\(\int x^m \sec ^3(a+2 \log (c x^{\frac {1}{2} \sqrt {-(1+m)^2}})) \, dx\) [260]

Optimal result

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

[Out]

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)

Rubi [C] (verified)

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

Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4605, 4601, 371} \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {8 e^{3 i a} x^{m+1} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{6 i} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),\frac {1}{2} \left (5-\frac {i (m+1)}{\sqrt {-(m+1)^2}}\right ),-e^{2 i a} \left (c x^{\frac {1}{2} \sqrt {-(m+1)^2}}\right )^{4 i}\right )}{1-i \left (-3 \sqrt {-(m+1)^2}+i m\right )} \]

[In]

Int[x^m*Sec[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))])/(1 - 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 4601

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[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]

Rule 4605

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(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])

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}}} \sec ^3(a+2 \log (x)) \, dx,x,c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )}{\sqrt {-(1+m)^2}} \\ & = \frac {\left (16 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 )}{1-i \left (i m-3 \sqrt {-(1+m)^2}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.80 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {x^{1+m} \left ((1+m) \cos \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )-\sqrt {-(1+m)^2} \sin \left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )}{2 (1+m)^2 \left (\cos \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )-\sin \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )^2 \left (\cos \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )+\sin \left (\frac {a}{2}+\log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right )\right )^2} \]

[In]

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

[Out]

(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 + Log[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)

Maple [F]

\[\int x^{m} {\sec \left (a +2 \ln \left (c \,x^{\frac {\sqrt {-\left (1+m \right )^{2}}}{2}}\right )\right )}^{3}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=-\frac {2 \, {\left (2 \, x^{2} x^{2 \, m} e^{\left (3 i \, a + 6 i \, \log \left (c\right )\right )} + e^{\left (5 i \, a + 10 i \, \log \left (c\right )\right )}\right )}}{{\left (m + 1\right )} x^{4} x^{4 \, m} + 2 \, {\left (m + 1\right )} x^{2} x^{2 \, m} e^{\left (2 i \, a + 4 i \, \log \left (c\right )\right )} + {\left (m + 1\right )} e^{\left (4 i \, a + 8 i \, \log \left (c\right )\right )}} \]

[In]

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

[Out]

-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)))

Sympy [F(-1)]

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

[In]

integrate(x**m*sec(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. 976 vs. \(2 (92) = 184\).

Time = 0.31 (sec) , antiderivative size = 976, normalized size of antiderivative = 8.87 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

2*((cos(a)*cos(2*log(c)) - sin(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(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)))*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)*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(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*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 = 11.64 (sec) , antiderivative size = 834, normalized size of antiderivative = 7.58 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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*a
bs(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*abs(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 + 1)) + x^(4*abs(m + 1))) + 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 = 31.92 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.60 \[ \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac {1}{2} \sqrt {-(1+m)^2}}\right )\right ) \, dx=\frac {\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{2{}\mathrm {i}}\,\left (m\,1{}\mathrm {i}+\sqrt {-{\left (m+1\right )}^2}+1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}-\frac {x^{m+1}\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{6{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,1{}\mathrm {i}-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\sqrt {-{\left (m+1\right )}^2}+m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,1{}\mathrm {i}\right )}{\sqrt {-{\left (m+1\right )}^2}}}{\left (m+1\right )\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^{\frac {\sqrt {-m^2-2\,m-1}}{2}}\right )}^{4{}\mathrm {i}}+1\right )}^2} \]

[In]

int(x^m/cos(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))^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
*exp(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)