\(\int x^m \csc ^3(a+2 \log (c x^{\frac {1}{2} \sqrt {-(1+m)^2}})) \, dx\) [302]

Optimal result

Integrand size = 31, antiderivative size = 110 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 (1+m)}-\frac {x^{1+m} \cot \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 \sqrt {-(1+m)^2}} \]

[Out]

1/2*x^(1+m)*csc(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))/(1+m)-1/2*x^(1+m)*cot(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))*
csc(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))/(-(1+m)^2)^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.21 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4606, 4602, 371} \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=-\frac {8 e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{6 i} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),\frac {1}{2} \left (5-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}\right )}{i m-3 \sqrt {-(m+1)^2}+i} \]

[In]

Int[x^m*Csc[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]]^3,x]

[Out]

(-8*E^((3*I)*a)*x^(1 + m)*(c*x^(Sqrt[-(1 + m)^2]/2))^(6*I)*Hypergeometric2F1[3, (3 - (I*(1 + m))/Sqrt[-(1 + m)
^2])/2, (5 - (I*(1 + m))/Sqrt[-(1 + m)^2])/2, E^((2*I)*a)*(c*x^(Sqrt[-(1 + m)^2]/2))^(4*I)])/(I + I*m - 3*Sqrt
[-(1 + m)^2])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4602

Int[Csc[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(-2*I)^p*E^(I*a*d*p), Int[(
e*x)^m*(x^(I*b*d*p)/(1 - E^(2*I*a*d)*x^(2*I*b*d))^p), x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rule 4606

Int[Csc[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Csc[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (2 x^{1+m} \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )^{-\frac {2 (1+m)}{\sqrt {-(1+m)^2}}}\right ) \text {Subst}\left (\int x^{-1+\frac {2 (1+m)}{\sqrt {-(1+m)^2}}} \csc ^3(a+2 \log (x)) \, dx,x,c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )}{\sqrt {-(1+m)^2}} \\ & = \frac {\left (16 i e^{3 i a} x^{1+m} \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )^{-\frac {2 (1+m)}{\sqrt {-(1+m)^2}}}\right ) \text {Subst}\left (\int \frac {x^{(-1+6 i)+\frac {2 (1+m)}{\sqrt {-(1+m)^2}}}}{\left (1-e^{2 i a} x^{4 i}\right )^3} \, dx,x,c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )}{\sqrt {-(1+m)^2}} \\ & = -\frac {8 e^{3 i a} x^{1+m} \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )^{6 i} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i (1+m)}{\sqrt {-(1+m)^2}}\right ),\frac {1}{2} \left (5-\frac {i (1+m)}{\sqrt {-(1+m)^2}}\right ),e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )^{4 i}\right )}{i+i m-3 \sqrt {-(1+m)^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \left (1+m+\sqrt {-(1+m)^2} \cot \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right ) \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 (1+m)^2} \]

[In]

Integrate[x^m*Csc[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]]^3,x]

[Out]

(x^(1 + m)*(1 + m + Sqrt[-(1 + m)^2]*Cot[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]])*Csc[a + 2*Log[c*x^(Sqrt[-(1 + m
)^2]/2)]])/(2*(1 + m)^2)

Maple [A] (verified)

Time = 217.82 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-\frac {\left (\left ({\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}-1\right ) \sqrt {-\left (1+m \right )^{2}}-2 \left (1+m \right ) \tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )\right ) x^{1+m} \left (1+{\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}\right )}{8 \left (1+m \right )^{2} {\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}}\)	\(123\)

[In]

int(x^m*csc(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))^3,x,method=_RETURNVERBOSE)

[Out]

