3.2.82 \(\int \frac {x}{(a+a \cos (x))^{3/2}} \, dx\) [182]
Optimal. Leaf size=150 \[ -\frac {1}{a \sqrt {a+a \cos (x)}}-\frac {i x \text {ArcTan}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {i \cos \left (\frac {x}{2}\right ) \text {PolyLog}\left (2,-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {i \cos \left (\frac {x}{2}\right ) \text {PolyLog}\left (2,i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}} \]
[Out]
-1/a/(a+a*cos(x))^(1/2)-I*x*arctan(exp(1/2*I*x))*cos(1/2*x)/a/(a+a*cos(x))^(1/2)+I*cos(1/2*x)*polylog(2,-I*exp
(1/2*I*x))/a/(a+a*cos(x))^(1/2)-I*cos(1/2*x)*polylog(2,I*exp(1/2*I*x))/a/(a+a*cos(x))^(1/2)+1/2*x*tan(1/2*x)/a
/(a+a*cos(x))^(1/2)
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Rubi [A]
time = 0.07, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3400, 4270,
4266, 2317, 2438} \begin {gather*} -\frac {i x \text {ArcTan}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}+\frac {i \text {Li}_2\left (-i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {i \text {Li}_2\left (i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {1}{a \sqrt {a \cos (x)+a}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a \cos (x)+a}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x/(a + a*Cos[x])^(3/2),x]
[Out]
-(1/(a*Sqrt[a + a*Cos[x]])) - (I*x*ArcTan[E^((I/2)*x)]*Cos[x/2])/(a*Sqrt[a + a*Cos[x]]) + (I*Cos[x/2]*PolyLog[
2, (-I)*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) - (I*Cos[x/2]*PolyLog[2, I*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) +
(x*Tan[x/2])/(2*a*Sqrt[a + a*Cos[x]])
Rule 2317
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2438
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rule 3400
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(2*a)^IntPart[n]
*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n])), Int[(c + d*x)^m*Sin[e/2
+ a*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
Rule 4266
Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Rule 4270
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(-b^2)*(c + d*x)*Cot[e + f*x]
*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]
Rubi steps
\begin {align*} \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx &=\frac {\cos \left (\frac {x}{2}\right ) \int x \sec ^3\left (\frac {x}{2}\right ) \, dx}{2 a \sqrt {a+a \cos (x)}}\\ &=-\frac {1}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}+\frac {\cos \left (\frac {x}{2}\right ) \int x \sec \left (\frac {x}{2}\right ) \, dx}{4 a \sqrt {a+a \cos (x)}}\\ &=-\frac {1}{a \sqrt {a+a \cos (x)}}-\frac {i x \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}-\frac {\cos \left (\frac {x}{2}\right ) \int \log \left (1-i