frac {\left (-c^{3}+3 c^{2} d -3 c \,d^{2}+d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}+\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}}{2}+\left (c d -3 d^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+c d -3 d^{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (2 c^{2}-6 c d +7 d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{3}}\right )}{f}\)	\(184\)
risch	\(\frac {a^{3} x \,c^{2}}{d^{3}}-\frac {3 a^{3} x c}{d^{2}}+\frac {7 a^{3} x}{2 d}+\frac {a^{3} {\mathrm e}^{i \left (f x +e \right )} c}{2 d^{2} f}-\frac {3 a^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 d f}+\frac {a^{3} {\mathrm e}^{-i \left (f x +e \right )} c}{2 d^{2} f}-\frac {3 a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 d f}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c^{2}}{\left (c +d \right ) f \,d^{3}}-\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right ) f \,d^{2}}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c^{2}}{\left (c +d \right ) f \,d^{3}}+\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right ) f \,d^{2}}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}-\frac {a^{3} \sin \left (2 f x +2 e \right )}{4 d f}\)	\(506\)

[In]

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

[Out]

2/f*a^3*((-c^3+3*c^2*d-3*c*d^2+d^3)/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
))+1/d^3*((1/2*tan(1/2*f*x+1/2*e)^3*d^2+(c*d-3*d^2)*tan(1/2*f*x+1/2*e)^2-1/2*d^2*tan(1/2*f*x+1/2*e)+c*d-3*d^2)
/(1+tan(1/2*f*x+1/2*e)^2)^2+1/2*(2*c^2-6*c*d+7*d^2)*arctan(tan(1/2*f*x+1/2*e))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 404, normalized size of antiderivative = 3.11 \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\left [-\frac {a^{3} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} f x - {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}, -\frac {a^{3} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} f x - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}\right ] \]

[In]

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

[Out]

[-1/2*(a^3*d^2*cos(f*x + e)*sin(f*x + e) - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*f*x - (a^3*c^2 - 2*a^3*c*d + a^
3*d^2)*sqrt(-(c - d)/(c + d))*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*((c^2 + c
*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*
sin(f*x + e) - c^2 - d^2)) - 2*(a^3*c*d - 3*a^3*d^2)*cos(f*x + e))/(d^3*f), -1/2*(a^3*d^2*cos(f*x + e)*sin(f*x
+ e) - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*f*x - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*sqrt((c - d)/(c + d))*arct
an(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) - 2*(a^3*c*d - 3*a^3*d^2)*cos(f*x + e))
/(d^3*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.78 \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\frac {\frac {{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} {\left (f x + e\right )}}{d^{3}} - \frac {4 \, {\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{3}} + \frac {2 \, {\left (a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{3} c - 6 \, a^{3} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{2}}}{2 \, f} \]

[In]

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

[Out]

1/2*((2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*(f*x + e)/d^3 - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(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)))/(sqrt(c^2 - d^2)
*d^3) + 2*(a^3*d*tan(1/2*f*x + 1/2*e)^3 + 2*a^3*c*tan(1/2*f*x + 1/2*e)^2 - 6*a^3*d*tan(1/2*f*x + 1/2*e)^2 - a^
3*d*tan(1/2*f*x + 1/2*e) + 2*a^3*c - 6*a^3*d)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*d^2))/f

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 3382, normalized size of antiderivative = 26.02 \[ \int \frac {(3+3 \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

((2*(a^3*c - 3*a^3*d))/d^2 + (a^3*tan(e/2 + (f*x)/2)^3)/d - (a^3*tan(e/2 + (f*x)/2))/d + (2*tan(e/2 + (f*x)/2)
^2*(a^3*c - 3*a^3*d))/d^2)/(f*(2*tan(e/2 + (f*x)/2)^2 + tan(e/2 + (f*x)/2)^4 + 1)) + (2*a^3*atan(((a^3*((8*(49
*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*
(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*
c^7*d^2))/d^6 + (a^3*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d
^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*
d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (
7*d^2)/2))/d^3 + (a^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))
/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*
d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (
8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 +
(8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)
/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2))/d^3)/((16*(2*a^9*c^7 - 14*a^9*c*d^6 - 8*a^9*c^6*d + 47*a^9*c^2*d^5 - 5
5*a^9*c^3*d^4 + 21*a^9*c^4*d^3 + 7*a^9*c^5*d^2))/d^5 + (16*tan(e/2 + (f*x)/2)*(8*a^9*c^8 - 98*a^9*c*d^7 - 72*a
^9*c^7*d + 462*a^9*c^2*d^6 - 926*a^9*c^3*d^5 + 1034*a^9*c^4*d^4 - 704*a^9*c^5*d^3 + 296*a^9*c^6*d^2))/d^6 - (a
^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 +
(f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^
3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c
^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)
/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 -
3*c*d + (7*d^2)/2)*1i)/d^3 + (a^3*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*
a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 -
116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*
d^6))/d^5 - (8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*
(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*
c*d + (7*d^2)/2)*1i)/d^3)*(c^2 - 3*c*d + (7*d^2)/2)*1i)/d^3))*(c^2 - 3*c*d + (7*d^2)/2))/(d^3*f) + (a^3*atan((
(a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6
*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 11
6*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*tan(e/2 + (f*x)/2)*
(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a
^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)
^(1/2))/(d^3*(c + d))))/(d^3*(c + d)))*1i)/(d^3*(c + d)) + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6
- 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^
8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^
6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (8*tan(e/2 + (f
*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 + (8*tan(e/2 +
(f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d)))*1i)/(d^3*(c
+ d)))/((16*(2*a^9*c^7 - 14*a^9*c*d^6 - 8*a^9*c^6*d + 47*a^9*c^2*d^5 - 55*a^9*c^3*d^4 + 21*a^9*c^4*d^3 + 7*a^9
*c^5*d^2))/d^5 + (16*tan(e/2 + (f*x)/2)*(8*a^9*c^8 - 98*a^9*c*d^7 - 72*a^9*c^7*d + 462*a^9*c^2*d^6 - 926*a^9*c
^3*d^5 + 1034*a^9*c^4*d^4 - 704*a^9*c^5*d^3 + 296*a^9*c^6*d^2))/d^6 - (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*
a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(
94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c
^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a^3*c
^3*d^7 - 8*a^3*c^4*d^6))/d^6 - (8*(14*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 + (a^3*(32*c^2*d^3 + (8
*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c + d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d))))/
(d^3*(c + d)) + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(49*a^6*c^2*d^6 - 84*a^6*c^3*d^5 + 64*a^6*c^4*d^4 - 24*a^6
*c^5*d^3 + 4*a^6*c^6*d^2))/d^5 + (8*tan(e/2 + (f*x)/2)*(94*a^6*c*d^8 - 144*a^6*c^2*d^7 + 19*a^6*c^3*d^6 + 116*
a^6*c^4*d^5 - 116*a^6*c^5*d^4 + 48*a^6*c^6*d^3 - 8*a^6*c^7*d^2))/d^6 + (a^3*(-(c + d)*(c - d)^5)^(1/2)*((8*(14
*a^3*c*d^8 - 16*a^3*c^2*d^7 + 2*a^3*c^3*d^6))/d^5 - (8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 24*a^3*c^2*d^8 + 24*a
^3*c^3*d^7 - 8*a^3*c^4*d^6))/d^6 + (a^3*(32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(-(c
+ d)*(c - d)^5)^(1/2))/(d^3*(c + d))))/(d^3*(c + d))))/(d^3*(c + d))))*(-(c + d)*(c - d)^5)^(1/2)*2i)/(d^3*f*
(c + d))