Integrand size = 31, antiderivative size = 110 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \sec \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 (1+m)}+\frac {x^{1+m} \sec \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \tan \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )}{2 \sqrt {-(1+m)^2}} \]
1/2*x^(1+m)*sec(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))/(1+m)+1/2*x^(1+m)*sec( a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))))*tan(a+2*ln(c*x^(1/2*(-(1+m)^2)^(1/2))) )/(-(1+m)^2)^(1/2)
Time = 1.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.80 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \left ((1+m) \cos \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )-\sqrt {-(1+m)^2} \sin \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )}{2 (1+m)^2 \left (\cos \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )-\sin \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )^2 \left (\cos \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )+\sin \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )^2} \]
(x^(1 + m)*((1 + m)*Cos[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]] - Sqrt[-(1 + m)^2]*Sin[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]]))/(2*(1 + m)^2*(Cos[a/2 + L og[c*x^(Sqrt[-(1 + m)^2]/2)]] - Sin[a/2 + Log[c*x^(Sqrt[-(1 + m)^2]/2)]])^ 2*(Cos[a/2 + Log[c*x^(Sqrt[-(1 + m)^2]/2)]] + Sin[a/2 + Log[c*x^(Sqrt[-(1 + m)^2]/2)]])^2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.45 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.37, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5020, 5016, 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.
| \(\displaystyle \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5020 |
| \(\displaystyle \frac {2 x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{-\frac {2 (m+1)}{\sqrt {-(m+1)^2}}} \int \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{\frac {2 (m+1)}{\sqrt {-(m+1)^2}}-1} \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )\right )d\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )}{\sqrt {-(m+1)^2}}\) |
| \(\Big \downarrow \) 5016 |
| \(\displaystyle \frac {16 e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{-\frac {2 (m+1)}{\sqrt {-(m+1)^2}}} \int \frac {\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{\frac {2 (m+1)}{\sqrt {-(m+1)^2}}-(1-6 i)}}{\left (e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}+1\right )^3}d\left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )}{\sqrt {-(m+1)^2}}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {8 e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{6 i} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),\frac {1}{2} \left (5-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),-e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}\right )}{\sqrt {-(m+1)^2} \left (\frac {m+1}{\sqrt {-(m+1)^2}}+3 i\right )}\) |
(8*E^((3*I)*a)*x^(1 + m)*(c*x^(Sqrt[-(1 + m)^2]/2))^(6*I)*Hypergeometric2F 1[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]))
3.3.60.3.1 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[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[2^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[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ .), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n)) Subst[Int[x ^((m + 1)/n - 1)*Sec[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])
\[\int x^{m} {\sec \left (a +2 \ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{3}d x\]
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=-\frac {2 \, {\left (2 \, x^{2} x^{2 \, m} e^{\left (3 i \, a + 6 i \, \log \left (c\right )\right )} + e^{\left (5 i \, a + 10 i \, \log \left (c\right )\right )}\right )}}{{\left (m + 1\right )} x^{4} x^{4 \, m} + 2 \, {\left (m + 1\right )} x^{2} x^{2 \, m} e^{\left (2 i \, a + 4 i \, \log \left (c\right )\right )} + {\left (m + 1\right )} e^{\left (4 i \, a + 8 i \, \log \left (c\right )\right )}} \]
-2*(2*x^2*x^(2*m)*e^(3*I*a + 6*I*log(c)) + 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. \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (92) = 184\).
Time = 0.31 (sec) , antiderivative size = 976, normalized size of antiderivative = 8.87 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]
2*((cos(a)*cos(2*log(c)) - sin(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(2*a)*cos(a) + sin(2*a)*sin(a))*cos(2*log(c)) + (c os(a)*sin(2*a) - cos(2*a)*sin(a))*sin(2*log(c)))*cos(4*log(c)) - ((cos(a)* sin(2*a) - cos(2*a)*sin(a))*cos(2*log(c)) - (cos(2*a)*cos(a) + sin(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(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)))*cos(8*log(c)) - ((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)))*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 = 11.64 (sec) , antiderivative size = 834, normalized size of antiderivative = 7.58 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]
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))) - 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*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))) + 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))) + 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*ab s(m + 1))) + 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 ...
Time = 31.92 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.60 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{2{}\mathrm {i}}\,\left (m\,1{}\mathrm {i}+\sqrt {-{\left (m+1\right )}^2}+1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}-\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{6{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,1{}\mathrm {i}-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\sqrt {-{\left (m+1\right )}^2}+m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}}{\left (m+1\right )\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{4{}\mathrm {i}}+1\right )}^2} \]
((x^(m + 1)*exp(a*1i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^2i*(m*1i + (-(m + 1)^2)^(1/2) + 1i))/(-(m + 1)^2)^(1/2) - (x^(m + 1)*exp(a*1i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^6i*(exp(a*2i)*1i - exp(a*2i)*(-(m + 1)^2)^(1/2) + m*e xp(a*2i)*1i))/(-(m + 1)^2)^(1/2))/((m + 1)*(exp(a*2i)*(c*x^((- 2*m - m^2 - 1)^(1/2)/2))^4i + 1)^2)