-1/8*((tan(1/2*a+ln(c*x^(1/2*(-(1+m)^2)^(1/2))))^2-1)*(-(1+m)^2)^(1/2)-2*(1+m)*tan(1/2*a+ln(c*x^(1/2*(-(1+m)^2
)^(1/2)))))*x^(1+m)*(1+tan(1/2*a+ln(c*x^(1/2*(-(1+m)^2)^(1/2))))^2)/(1+m)^2/tan(1/2*a+ln(c*x^(1/2*(-(1+m)^2)^(
1/2))))^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=-\frac {2 \, {\left (2 i \, x^{2} x^{2 \, m} e^{\left (3 i \, a + 6 i \, \log \left (c\right )\right )} - i \, e^{\left (5 i \, a + 10 i \, \log \left (c\right )\right )}\right )}}{{\left (m + 1\right )} x^{4} x^{4 \, m} - 2 \, {\left (m + 1\right )} x^{2} x^{2 \, m} e^{\left (2 i \, a + 4 i \, \log \left (c\right )\right )} + {\left (m + 1\right )} e^{\left (4 i \, a + 8 i \, \log \left (c\right )\right )}} \]

[In]

integrate(x^m*csc(a+2*log(c*x^(1/2*(-(1+m)^2)^(1/2))))^3,x, algorithm="fricas")

[Out]

-2*(2*I*x^2*x^(2*m)*e^(3*I*a + 6*I*log(c)) - I*e^(5*I*a + 10*I*log(c)))/((m + 1)*x^4*x^(4*m) - 2*(m + 1)*x^2*x
^(2*m)*e^(2*I*a + 4*I*log(c)) + (m + 1)*e^(4*I*a + 8*I*log(c)))

Sympy [F(-1)]

Timed out. \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(x**m*csc(a+2*ln(c*x**(1/2*(-(1+m)**2)**(1/2))))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (92) = 184\).

Time = 0.31 (sec) , antiderivative size = 974, normalized size of antiderivative = 8.85 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]

[In]

integrate(x^m*csc(a+2*log(c*x^(1/2*(-(1+m)^2)^(1/2))))^3,x, algorithm="maxima")

[Out]

