3.260 \(\int x^m \sec ^3(a+2 \log (c x^{\frac{1}{2} \sqrt{-(1+m)^2}})) \, dx\)
Optimal. Leaf size=110 \[ \frac{x^{m+1} \sec \left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )}{2 (m+1)}+\frac{x^{m+1} \tan \left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right ) \sec \left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )}{2 \sqrt{-(m+1)^2}} \]
[Out]
(x^(1 + m)*Sec[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]])/(2*(1 + m)) + (x^(1 + m)*Sec[a + 2*Log[c*x^(Sqrt[-(1 + m)
^2]/2)]]*Tan[a + 2*Log[c*x^(Sqrt[-(1 + m)^2]/2)]])/(2*Sqrt[-(1 + m)^2])
________________________________________________________________________________________
Rubi [C] time = 0.217478, antiderivative size = 146, normalized size of antiderivative =
1.33, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules
used = {4509, 4505, 364} \[ \frac{8 e^{3 i a} x^{m+1} \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )^{6 i} \, _2F_1\left (3,\frac{1}{2} \left (3-\frac{i (m+1)}{\sqrt{-(m+1)^2}}\right );\frac{1}{2} \left (5-\frac{i (m+1)}{\sqrt{-(m+1)^2}}\right );-e^{2 i a} \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )^{4 i}\right )}{1-i \left (-3 \sqrt{-(m+1)^2}+i m\right )} \]
Warning: Unable to verify antiderivative.
[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 4509
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])
Rule 4505
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 364
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])
Rubi steps
\begin{align*} \int x^m \sec ^3\left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(1+m)^2}}\right )\right ) \, dx &=\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 ) \operatorname{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 ) \operatorname{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} \, _2F_1\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] time = 2.11106, size = 198, normalized size = 1.8 \[ \frac{x^{m+1} \left ((m+1) \cos \left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )-\sqrt{-(m+1)^2} \sin \left (a+2 \log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )\right )}{2 (m+1)^2 \left (\cos \left (\frac{a}{2}+\log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )-\sin \left (\frac{a}{2}+\log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )\right )^2 \left (\sin \left (\frac{a}{2}+\log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )+\cos \left (\frac{a}{2}+\log \left (c x^{\frac{1}{2} \sqrt{-(m+1)^2}}\right )\right )\right )^2} \]
Antiderivative was successfully verified.
[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] time = 0.228, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sec \left ( a+2\,\ln \left ( c{x}^{1/2\,\sqrt{- \left ( 1+m \right ) ^{2}}} \right ) \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
Maxima [B] time = 1.42587, size = 1318, normalized size = 11.98 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)))))
________________________________________________________________________________________
Fricas [C] time = 0.476942, size = 231, normalized size = 2.1 \begin{align*} -\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**m*sec(a+2*ln(c*x**(1/2*(-(1+m)**2)**(1/2))))**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [C] time = 15.4405, size = 1126, normalized size = 10.24 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)))