3.150 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=160 \[ \frac{3 \csc (e+f x)}{8 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^2(e+f x) \csc (e+f x)}{4 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]
[Out]
(3*Csc[e + f*x])/(8*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (Cot[e + f*x]^2*Csc[e + f*x
])/(4*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (3*ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(8
*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
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Rubi [A] time = 0.201946, antiderivative size = 160, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used =
{3959, 2611, 3770} \[ \frac{3 \csc (e+f x)}{8 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^2(e+f x) \csc (e+f x)}{4 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2)),x]
[Out]
(3*Csc[e + f*x])/(8*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (Cot[e + f*x]^2*Csc[e + f*x
])/(4*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (3*ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(8
*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
Rule 3959
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(m_), x_Symbol] :> Dist[((-(a*c))^(m + 1/2)*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])
, Int[Csc[e + f*x]*Cot[e + f*x]^(2*m), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && IntegerQ[m + 1/2]
Rule 2611
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2}} \, dx &=\frac{\tan (e+f x) \int \cot ^4(e+f x) \csc (e+f x) \, dx}{a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\cot ^2(e+f x) \csc (e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{(3 \tan (e+f x)) \int \cot ^2(e+f x) \csc (e+f x) \, dx}{4 a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^2(e+f x) \csc (e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{(3 \tan (e+f x)) \int \csc (e+f x) \, dx}{8 a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^2(e+f x) \csc (e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{8 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 1.39122, size = 84, normalized size = 0.52 \[ \frac{(1-5 \cos (2 (e+f x))) \csc ^3(e+f x)-12 \tan (e+f x) \tanh ^{-1}\left (e^{i (e+f x)}\right )}{16 a^2 c^2 f \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2)),x]
[Out]
((1 - 5*Cos[2*(e + f*x)])*Csc[e + f*x]^3 - 12*ArcTanh[E^(I*(e + f*x))]*Tan[e + f*x])/(16*a^2*c^2*f*Sqrt[a*(1 +
Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
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Maple [A] time = 0.277, size = 173, normalized size = 1.1 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -5\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-6\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\,\cos \left ( fx+e \right ) +3\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x)
[Out]
-1/8/f/a^3*(-1+cos(f*x+e))^3*(3*cos(f*x+e)^4*ln(-(-1+cos(f*x+e))/sin(f*x+e))-5*cos(f*x+e)^3-6*ln(-(-1+cos(f*x+
e))/sin(f*x+e))*cos(f*x+e)^2+3*cos(f*x+e)+3*ln(-(-1+cos(f*x+e))/sin(f*x+e)))*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(
1/2)/sin(f*x+e)^5/cos(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)
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Maxima [B] time = 2.93168, size = 2240, normalized size = 14. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
1/8*(3*(2*(4*cos(6*f*x + 6*e) - 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) - 1)*cos(8*f*x + 8*e) - cos(8*f*x + 8*
e)^2 + 8*(6*cos(4*f*x + 4*e) - 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) - 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*
f*x + 2*e) - 1)*cos(4*f*x + 4*e) - 36*cos(4*f*x + 4*e)^2 - 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) - 3*s
in(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) - sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) - 2*sin(2
*f*x + 2*e))*sin(6*f*x + 6*e) - 16*sin(6*f*x + 6*e)^2 - 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x
+ 2*e) - 16*sin(2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) - 3*(2*(4*cos
(6*f*x + 6*e) - 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) - 1)*cos(8*f*x + 8*e) - cos(8*f*x + 8*e)^2 + 8*(6*cos(
