3.116 \(\int \frac{1}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=274 \[ -\frac{3 \tan (e+f x)}{4 c^2 f (1-\sec (e+f x)) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 c^2 f (1-\sec (e+f x))^2 \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{7 \tan (e+f x) \log (1-\sec (e+f x))}{8 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sec (e+f x)+1)}{8 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

(Log[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]]) + (7*Log[1 - Sec[e
+ f*x]]*Tan[e + f*x])/(8*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[1 + Sec[e + f*x]]*Tan
[e + f*x])/(8*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(4*c^2*f*(1 - Sec[e + f*
x])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (3*Tan[e + f*x])/(4*c^2*f*(1 - Sec[e + f*x])*Sqrt[a
 + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.160944, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3912, 72} \[ -\frac{3 \tan (e+f x)}{4 c^2 f (1-\sec (e+f x)) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 c^2 f (1-\sec (e+f x))^2 \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{7 \tan (e+f x) \log (1-\sec (e+f x))}{8 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\sec (e+f x)+1)}{8 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \log (\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(Log[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]]) + (7*Log[1 - Sec[e
+ f*x]]*Tan[e + f*x])/(8*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (Log[1 + Sec[e + f*x]]*Tan
[e + f*x])/(8*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(4*c^2*f*(1 - Sec[e + f*
x])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (3*Tan[e + f*x])/(4*c^2*f*(1 - Sec[e + f*x])*Sqrt[a
 + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3912

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{x (a+a x) (c-c x)^3} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \left (-\frac{1}{2 a c^3 (-1+x)^3}+\frac{3}{4 a c^3 (-1+x)^2}-\frac{7}{8 a c^3 (-1+x)}+\frac{1}{a c^3 x}-\frac{1}{8 a c^3 (1+x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\log (\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)}}+\frac{7 \log (1-\sec (e+f x)) \tan (e+f x)}{8 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\log (1+\sec (e+f x)) \tan (e+f x)}{8 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 c^2 f (1-\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x)}{4 c^2 f (1-\sec (e+f x)) \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.87856, size = 194, normalized size = 0.71 \[ \frac{\tan (e+f x) \left (21 \log \left (1-e^{i (e+f x)}\right )+3 \log \left (1+e^{i (e+f x)}\right )+\left (-28 \log \left (1-e^{i (e+f x)}\right )-4 \log \left (1+e^{i (e+f x)}\right )+16 i f x-10\right ) \cos (e+f x)+\left (7 \log \left (1-e^{i (e+f x)}\right )+\log \left (1+e^{i (e+f x)}\right )-4 i f x\right ) \cos (2 (e+f x))-12 i f x+8\right )}{8 c^2 f (\cos (e+f x)-1)^2 \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

((8 - (12*I)*f*x + 21*Log[1 - E^(I*(e + f*x))] + Cos[e + f*x]*(-10 + (16*I)*f*x - 28*Log[1 - E^(I*(e + f*x))]
- 4*Log[1 + E^(I*(e + f*x))]) + 3*Log[1 + E^(I*(e + f*x))] + Cos[2*(e + f*x)]*((-4*I)*f*x + 7*Log[1 - E^(I*(e
+ f*x))] + Log[1 + E^(I*(e + f*x))]))*Tan[e + f*x])/(8*c^2*f*(-1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])]*
Sqrt[c - c*Sec[e + f*x]])

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Maple [A]  time = 0.303, size = 229, normalized size = 0.8 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{16\,af\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 28\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-16\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-9\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-56\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +32\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +28\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -16\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +7 \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/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/16/f/a*(-1+cos(f*x+e))*(28*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-16*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^
2-9*cos(f*x+e)^2-56*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+32*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*cos(f*x+e)
+28*ln(-(-1+cos(f*x+e))/sin(f*x+e))-16*ln(2/(1+cos(f*x+e)))+7)*(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)

