3.106 \(\int \frac{(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac{a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 a^3 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[Out]

(-2*a^3*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) + (a^3*Log[1 - Cos[e + f*x]]*Tan
[e + f*x])/(c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.180779, antiderivative size = 100, 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 = {3910, 3911, 31} \[ \frac{a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 a^3 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(-2*a^3*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) + (a^3*Log[1 - Cos[e + f*x]]*Tan
[e + f*x])/(c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3910

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-8*a^3*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*
d, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)]

Rule 3911

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> -Dis
t[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((b + a*x)^(m - 1/2)*(d
+ c*x)^(n - 1/2))/x^(m + n), x], x, Sin[e + f*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] && EqQ[m + n, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac{2 a^3 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sec (e+f x)}}{\sqrt{c-c \sec (e+f x)}} \, dx}{c^2}\\ &=-\frac{2 a^3 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac{\left (a^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{2 a^3 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac{a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.36979, size = 155, normalized size = 1.55 \[ \frac{a^2 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (6 \log \left (1-e^{i (e+f x)}\right )+\left (-8 \log \left (1-e^{i (e+f x)}\right )+4 i f x-8\right ) \cos (e+f x)+\left (2 \log \left (1-e^{i (e+f x)}\right )-i f x\right ) \cos (2 (e+f x))-3 i f x+4\right )}{2 c^2 f (\cos (e+f x)-1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a^2*(4 - (3*I)*f*x + Cos[e + f*x]*(-8 + (4*I)*f*x - 8*Log[1 - E^(I*(e + f*x))]) + 6*Log[1 - E^(I*(e + f*x))]
+ Cos[2*(e + f*x)]*((-I)*f*x + 2*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(2*c^
2*f*(-1 + Cos[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.266, size = 229, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }{2\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \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}-2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-8\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\cos \left ( fx+e \right ) -2\,\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{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x)

[Out]

-1/2/f*a^2*(-1+cos(f*x+e))*(4*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-2*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2
-8*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-3*cos(f*x+e)^2+4*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+4*ln(-(-1+cos(f
*x+e))/sin(f*x+e))-2*cos(f*x+e)-2*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)/cos
(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)

________________________________________________________________________________________

Maxima [A]  time = 1.51431, size = 188, normalized size = 1.88 \begin{align*} \frac{\frac{4 \, \sqrt{-a} a^{2} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac{5}{2}}} - \frac{2 \, \sqrt{-a} a^{2} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac{5}{2}}} - \frac{{\left (\sqrt{-a} a^{2} \sqrt{c} - \frac{2 \, \sqrt{-a} a^{2} \sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}{c^{3} \sin \left (f x + e\right )^{4}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/2*(4*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1))/c^(5/2) - 2*sqrt(-a)*a^2*log(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1)/c^(5/2) - (sqrt(-a)*a^2*sqrt(c) - 2*sqrt(-a)*a^2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)
*(cos(f*x + e) + 1)^4/(c^3*sin(f*x + e)^4))/f

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\right )} \sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{c^{3} \sec \left (f x + e\right )^{3} - 3 \, c^{3} \sec \left (f x + e\right )^{2} + 3 \, c^{3} \sec \left (f x + e\right ) - c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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((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((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out