3.122 \(\int \frac{1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{\cot (e+f x)}{2 a c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{a c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

Cot[e + f*x]/(2*a*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[Sin[e + f*x]]*Tan[e + f*x])/(a
*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.121461, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3905, 3473, 3475} \[ \frac{\cot (e+f x)}{2 a c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{a c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

Cot[e + f*x]/(2*a*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[Sin[e + f*x]]*Tan[e + f*x])/(a
*c*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))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx &=-\frac{\tan (e+f x) \int \cot ^3(e+f x) \, dx}{a c \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\cot (e+f x)}{2 a c 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 c \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\cot (e+f x)}{2 a c 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 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.47835, size = 121, normalized size = 1.2 \[ \frac{\tan (e+f x) \sec ^2(e+f x) \left (\log \left (1-e^{2 i (e+f x)}\right )+\left (i f x-\log \left (1-e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))-i f x+1\right )}{2 c f (\sec (e+f x)-1) (a (\sec (e+f x)+1))^{3/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

((1 - I*f*x + Cos[2*(e + f*x)]*(I*f*x - Log[1 - E^((2*I)*(e + f*x))]) + Log[1 - E^((2*I)*(e + f*x))])*Sec[e +
f*x]^2*Tan[e + f*x])/(2*c*f*(-1 + Sec[e + f*x])*(a*(1 + Sec[e + f*x]))^(3/2)*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]  time = 0.264, size = 175, normalized size = 1.7 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{4\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) } \left ( 4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +4\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -1 \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{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f/a^2*(-1+cos(f*x+e))^2*(4*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-4*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^
2-cos(f*x+e)^2-4*ln(-(-1+cos(f*x+e))/sin(f*x+e))+4*ln(2/(1+cos(f*x+e)))-1)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/
2)/sin(f*x+e)^3/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/cos(f*x+e)

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Maxima [B]  time = 1.83737, size = 656, normalized size = 6.5 \begin{align*} -\frac{{\left ({\left (f x + e\right )} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \,{\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} +{\left (f x + e\right )} \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \,{\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + f x +{\left (2 \,{\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \cos \left (4 \, f x + 4 \, e\right ) - \cos \left (4 \, f x + 4 \, e\right )^{2} - 4 \, \cos \left (2 \, f x + 2 \, e\right )^{2} - \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 4 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) - 1\right ) + 2 \,{\left (f x - 2 \,{\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) + e + \sin \left (2 \, f x + 2 \, e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) - 4 \,{\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 2 \,{\left (2 \,{\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) + e + 2 \, \sin \left (2 \, f x + 2 \, e\right )\right )} \sqrt{a} \sqrt{c}}{{\left (a^{2} c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} - 4 \, a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, a^{2} c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} c^{2} - 2 \,{\left (2 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2} c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((f*x + e)*cos(4*f*x + 4*e)^2 + 4*(f*x + e)*cos(2*f*x + 2*e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 4*(f*x + e)*s
in(2*f*x + 2*e)^2 + f*x + (2*(2*cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 - 4*cos(2*f*x + 2*
e)^2 - sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) - 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) -
1)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) - 1) + 2*(f*x - 2*(f*x + e)*cos(2*f*x + 2*e) + e + sin(2*f*x + 2
*e))*cos(4*f*x + 4*e) - 4*(f*x + e)*cos(2*f*x + 2*e) - 2*(2*(f*x + e)*sin(2*f*x + 2*e) + cos(2*f*x + 2*e))*sin
(4*f*x + 4*e) + e + 2*sin(2*f*x + 2*e))*sqrt(a)*sqrt(c)/((a^2*c^2*cos(4*f*x + 4*e)^2 + 4*a^2*c^2*cos(2*f*x + 2
*e)^2 + a^2*c^2*sin(4*f*x + 4*e)^2 - 4*a^2*c^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*a^2*c^2*sin(2*f*x + 2*e)^
2 - 4*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2 - 2*(2*a^2*c^2*cos(2*f*x + 2*e) - a^2*c^2)*cos(4*f*x + 4*e))*f)

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Fricas [B]  time = 2.43781, size = 1242, normalized size = 12.3 \begin{align*} \left [-\frac{9 \, \sqrt{-a c}{\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac{8 \,{\left ({\left (256 \, \cos \left (f x + e\right )^{5} - 512 \, \cos \left (f x + e\right )^{3} + 175 \, \cos \left (f x + e\right )\right )} \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 )}} -{\left (256 \, a c \cos \left (f x + e\right )^{4} - 512 \, a c \cos \left (f x + e\right )^{2} + 337 \, 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 ) +{\left (16 \, \cos \left (f x + e\right )^{3} - 25 \, \cos \left (f x + e\right )\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 )}}}{18 \,{\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, -\frac{18 \, \sqrt{a c}{\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac{{\left (16 \, \cos \left (f x + e\right )^{3} - 7 \, \cos \left (f x + e\right )\right )} \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 )}}}{{\left (16 \, a c \cos \left (f x + e\right )^{2} - 25 \, a c\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) +{\left (16 \, \cos \left (f x + e\right )^{3} - 25 \, \cos \left (f x + e\right )\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 )}}}{18 \,{\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/18*(9*sqrt(-a*c)*(cos(f*x + e)^2 - 1)*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) +
(16*cos(f*x + e)^3 - 25*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^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e)), -1/18*(18*sqrt(a*c)*(cos(f*x + e)^2 - 1)*arcta
n((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) + (16*cos(f*x + e)^3 - 25*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^2*c^2*f*cos(f*
x + e)^2 - a^2*c^2*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(1/(a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(3/2),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(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out