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

Optimal. Leaf size=145 \[ \frac{a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a^2 \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.431088, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3954, 3952} \[ \frac{a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a^2 \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}} \]

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]

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

Rule 3954

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

Rule 3952

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)])/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)], x_Symbol] :> Simp[(a*c*Log[1 + (b*Csc[e + f*x])/a]*Cot[e + f*x])/(b*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c +
d*Csc[e + f*x]]), x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

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{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac{a \int \frac{\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c}\\ &=-\frac{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{a^2 \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac{a^2 \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{\sqrt{c-c \sec (e+f x)}} \, dx}{c^2}\\ &=-\frac{a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{a^2 \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac{a^3 \log (1-\sec (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 = 1.32972, size = 182, normalized size = 1.26 \[ -\frac{a^2 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (-6 \log \left (1-e^{i (e+f x)}\right )+3 \log \left (1+e^{2 i (e+f x)}\right )+\left (8 \log \left (1-e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+\left (\log \left (1+e^{2 i (e+f x)}\right )-2 \log \left (1-e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+4\right )}{2 c^2 f (\cos (e+f x)-1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(4 - 6*Log[1 - E^(I*(e + f*x))] + Cos[e + f*x]*(8*Log[1 - E^(I*(e + f*x))] - 4*Log[1 + E^((2*I)*(e + f*x
))]) + 3*Log[1 + E^((2*I)*(e + f*x))] + Cos[2*(e + f*x)]*(-2*Log[1 - E^(I*(e + f*x))] + Log[1 + E^((2*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 = 0.276, size = 366, normalized size = 2.5 \begin{align*}{\frac{{a}^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }{2\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) } \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-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\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -4\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +8\,\cos \left ( fx+e \right ) \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 ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\cos \left ( fx+e \right ) +3 \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]

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

________________________________________________________________________________________

Maxima [A]  time = 1.54619, size = 228, normalized size = 1.57 \begin{align*} -\frac{\frac{2 \, \sqrt{-a} a^{2} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{\frac{5}{2}}} + \frac{2 \, \sqrt{-a} a^{2} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac{5}{2}}} - \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{{\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(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/2*(2*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^(5/2) + 2*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f
*x + e) + 1) - 1)/c^(5/2) - 4*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 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 )^{3} + 2 \, a^{2} \sec \left (f x + e\right )^{2} + a^{2} \sec \left (f x + e\right )\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(sec(f*x+e)*(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)^3 + 2*a^2*sec(f*x + e)^2 + a^2*sec(f*x + e))*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(sec(f*x+e)*(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(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