3.6.79 \(\int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx\) [579]

3.6.79.1 Optimal result

Integrand size = 14, antiderivative size = 461 \[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=-\frac {i x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {i x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {3 x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {3 x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {3 \operatorname {PolyLog}\left (4,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{4 \sqrt {4 a^2-b^2}}-\frac {3 \operatorname {PolyLog}\left (4,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{4 \sqrt {4 a^2-b^2}} \]

output

-I*x^3*ln(1-I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+I*x^ 
3*ln(1-I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)-3/2*x^2*p 
olylog(2,I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+3/2*x^2 
*polylog(2,I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)-3/2*I 
*x*polylog(3,I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+3/2 
*I*x*polylog(3,I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+3 
/4*polylog(4,I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)-3/4 
*polylog(4,I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)
 

3.6.79.2 Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.74 \[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\frac {-4 i x^3 \log \left (1+\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )+4 i x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-6 x^2 \operatorname {PolyLog}\left (2,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )+6 x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-6 i x \operatorname {PolyLog}\left (3,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )+6 i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )+3 \operatorname {PolyLog}\left (4,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )-3 \operatorname {PolyLog}\left (4,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{4 \sqrt {4 a^2-b^2}} \]

input

Integrate[x^3/(a + b*Cos[x]*Sin[x]),x]
 

output

((-4*I)*x^3*Log[1 + (I*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] + (4*I)* 
x^3*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] - 6*x^2*PolyLog[2 
, ((-I)*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] + 6*x^2*PolyLog[2, (I*b 
*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] - (6*I)*x*PolyLog[3, ((-I)*b*E^(( 
2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] + (6*I)*x*PolyLog[3, (I*b*E^((2*I)*x) 
)/(2*a + Sqrt[4*a^2 - b^2])] + 3*PolyLog[4, ((-I)*b*E^((2*I)*x))/(-2*a + S 
qrt[4*a^2 - b^2])] - 3*PolyLog[4, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^ 
2])])/(4*Sqrt[4*a^2 - b^2])
 

3.6.79.3 Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 425, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5095, 3042, 3804, 27, 2694, 27, 2620, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3}{a+b \sin (x) \cos (x)} \, dx\)

\(\Big \downarrow \) 5095

\(\displaystyle \int \frac {x^3}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^3}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3804

\(\displaystyle 2 \int \frac {2 e^{2 i x} x^3}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int \frac {e^{2 i x} x^3}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\)

\(\Big \downarrow \) 2694

\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x^3}{2 \left (2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x^3}{2 \left (2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x^3}{2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x^3}{2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 4 \left (\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {3 \int x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {3 \int x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle 4 \left (\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-i \int x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-i \int x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 4 \left (\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{2} i \int \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{2} i \int \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle 4 \left (\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )de^{2 i x}-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )de^{2 i x}-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle 4 \left (\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{4} \operatorname {PolyLog}\left (4,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^3 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {3 \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-i \left (\frac {1}{4} \operatorname {PolyLog}\left (4,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )\right )\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

input

Int[x^3/(a + b*Cos[x]*Sin[x]),x]
 

output

4*(((-1/2*I)*b*((x^3*Log[1 - (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])]) 
/(2*b) - (3*((I/2)*x^2*PolyLog[2, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^ 
2])] - I*((-1/2*I)*x*PolyLog[3, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2] 
)] + PolyLog[4, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])]/4)))/(2*b)))/ 
Sqrt[4*a^2 - b^2] + ((I/2)*b*((x^3*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4 
*a^2 - b^2])])/(2*b) - (3*((I/2)*x^2*PolyLog[2, (I*b*E^((2*I)*x))/(2*a + S 
qrt[4*a^2 - b^2])] - I*((-1/2*I)*x*PolyLog[3, (I*b*E^((2*I)*x))/(2*a + Sqr 
t[4*a^2 - b^2])] + PolyLog[4, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] 
/4)))/(2*b)))/Sqrt[4*a^2 - b^2])
 

3.6.79.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5095

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + Cos[(c_.) + (d_.)*(x_)]*(b_.)*Sin[(c 
_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[(e + f*x)^m*(a + b*(Sin[2*c + 2* 
d*x]/2))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

3.6.79.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2281 vs. \(2 (389 ) = 778\).