e^{\frac {i x}{2}}\right ) \, dx}{2 a \sqrt {a+a \cos (x)}}+\frac {\cos \left (\frac {x}{2}\right ) \int \log \left (1+i e^{\frac {i x}{2}}\right ) \, dx}{2 a \sqrt {a+a \cos (x)}}\\ &=-\frac {1}{a \sqrt {a+a \cos (x)}}-\frac {i x \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}+\frac {\left (i \cos \left (\frac {x}{2}\right )\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {\left (i \cos \left (\frac {x}{2}\right )\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}\\ &=-\frac {1}{a \sqrt {a+a \cos (x)}}-\frac {i x \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {i \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {i \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 165, normalized size = 1.10 \begin {gather*} \frac {\sec \left (\frac {x}{2}\right ) \left (-4 \cos \left (\frac {x}{2}\right )+x \log \left (1-i e^{\frac {i x}{2}}\right )+x \cos (x) \log \left (1-i e^{\frac {i x}{2}}\right )-x \log \left (1+i e^{\frac {i x}{2}}\right )-x \cos (x) \log \left (1+i e^{\frac {i x}{2}}\right )+2 i (1+\cos (x)) \text {PolyLog}\left (2,-i e^{\frac {i x}{2}}\right )-2 i (1+\cos (x)) \text {PolyLog}\left (2,i e^{\frac {i x}{2}}\right )+2 x \sin \left (\frac {x}{2}\right )\right )}{4 a \sqrt {a (1+\cos (x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x/(a + a*Cos[x])^(3/2),x]
[Out]
(Sec[x/2]*(-4*Cos[x/2] + x*Log[1 - I*E^((I/2)*x)] + x*Cos[x]*Log[1 - I*E^((I/2)*x)] - x*Log[1 + I*E^((I/2)*x)]
- x*Cos[x]*Log[1 + I*E^((I/2)*x)] + (2*I)*(1 + Cos[x])*PolyLog[2, (-I)*E^((I/2)*x)] - (2*I)*(1 + Cos[x])*Poly
Log[2, I*E^((I/2)*x)] + 2*x*Sin[x/2]))/(4*a*Sqrt[a*(1 + Cos[x])])
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (a +a \cos \left (x \right )\right )^{\frac {3}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(a+a*cos(x))^(3/2),x)
[Out]
int(x/(a+a*cos(x))^(3/2),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(a+a*cos(x))^(3/2),x, algorithm="maxima")
[Out]
1/3*(8*x*cos(3/2*x)*sin(3*x)^3 - 8*x*cos(3*x)^3*sin(3/2*x) - 8*((3*x*sin(3/2*x) + 2*cos(3/2*x))*cos(2*x) - (3*
x*sin(x) - 2*cos(x))*cos(3/2*x) - (3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(2*x) + (3*x*cos(x) + 3*x - 2*sin(x))*sin
(3/2*x))*cos(3*x)^2 - 48*cos(2*x)^2*cos(3/2*x) - 24*((3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(3*x) + 3*(3*x*sin(3/2
*x) - 2*cos(3/2*x))*cos(2*x) - (9*x*sin(x) + 6*cos(x) + 2)*cos(3/2*x) - 27*(a^2*cos(3*x)^2 + 9*a^2*cos(2*x)^2
+ 9*a^2*cos(x)^2 + a^2*sin(3*x)^2 + 9*a^2*sin(2*x)^2 + 18*a^2*sin(2*x)*sin(x) + 9*a^2*sin(x)^2 + 6*a^2*cos(x)
+ a^2 + 2*(3*a^2*cos(2*x) + 3*a^2*cos(x) + a^2)*cos(3*x) + 6*(3*a^2*cos(x) + a^2)*cos(2*x) + 6*(a^2*sin(2*x) +
