Integrand size = 31, antiderivative size = 110 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 (1+m)}-\frac {x^{1+m} \cot \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 \sqrt {-(1+m)^2}} \]
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)
Time = 1.51 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \left (1+m+\sqrt {-(1+m)^2} \cot \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right ) \csc \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 (1+m)^2} \]
(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)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.43 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5021, 5017, 888}
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.
\(\displaystyle \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )\right ) \, dx\) |
\(\Big \downarrow \) 5021 |
\(\displaystyle \frac {2 x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{-\frac {2 (m+1)}{\sqrt {-(m+1)^2}}} \int \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{\frac {2 (m+1)}{\sqrt {-(m+1)^2}}-1} \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )\right )d\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )}{\sqrt {-(m+1)^2}}\) |
\(\Big \downarrow \) 5017 |
\(\displaystyle \frac {16 i e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{-\frac {2 (m+1)}{\sqrt {-(m+1)^2}}} \int \frac {\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{\frac {2 (m+1)}{\sqrt {-(m+1)^2}}-(1-6 i)}}{\left (1-e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}\right )^3}d\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )}{\sqrt {-(m+1)^2}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {8 i e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{6 i} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),\frac {1}{2} \left (5-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}\right )}{\sqrt {-(m+1)^2} \left (\frac {m+1}{\sqrt {-(m+1)^2}}+3 i\right )}\) |
((8*I)*E^((3*I)*a)*x^(1 + m)*(c*x^(Sqrt[-(1 + m)^2]/2))^(6*I)*Hypergeometr ic2F1[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)])/(Sqrt[-(1 + m)^2 ]*(3*I + (1 + m)/Sqrt[-(1 + m)^2]))
3.4.2.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[Csc[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(-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]
Int[Csc[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_ .), x_Symbol] :> Simp[(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])
Time = 217.82 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(-\frac {\left (\left ({\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}-1\right ) \sqrt {-\left (1+m \right )^{2}}-2 \left (1+m \right ) \tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )\right ) x^{1+m} \left (1+{\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}\right )}{8 \left (1+m \right )^{2} {\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{2}}\) | \(123\) |
-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
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=-\frac {2 \, {\left (2 i \, x^{2} x^{2 \, m} e^{\left (3 i \, a + 6 i \, \log \left (c\right )\right )} - i \, e^{\left (5 i \, a + 10 i \, \log \left (c\right )\right )}\right )}}{{\left (m + 1\right )} x^{4} x^{4 \, m} - 2 \, {\left (m + 1\right )} x^{2} x^{2 \, m} e^{\left (2 i \, a + 4 i \, \log \left (c\right )\right )} + {\left (m + 1\right )} e^{\left (4 i \, a + 8 i \, \log \left (c\right )\right )}} \]
-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)))
Timed out. \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Timed out} \]
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 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]
2*((cos(2*log(c))*sin(a) + cos(a)*sin(2*log(c)))*x*e^(m*log(x) + 14*arctan 2(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)) - (c os(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)*si n(a))*sin(2*log(c)))*sin(4*log(c)))*x*e^(m*log(x) + 10*arctan2(sin(1/2*m*l og(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*log(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)*cos(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*l og(c)) - sin(2*a)*sin(4*log(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*arcta n2(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*...
Result contains complex when optimal does not.
Time = 11.51 (sec) , antiderivative size = 839, normalized size of antiderivative = 7.63 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]
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)*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*a bs(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*a bs(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...
Time = 32.42 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.55 \[ \int x^m \csc ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{6{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}+{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\sqrt {-{\left (m+1\right )}^2}\,1{}\mathrm {i}+m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )}{\sqrt {-{\left (m+1\right )}^2}}+\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{2{}\mathrm {i}}\,\left (m+1-\sqrt {-{\left (m+1\right )}^2}\,1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}}{\left (m+1\right )\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{4{}\mathrm {i}}-1\right )}^2} \]
((x^(m + 1)*exp(a*1i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^6i*(exp(a*2i) + ex p(a*2i)*(-(m + 1)^2)^(1/2)*1i + m*exp(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)