Time = 1.02 (sec) , antiderivative size = 2282, normalized size of antiderivative = 4.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(2282\)

input

int(x^3/(a+b*cos(x)*sin(x)),x,method=_RETURNVERBOSE)
 

output

-4*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*a^2*x^4+6*I/(8*a^2-2* 
b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(4,b*exp(2*I*x)/(-2*I*a+(-(2 
*a-b)*(2*a+b))^(1/2)))*a^2-3/2*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^ 
(1/2))*polylog(4,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*b^2+12/(8 
*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,b*exp(2*I*x)/(-2*I 
*a-(-(2*a-b)*(2*a+b))^(1/2)))*a^2*x-3/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a 
+b))^(1/2))*polylog(3,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2)))*b^2* 
x+3/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/ 
2)*polylog(4,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2)))*a+6*I/(8*a^2- 
2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(4,b*exp(2*I*x)/(-2*I*a-(- 
(2*a-b)*(2*a+b))^(1/2)))*a^2+6*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^ 
(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*polylog(3,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*( 
2*a+b))^(1/2)))*a*x+4*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*(- 
(2*a-b)*(2*a+b))^(1/2)*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)) 
)*a*x^3-3/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,b*exp( 
2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*b^2*x-3/(8*a^2-2*b^2)/(-2*I*a+(- 
(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*polylog(4,b*exp(2*I*x)/(- 
2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a+I/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+ 
b))^(1/2))*b^2*x^4+I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*b^2*x 
^4-6*I/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b...
 

3.6.79.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3308 vs. \(2 (367) = 734\).

Time = 0.98 (sec) , antiderivative size = 3308, normalized size of antiderivative = 7.18 \[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\text {Too large to display} \]

input

integrate(x^3/(a+b*cos(x)*sin(x)),x, algorithm="fricas")
 

output

-1/2*(b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) - (b 
*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2 
)/b^2) + 2*I*a)/b) - b)/b) + b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a* 
cos(x) - 2*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sq 
rt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) - b*x^3*sqrt(-(4*a^2 - 
b^2)/b^2)*log(-((2*I*a*cos(x) - 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt( 
-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) 
 - b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos(x) + 2*a*sin(x) + (b*c 
os(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2) 
/b^2) + 2*I*a)/b) - b)/b) + b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*co 
s(x) - 2*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt 
((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) + b*x^3*sqrt(-(4*a^2 - b^ 
2)/b^2)*log(-((-2*I*a*cos(x) + 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt(- 
(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) - 
 b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) + (b*cos( 
x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^ 
2) - 2*I*a)/b) - b)/b) - b*x^3*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos( 
x) - 2*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(- 
(b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) + 3*I*b*x^2*sqrt(-(4*a^2 - 
 b^2)/b^2)*dilog(((2*I*a*cos(x) + 2*a*sin(x) - (b*cos(x) - I*b*sin(x))*...
 

3.6.79.6 Sympy [F]

\[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x^{3}}{a + b \sin {\left (x \right )} \cos {\left (x \right )}}\, dx \]

input

integrate(x**3/(a+b*cos(x)*sin(x)),x)
 

output

Integral(x**3/(a + b*sin(x)*cos(x)), x)
 

3.6.79.7 Maxima [F]

\[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x^{3}}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \]

input

integrate(x^3/(a+b*cos(x)*sin(x)),x, algorithm="maxima")
 

output

integrate(x^3/(b*cos(x)*sin(x) + a), x)
 

3.6.79.8 Giac [F]

\[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x^{3}}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \]

input

integrate(x^3/(a+b*cos(x)*sin(x)),x, algorithm="giac")
 

output

integrate(x^3/(b*cos(x)*sin(x) + a), x)
 

3.6.79.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x^3}{a+b\,\cos \left (x\right )\,\sin \left (x\right )} \,d x \]

input

int(x^3/(a + b*cos(x)*sin(x)),x)
 

output

int(x^3/(a + b*cos(x)*sin(x)), x)