2*((cos(2*log(c))*sin(a) + cos(a)*sin(2*log(c)))*x*e^(m*log(x) + 14*arctan2(sin(1/2*m*log(x)), cos(1/2*m*log(x
))) + 14*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) + 2*(((cos(a)*sin(2*a) - cos(2*a)*sin(a))*cos(2*log(c)) -
(cos(2*a)*cos(a) + sin(2*a)*sin(a))*sin(2*log(c)))*cos(4*log(c)) + ((cos(2*a)*cos(a) + sin(2*a)*sin(a))*cos(2*
log(c)) + (cos(a)*sin(2*a) - cos(2*a)*sin(a))*sin(2*log(c)))*sin(4*log(c)))*x*e^(m*log(x) + 10*arctan2(sin(1/2
*m*log(x)), cos(1/2*m*log(x))) + 10*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) - (((cos(a)*sin(4*a) - cos(4*a)
*sin(a))*cos(2*log(c)) - (cos(4*a)*cos(a) + sin(4*a)*sin(a))*sin(2*log(c)))*cos(8*log(c)) + ((cos(4*a)*cos(a)
+ sin(4*a)*sin(a))*cos(2*log(c)) + (cos(a)*sin(4*a) - cos(4*a)*sin(a))*sin(2*log(c)))*sin(8*log(c)))*x*e^(m*lo
g(x) + 6*arctan2(sin(1/2*m*log(x)), cos(1/2*m*log(x))) + 6*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))))/((cos(4
*a)^2 + sin(4*a)^2)*cos(8*log(c))^2 + (cos(4*a)^2 + sin(4*a)^2)*sin(8*log(c))^2 + ((cos(4*a)^2 + sin(4*a)^2)*c
os(8*log(c))^2 + (cos(4*a)^2 + sin(4*a)^2)*sin(8*log(c))^2)*m + (m + 1)*e^(16*arctan2(sin(1/2*m*log(x)), cos(1
/2*m*log(x))) + 16*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) - 4*((cos(2*a)*cos(4*log(c)) - sin(2*a)*sin(4*lo
g(c)))*m + cos(2*a)*cos(4*log(c)) - sin(2*a)*sin(4*log(c)))*e^(12*arctan2(sin(1/2*m*log(x)), cos(1/2*m*log(x))
) + 12*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) + 2*(2*(cos(2*a)^2 + sin(2*a)^2)*cos(4*log(c))^2 + 2*(cos(2*
a)^2 + sin(2*a)^2)*sin(4*log(c))^2 + (2*(cos(2*a)^2 + sin(2*a)^2)*cos(4*log(c))^2 + 2*(cos(2*a)^2 + sin(2*a)^2
)*sin(4*log(c))^2 + cos(4*a)*cos(8*log(c)) - sin(4*a)*sin(8*log(c)))*m + cos(4*a)*cos(8*log(c)) - sin(4*a)*sin
(8*log(c)))*e^(8*arctan2(sin(1/2*m*log(x)), cos(1/2*m*log(x))) + 8*arctan2(sin(1/2*log(x)), cos(1/2*log(x))))
- 4*((((cos(4*a)*cos(2*a) + sin(4*a)*sin(2*a))*cos(4*log(c)) + (cos(2*a)*sin(4*a) - cos(4*a)*sin(2*a))*sin(4*l
og(c)))*cos(8*log(c)) - ((cos(2*a)*sin(4*a) - cos(4*a)*sin(2*a))*cos(4*log(c)) - (cos(4*a)*cos(2*a) + sin(4*a)
*sin(2*a))*sin(4*log(c)))*sin(8*log(c)))*m + ((cos(4*a)*cos(2*a) + sin(4*a)*sin(2*a))*cos(4*log(c)) + (cos(2*a
)*sin(4*a) - cos(4*a)*sin(2*a))*sin(4*log(c)))*cos(8*log(c)) - ((cos(2*a)*sin(4*a) - cos(4*a)*sin(2*a))*cos(4*
log(c)) - (cos(4*a)*cos(2*a) + sin(4*a)*sin(2*a))*sin(4*log(c)))*sin(8*log(c)))*e^(4*arctan2(sin(1/2*m*log(x))
, cos(1/2*m*log(x))) + 4*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.51 (sec) , antiderivative size = 839, normalized size of antiderivative = 7.63 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]

[In]

integrate(x^m*csc(a+2*log(c*x^(1/2*(-(1+m)^2)^(1/2))))^3,x, algorithm="giac")

[Out]