a^2*sin(x))*sin(3*x))*integrate(1/3*(x*cos(4*x)*cos(3/2*x) + 4*x*cos(3*x)*cos(3/2*x) + 6*x*cos(2*x)*cos(3/2*x
) + x*sin(4*x)*sin(3/2*x) + 4*x*sin(3*x)*sin(3/2*x) + 6*x*sin(2*x)*sin(3/2*x) + 4*x*sin(3/2*x)*sin(x) + (4*x*c
os(x) + x)*cos(3/2*x))/(a^2*cos(4*x)^2 + 16*a^2*cos(3*x)^2 + 36*a^2*cos(2*x)^2 + 16*a^2*cos(x)^2 + a^2*sin(4*x
)^2 + 16*a^2*sin(3*x)^2 + 36*a^2*sin(2*x)^2 + 48*a^2*sin(2*x)*sin(x) + 16*a^2*sin(x)^2 + 8*a^2*cos(x) + a^2 +
2*(4*a^2*cos(3*x) + 6*a^2*cos(2*x) + 4*a^2*cos(x) + a^2)*cos(4*x) + 8*(6*a^2*cos(2*x) + 4*a^2*cos(x) + a^2)*co
s(3*x) + 12*(4*a^2*cos(x) + a^2)*cos(2*x) + 4*(2*a^2*sin(3*x) + 3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(4*x) + 16*(
3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(3*x)), x) - (3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(3*x) - 3*(3*x*cos(3/2*x) +
2*sin(3/2*x))*sin(2*x) + 3*(3*x*cos(x) + x - 2*sin(x))*sin(3/2*x))*cos(4/3*arctan2(sin(3/2*x), cos(3/2*x)))^2
- 24*((3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(3*x) + 3*(3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(2*x) - (9*x*sin(x) + 6*
cos(x) + 2)*cos(3/2*x) - 27*(a^2*cos(3*x)^2 + 9*a^2*cos(2*x)^2 + 9*a^2*cos(x)^2 + a^2*sin(3*x)^2 + 9*a^2*sin(2
*x)^2 + 18*a^2*sin(2*x)*sin(x) + 9*a^2*sin(x)^2 + 6*a^2*cos(x) + a^2 + 2*(3*a^2*cos(2*x) + 3*a^2*cos(x) + a^2)
*cos(3*x) + 6*(3*a^2*cos(x) + a^2)*cos(2*x) + 6*(a^2*sin(2*x) + a^2*sin(x))*sin(3*x))*integrate(1/3*(x*cos(4*x
)*cos(3/2*x) + 4*x*cos(3*x)*cos(3/2*x) + 6*x*cos(2*x)*cos(3/2*x) + x*sin(4*x)*sin(3/2*x) + 4*x*sin(3*x)*sin(3/
2*x) + 6*x*sin(2*x)*sin(3/2*x) + 4*x*sin(3/2*x)*sin(x) + (4*x*cos(x) + x)*cos(3/2*x))/(a^2*cos(4*x)^2 + 16*a^2
*cos(3*x)^2 + 36*a^2*cos(2*x)^2 + 16*a^2*cos(x)^2 + a^2*sin(4*x)^2 + 16*a^2*sin(3*x)^2 + 36*a^2*sin(2*x)^2 + 4
8*a^2*sin(2*x)*sin(x) + 16*a^2*sin(x)^2 + 8*a^2*cos(x) + a^2 + 2*(4*a^2*cos(3*x) + 6*a^2*cos(2*x) + 4*a^2*cos(
x) + a^2)*cos(4*x) + 8*(6*a^2*cos(2*x) + 4*a^2*cos(x) + a^2)*cos(3*x) + 12*(4*a^2*cos(x) + a^2)*cos(2*x) + 4*(
2*a^2*sin(3*x) + 3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(4*x) + 16*(3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(3*x)), x) -
(3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(3*x) - 3*(3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(2*x) + 3*(3*x*cos(x) + x - 2*
sin(x))*sin(3/2*x))*cos(2/3*arctan2(sin(3/2*x), cos(3/2*x)))^2 - 8*(x*cos(3*x)*sin(3/2*x) + (3*x*sin(3/2*x) -
2*cos(3/2*x))*cos(2*x) - (3*x*sin(x) + 2*cos(x))*cos(3/2*x) - (3*x*cos(3/2*x) - 2*sin(3/2*x))*sin(2*x) + (3*x*
cos(x) + x + 2*sin(x))*sin(3/2*x))*sin(3*x)^2 - 48*cos(3/2*x)*sin(2*x)^2 - 24*((3*x*sin(3/2*x) - 2*cos(3/2*x))
*cos(3*x) + 3*(3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(2*x) - (9*x*sin(x) + 6*cos(x) + 2)*cos(3/2*x) - 27*(a^2*cos(