4*f*x + 4*e) - 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) - 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) - 1)*
cos(4*f*x + 4*e) - 36*cos(4*f*x + 4*e)^2 - 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) - 3*sin(4*f*x + 4*e)
+ 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) - sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) - 2*sin(2*f*x + 2*e))*sin
(6*f*x + 6*e) - 16*sin(6*f*x + 6*e)^2 - 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) - 16*sin(
2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) - 2*(5*sin(7*f*x + 7*e) + 3*s
in(5*f*x + 5*e) + 3*sin(3*f*x + 3*e) + 5*sin(f*x + e))*cos(8*f*x + 8*e) - 20*(2*sin(6*f*x + 6*e) - 3*sin(4*f*x
+ 4*e) + 2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 8*(3*sin(5*f*x + 5*e) + 3*sin(3*f*x + 3*e) + 5*sin(f*x + e))*
cos(6*f*x + 6*e) + 12*(3*sin(4*f*x + 4*e) - 2*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) - 12*(3*sin(3*f*x + 3*e) + 5*
sin(f*x + e))*cos(4*f*x + 4*e) + 2*(5*cos(7*f*x + 7*e) + 3*cos(5*f*x + 5*e) + 3*cos(3*f*x + 3*e) + 5*cos(f*x +
e))*sin(8*f*x + 8*e) + 10*(4*cos(6*f*x + 6*e) - 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) - 1)*sin(7*f*x + 7*e)
- 8*(3*cos(5*f*x + 5*e) + 3*cos(3*f*x + 3*e) + 5*cos(f*x + e))*sin(6*f*x + 6*e) - 6*(6*cos(4*f*x + 4*e) - 4*c
os(2*f*x + 2*e) + 1)*sin(5*f*x + 5*e) + 12*(3*cos(3*f*x + 3*e) + 5*cos(f*x + e))*sin(4*f*x + 4*e) + 6*(4*cos(2
*f*x + 2*e) - 1)*sin(3*f*x + 3*e) - 24*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) - 40*cos(f*x + e)*sin(2*f*x + 2*e) +
40*cos(2*f*x + 2*e)*sin(f*x + e) - 10*sin(f*x + e))*sqrt(a)*sqrt(c)/((a^3*c^3*cos(8*f*x + 8*e)^2 + 16*a^3*c^3*
cos(6*f*x + 6*e)^2 + 36*a^3*c^3*cos(4*f*x + 4*e)^2 + 16*a^3*c^3*cos(2*f*x + 2*e)^2 + a^3*c^3*sin(8*f*x + 8*e)^
2 + 16*a^3*c^3*sin(6*f*x + 6*e)^2 + 36*a^3*c^3*sin(4*f*x + 4*e)^2 - 48*a^3*c^3*sin(4*f*x + 4*e)*sin(2*f*x + 2*
e) + 16*a^3*c^3*sin(2*f*x + 2*e)^2 - 8*a^3*c^3*cos(2*f*x + 2*e) + a^3*c^3 - 2*(4*a^3*c^3*cos(6*f*x + 6*e) - 6*
a^3*c^3*cos(4*f*x + 4*e) + 4*a^3*c^3*cos(2*f*x + 2*e) - a^3*c^3)*cos(8*f*x + 8*e) - 8*(6*a^3*c^3*cos(4*f*x + 4
*e) - 4*a^3*c^3*cos(2*f*x + 2*e) + a^3*c^3)*cos(6*f*x + 6*e) - 12*(4*a^3*c^3*cos(2*f*x + 2*e) - a^3*c^3)*cos(4
*f*x + 4*e) - 4*(2*a^3*c^3*sin(6*f*x + 6*e) - 3*a^3*c^3*sin(4*f*x + 4*e) + 2*a^3*c^3*sin(2*f*x + 2*e))*sin(8*f
*x + 8*e) - 16*(3*a^3*c^3*sin(4*f*x + 4*e) - 2*a^3*c^3*sin(2*f*x + 2*e))*sin(6*f*x + 6*e))*f)
________________________________________________________________________________________
Fricas [A] time = 0.771649, size = 1177, normalized size = 7.36 \begin{align*} \left [-\frac{3 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt{-a c} \log \left (-\frac{4 \,{\left (2 \, \sqrt{-a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} +{\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \,{\left (5 \, \cos \left (f x + e\right )^{4} - 3 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{16 \,{\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} f\right )} \sin \left (f x + e\right )}, \frac{3 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) +{\left (5 \, \cos \left (f x + e\right )^{4} - 3 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \,{\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[-1/16*(3*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-a*c)*log(-4*(2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/co
s(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/
((cos(f*x + e)^2 - 1)*sin(f*x + e)))*sin(f*x + e) - 2*(5*cos(f*x + e)^4 - 3*cos(f*x + e)^2)*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^3*f*cos(f*x + e)^4 - 2*a^3*c^3*f*cos(f*
x + e)^2 + a^3*c^3*f)*sin(f*x + e)), 1/8*(3*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(a*c)*arctan(sqrt(a*c)
*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*c*sin(f*x + e)))*sin(f*x +
e) + (5*cos(f*x + e)^4 - 3*cos(f*x + e)^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/
cos(f*x + e)))/((a^3*c^3*f*cos(f*x + e)^4 - 2*a^3*c^3*f*cos(f*x + e)^2 + a^3*c^3*f)*sin(f*x + e))]
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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(sec(f*x+e)/(a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [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(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out