I*c^(6*I)*m*x*x^m*x^abs(m + 1)*e^(3*I*a)/(c^(8*I)*m^2*e^(4*I*a) + 2*c^(8*I)*m*e^(4*I*a) + c^(8*I)*e^(4*I*a) -
2*c^(4*I)*m^2*x^(2*abs(m + 1))*e^(2*I*a) - 4*c^(4*I)*m*x^(2*abs(m + 1))*e^(2*I*a) - 2*c^(4*I)*x^(2*abs(m + 1))
*e^(2*I*a) + m^2*x^(4*abs(m + 1)) + 2*m*x^(4*abs(m + 1)) + x^(4*abs(m + 1))) - I*c^(6*I)*x*x^m*x^abs(m + 1)*ab
s(m + 1)*e^(3*I*a)/(c^(8*I)*m^2*e^(4*I*a) + 2*c^(8*I)*m*e^(4*I*a) + c^(8*I)*e^(4*I*a) - 2*c^(4*I)*m^2*x^(2*abs
(m + 1))*e^(2*I*a) - 4*c^(4*I)*m*x^(2*abs(m + 1))*e^(2*I*a) - 2*c^(4*I)*x^(2*abs(m + 1))*e^(2*I*a) + m^2*x^(4*
abs(m + 1)) + 2*m*x^(4*abs(m + 1)) + x^(4*abs(m + 1))) + I*c^(6*I)*x*x^m*x^abs(m + 1)*e^(3*I*a)/(c^(8*I)*m^2*e
^(4*I*a) + 2*c^(8*I)*m*e^(4*I*a) + c^(8*I)*e^(4*I*a) - 2*c^(4*I)*m^2*x^(2*abs(m + 1))*e^(2*I*a) - 4*c^(4*I)*m*
x^(2*abs(m + 1))*e^(2*I*a) - 2*c^(4*I)*x^(2*abs(m + 1))*e^(2*I*a) + m^2*x^(4*abs(m + 1)) + 2*m*x^(4*abs(m + 1)
) + x^(4*abs(m + 1))) - I*c^(2*I)*m*x*x^m*x^(3*abs(m + 1))*e^(I*a)/(c^(8*I)*m^2*e^(4*I*a) + 2*c^(8*I)*m*e^(4*I
*a) + c^(8*I)*e^(4*I*a) - 2*c^(4*I)*m^2*x^(2*abs(m + 1))*e^(2*I*a) - 4*c^(4*I)*m*x^(2*abs(m + 1))*e^(2*I*a) -
2*c^(4*I)*x^(2*abs(m + 1))*e^(2*I*a) + m^2*x^(4*abs(m + 1)) + 2*m*x^(4*abs(m + 1)) + x^(4*abs(m + 1))) - I*c^(
2*I)*x*x^m*x^(3*abs(m + 1))*abs(m + 1)*e^(I*a)/(c^(8*I)*m^2*e^(4*I*a) + 2*c^(8*I)*m*e^(4*I*a) + c^(8*I)*e^(4*I
*a) - 2*c^(4*I)*m^2*x^(2*abs(m + 1))*e^(2*I*a) - 4*c^(4*I)*m*x^(2*abs(m + 1))*e^(2*I*a) - 2*c^(4*I)*x^(2*abs(m
+ 1))*e^(2*I*a) + m^2*x^(4*abs(m + 1)) + 2*m*x^(4*abs(m + 1)) + x^(4*abs(m + 1))) - I*c^(2*I)*x*x^m*x^(3*abs(
m + 1))*e^(I*a)/(c^(8*I)*m^2*e^(4*I*a) + 2*c^(8*I)*m*e^(4*I*a) + c^(8*I)*e^(4*I*a) - 2*c^(4*I)*m^2*x^(2*abs(m
+ 1))*e^(2*I*a) - 4*c^(4*I)*m*x^(2*abs(m + 1))*e^(2*I*a) - 2*c^(4*I)*x^(2*abs(m + 1))*e^(2*I*a) + m^2*x^(4*abs
(m + 1)) + 2*m*x^(4*abs(m + 1)) + x^(4*abs(m + 1)))

Mupad [B] (verification not implemented)

Time = 32.42 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.55 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{6{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}+{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\sqrt {-{\left (m+1\right )}^2}\,1{}\mathrm {i}+m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )}{\sqrt {-{\left (m+1\right )}^2}}+\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{2{}\mathrm {i}}\,\left (m+1-\sqrt {-{\left (m+1\right )}^2}\,1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}}{\left (m+1\right )\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{4{}\mathrm {i}}-1\right )}^2} \]

[In]

int(x^m/sin(a + 2*log(c*x^((-(m + 1)^2)^(1/2)/2)))^3,x)

[Out]

((x^(m + 1)*exp(a*1i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^6i*(exp(a*2i) + exp(a*2i)*(-(m + 1)^2)^(1/2)*1i + m*ex
p(a*2i)))/(-(m + 1)^2)^(1/2) + (x^(m + 1)*exp(a*1i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^2i*(m - (-(m + 1)^2)^(1/
2)*1i + 1))/(-(m + 1)^2)^(1/2))/((m + 1)*(exp(a*2i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^4i - 1)^2)