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Maxima [B]  time = 2.54955, size = 2978, normalized size = 10.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-1/4*(4*(f*x + e)*cos(4*f*x + 4*e)^2 + 144*(f*x + e)*cos(2*f*x + 2*e)^2 + 64*(f*x + e)*cos(3/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + 64*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*(f*x
 + e)*sin(4*f*x + 4*e)^2 + 144*(f*x + e)*sin(2*f*x + 2*e)^2 + 64*(f*x + e)*sin(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e)))^2 + 64*(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*f*x - (2*(6*cos(
2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 36*cos(2*f*x + 2*e)^2 - 8*(cos(4*f*x + 4*e) + 6*cos(
2*f*x + 2*e) - 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)*cos(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) + 16*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(cos(4*f*x + 4*e) + 6*cos(2*
f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*cos(1/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e)))^2 + sin(4*f*x + 4*e)^2 + 12*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*sin(2*f*x + 2*e)^2 - 8*(si
n(4*f*x + 4*e) + 6*sin(2*f*x + 2*e) - 4*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(sin(4*f*
x + 4*e) + 6*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 7*(2*(6*cos(2*f*x + 2*e) + 1)*cos(4*f
*x + 4*e) + cos(4*f*x + 4*e)^2 + 36*cos(2*f*x + 2*e)^2 - 8*(cos(4*f*x + 4*e) + 6*cos(2*f*x + 2*e) - 4*cos(1/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*co
s(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(cos(4*f*x + 4*e) + 6*cos(2*f*x + 2*e) + 1)*cos(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(
4*f*x + 4*e)^2 + 12*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*sin(2*f*x + 2*e)^2 - 8*(sin(4*f*x + 4*e) + 6*sin(2*
f*x + 2*e) - 4*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 16*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(sin(4*f*x + 4*e) + 6*sin(2*f*x +
2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e)))^2 + 12*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 1) + 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) + e - 2*sin(2*f*x + 2*e)
)*cos(4*f*x + 4*e) + 48*(f*x + e)*cos(2*f*x + 2*e) - 2*(16*f*x + 16*(f*x + e)*cos(4*f*x + 4*e) + 96*(f*x + e)*
cos(2*f*x + 2*e) - 64*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*e + 5*sin(4*f*x + 4*
e) - 2*sin(2*f*x + 2*e))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*(16*f*x + 16*(f*x + e)*cos(4
*f*x + 4*e) + 96*(f*x + e)*cos(2*f*x + 2*e) + 16*e + 5*sin(4*f*x + 4*e) - 2*sin(2*f*x + 2*e))*cos(1/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*(3*(f*x + e)*sin(2*f*x + 2*e) + cos(2*f*x + 2*e))*sin(4*f*x + 4*e) -
 2*(16*(f*x + e)*sin(4*f*x + 4*e) + 96*(f*x + e)*sin(2*f*x + 2*e) - 64*(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) - 5*cos(4*f*x + 4*e) + 2*cos(2*f*x + 2*e) - 5)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) - 2*(16*(f*x + e)*sin(4*f*x + 4*e) + 96*(f*x + e)*sin(2*f*x + 2*e) - 5*cos(4*f*x + 4*e) + 2*cos(
2*f*x + 2*e) - 5)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*e - 16*sin(2*f*x + 2*e))/((c^2*cos(
4*f*x + 4*e)^2 + 36*c^2*cos(2*f*x + 2*e)^2 + 16*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 1
6*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + c^2*sin(4*f*x + 4*e)^2 + 12*c^2*sin(4*f*x + 4*e
)*sin(2*f*x + 2*e) + 36*c^2*sin(2*f*x + 2*e)^2 + 16*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2
 + 16*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*c^2*cos(2*f*x + 2*e) + c^2 + 2*(6*c^2*co
s(2*f*x + 2*e) + c^2)*cos(4*f*x + 4*e) - 8*(c^2*cos(4*f*x + 4*e) + 6*c^2*cos(2*f*x + 2*e) - 4*c^2*cos(1/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^2*
cos(4*f*x + 4*e) + 6*c^2*cos(2*f*x + 2*e) + c^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^2
*sin(4*f*x + 4*e) + 6*c^2*sin(2*f*x + 2*e) - 4*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^2*sin(4*f*x + 4*e) + 6*c^2*sin(2*f*x + 2*e))*sin(1/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)*f)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{a c^{3} \sec \left (f x + e\right )^{4} - 2 \, a c^{3} \sec \left (f x + e\right )^{3} + 2 \, a c^{3} \sec \left (f x + e\right ) - a c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a*c^3*sec(f*x + e)^4 - 2*a*c^3*sec(f*x + e)^3 +
2*a*c^3*sec(f*x + e) - a*c^3), x)

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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/(c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(1/2),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/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out