3.130 \(\int \frac{1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=151 \[ -\frac{\cot ^3(e+f x)}{4 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\cot (e+f x)}{2 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]
[Out]
Cot[e + f*x]/(2*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Cot[e + f*x]^3/(4*a^2*c^2*f*Sqr
t[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[Sin[e + f*x]]*Tan[e + f*x])/(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 = 0.13111, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used =
{3905, 3473, 3475} \[ -\frac{\cot ^3(e+f x)}{4 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\cot (e+f x)}{2 a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{a^2 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2)),x]
[Out]
Cot[e + f*x]/(2*a^2*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Cot[e + f*x]^3/(4*a^2*c^2*f*Sqr
t[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[Sin[e + f*x]]*Tan[e + f*x])/(a^2*c^2*f*Sqrt[a + a*Sec[e
+ f*x]]*Sqrt[c - c*Sec[e + f*x]])
Rule 3905
Int[(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[Cot[e + f*x]^(2*m)
, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m + 1/2]
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Rule 3475
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2}} \, dx &=\frac{\tan (e+f x) \int \cot ^5(e+f x) \, dx}{a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\cot ^3(e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x) \int \cot ^3(e+f x) \, dx}{a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\cot (e+f x)}{2 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^3(e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \int \cot (e+f x) \, dx}{a^2 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\cot (e+f x)}{2 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\cot ^3(e+f x)}{4 a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\log (\sin (e+f x)) \tan (e+f x)}{a^2 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 2.10466, size = 149, normalized size = 0.99 \[ \frac{\csc ^3(e+f x) \sec (e+f x) \left (3 \log \left (1-e^{2 i (e+f x)}\right )+\left (-4 \log \left (1-e^{2 i (e+f x)}\right )+4 i f x-4\right ) \cos (2 (e+f x))+\left (\log \left (1-e^{2 i (e+f x)}\right )-i f x\right ) \cos (4 (e+f x))-3 i f x+2\right )}{8 a^2 c^2 f \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2)),x]
[Out]
(Csc[e + f*x]^3*(2 - (3*I)*f*x + Cos[2*(e + f*x)]*(-4 + (4*I)*f*x - 4*Log[1 - E^((2*I)*(e + f*x))]) + 3*Log[1
- E^((2*I)*(e + f*x))] + Cos[4*(e + f*x)]*((-I)*f*x + Log[1 - E^((2*I)*(e + f*x))]))*Sec[e + f*x])/(8*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 = 0.276, size = 237, normalized size = 1.6 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}}{32\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 32\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -32\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -13\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-64\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+64\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+32\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -32\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +11 \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(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x)
[Out]
-1/32/f/a^3*(-1+cos(f*x+e))^3*(32*cos(f*x+e)^4*ln(-(-1+cos(f*x+e))/sin(f*x+e))-32*cos(f*x+e)^4*ln(2/(1+cos(f*x
+e)))-13*cos(f*x+e)^4-64*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+64*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2-6*c
os(f*x+e)^2+32*ln(-(-1+cos(f*x+e))/sin(f*x+e))-32*ln(2/(1+cos(f*x+e)))+11)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/
2)/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)^5/cos(f*x+e)^2
________________________________________________________________________________________
Maxima [B] time = 2.48675, size = 1871, normalized size = 12.39 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
-((f*x + e)*cos(8*f*x + 8*e)^2 + 16*(f*x + e)*cos(6*f*x + 6*e)^2 + 36*(f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x +
e)*cos(2*f*x + 2*e)^2 + (f*x + e)*sin(8*f*x + 8*e)^2 + 16*(f*x + e)*sin(6*f*x + 6*e)^2 + 36*(f*x + e)*sin(4*f
*x + 4*e)^2 + 16*(f*x + e)*sin(2*f*x + 2*e)^2 + f*x + (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*c
os(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)*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) - 1) + 2*(f*x - 4*(f*x + e)*cos(6*f*x + 6*e) + 6*(f*x + e)*cos(4*f*x + 4*
e) - 4*(f*x + e)*cos(2*f*x + 2*e) + e + 2*sin(6*f*x + 6*e) - 2*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(8*f*
x + 8*e) - 8*(f*x + 6*(f*x + e)*cos(4*f*x + 4*e) - 4*(f*x + e)*cos(2*f*x + 2*e) + e + sin(4*f*x + 4*e))*cos(6*
f*x + 6*e) + 4*(3*f*x - 12*(f*x + e)*cos(2*f*x + 2*e) + 3*e + 2*sin(2*f*x + 2*e))*cos(4*f*x + 4*e) - 8*(f*x +
e)*cos(2*f*x + 2*e) - 4*(2*(f*x + e)*sin(6*f*x + 6*e) - 3*(f*x + e)*sin(4*f*x + 4*e) + 2*(f*x + e)*sin(2*f*x +
2*e) + cos(6*f*x + 6*e) - cos(4*f*x + 4*e) + cos(2*f*x + 2*e))*sin(8*f*x + 8*e) - 4*(12*(f*x + e)*sin(4*f*x +
4*e) - 8*(f*x + e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e) - 1)*sin(6*f*x + 6*e) - 4*(12*(f*x + e)*sin(2*f*x +
2*e) + 2*cos(2*f*x + 2*e) + 1)*sin(4*f*x + 4*e) + e + 4*sin(2*f*x + 2*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)
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Fricas [A] time = 3.18785, size = 1446, normalized size = 9.58 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[-1/324*(162*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-a*c)*log(-8*((256*cos(f*x + e)^5 - 512*cos(f*x + e)
^3 + 175*cos(f*x + e))*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)) - (256*a*c*cos(f*x + e)^4 - 512*a*c*cos(f*x + e)^2 + 337*a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x
+ e)))*sin(f*x + e) + (832*cos(f*x + e)^5 - 1988*cos(f*x + e)^3 + 1075*cos(f*x + e))*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)), -1/324*(324*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(a*c)*arctan((16*cos(f*x
+ e)^3 - 7*cos(f*x + e))*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))/((16*a*c*cos(f*x + e)^2 - 25*a*c)*sin(f*x + e)))*sin(f*x + e) + (832*cos(f*x + e)^5 - 1988*cos(f*x + e)^
3 + 1075*cos(f*x + e))*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))]
________________________________________________________________________________________
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(1/(a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
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(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out