3*x)^2 + 9*a^2*cos(2*x)^2 + 9*a^2*cos(x)^2 + a^2*sin(3*x)^2 + 9*a^2*sin(2*x)^2 + 18*a^2*sin(2*x)*sin(x) + 9*a^
2*sin(x)^2 + 6*a^2*cos(x) + a^2 + 2*(3*a^2*cos(2*x) + 3*a^2*cos(x) + a^2)*cos(3*x) + 6*(3*a^2*cos(x) + a^2)*co
s(2*x) + 6*(a^2*sin(2*x) + a^2*sin(x))*sin(3*x))*integrate(1/3*(x*cos(4*x)*cos(3/2*x) + 4*x*cos(3*x)*cos(3/2*x
) + 6*x*cos(2*x)*cos(3/2*x) + x*sin(4*x)*sin(3/2*x) + 4*x*sin(3*x)*sin(3/2*x) + 6*x*sin(2*x)*sin(3/2*x) + 4*x*
sin(3/2*x)*sin(x) + (4*x*cos(x) + x)*cos(3/2*x))/(a^2*cos(4*x)^2 + 16*a^2*cos(3*x)^2 + 36*a^2*cos(2*x)^2 + 16*
a^2*cos(x)^2 + a^2*sin(4*x)^2 + 16*a^2*sin(3*x)^2 + 36*a^2*sin(2*x)^2 + 48*a^2*sin(2*x)*sin(x) + 16*a^2*sin(x)
^2 + 8*a^2*cos(x) + a^2 + 2*(4*a^2*cos(3*x) + 6*a^2*cos(2*x) + 4*a^2*cos(x) + a^2)*cos(4*x) + 8*(6*a^2*cos(2*x
) + 4*a^2*cos(x) + a^2)*cos(3*x) + 12*(4*a^2*cos(x) + a^2)*cos(2*x) + 4*(2*a^2*sin(3*x) + 3*a^2*sin(2*x) + 2*a
^2*sin(x))*sin(4*x) + 16*(3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(3*x)), x) - (3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(3
*x) - 3*(3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(2*x) + 3*(3*x*cos(x) + x - 2*sin(x))*sin(3/2*x))*sin(4/3*arctan2(s
in(3/2*x), cos(3/2*x)))^2 - 24*((3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(3*x) + 3*(3*x*sin(3/2*x) - 2*cos(3/2*x))*c
os(2*x) - (9*x*sin(x) + 6*cos(x) + 2)*cos(3/2*x) - 27*(a^2*cos(3*x)^2 + 9*a^2*cos(2*x)^2 + 9*a^2*cos(x)^2 + a^
2*sin(3*x)^2 + 9*a^2*sin(2*x)^2 + 18*a^2*sin(2*x)*sin(x) + 9*a^2*sin(x)^2 + 6*a^2*cos(x) + a^2 + 2*(3*a^2*cos(
2*x) + 3*a^2*cos(x) + a^2)*cos(3*x) + 6*(3*a^2*cos(x) + a^2)*cos(2*x) + 6*(a^2*sin(2*x) + a^2*sin(x))*sin(3*x)
)*integrate(1/3*(x*cos(4*x)*cos(3/2*x) + 4*x*cos(3*x)*cos(3/2*x) + 6*x*cos(2*x)*cos(3/2*x) + x*sin(4*x)*sin(3/
2*x) + 4*x*sin(3*x)*sin(3/2*x) + 6*x*sin(2*x)*sin(3/2*x) + 4*x*sin(3/2*x)*sin(x) + (4*x*cos(x) + x)*cos(3/2*x)
)/(a^2*cos(4*x)^2 + 16*a^2*cos(3*x)^2 + 36*a^2*...
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(a+a*cos(x))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(a*cos(x) + a)*x/(a^2*cos(x)^2 + 2*a^2*cos(x) + a^2), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a \left (\cos {\left (x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(a+a*cos(x))**(3/2),x)
[Out]
Integral(x/(a*(cos(x) + 1))**(3/2), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(a+a*cos(x))^(3/2),x, algorithm="giac")
[Out]
integrate(x/(a*cos(x) + a)^(3/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (a+a\,\cos \left (x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(a + a*cos(x))^(3/2),x)
[Out]
int(x/(a + a*cos(x))^(3/2), x)
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