∫ 1_____________ (a+a sec(e+fx))3∕2(c−c sec(e+fx))5∕2 dx

3.123 \(\int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=347 \[ -\frac {\tan (e+f x)}{2 a c^2 f (1-\sec (e+f x)) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (\sec (e+f x)+1) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1-\sec (e+f x))^2 \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {11 \tan (e+f x) \log (1-\sec (e+f x))}{16 a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {5 \tan (e+f x) \log (\sec (e+f x)+1)}{16 a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \log (\cos (e+f x))}{a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[Out]

ln(cos(f*x+e))*tan(f*x+e)/a/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+11/16*ln(1-sec(f*x+e))*tan(f*x

+e)/a/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+5/16*ln(1+sec(f*x+e))*tan(f*x+e)/a/c^2/f/(a+a*sec(f*

x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/a/c^2/f/(1-sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x

+e))^(1/2)-1/2*tan(f*x+e)/a/c^2/f/(1-sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/

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

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Rubi [A] time = 0.18, antiderivative size = 347, normalized size of antiderivative = 1.00, 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, 88} \[ -\frac {\tan (e+f x)}{2 a c^2 f (1-\sec (e+f x)) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (\sec (e+f x)+1) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1-\sec (e+f x))^2 \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {11 \tan (e+f x) \log (1-\sec (e+f x))}{16 a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {5 \tan (e+f x) \log (\sec (e+f x)+1)}{16 a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \log (\cos (e+f x))}{a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(Log[Cos[e + f*x]]*Tan[e + f*x])/(a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (11*Log[1 - Sec

[e + f*x]]*Tan[e + f*x])/(16*a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (5*Log[1 + Sec[e + f

*x]]*Tan[e + f*x])/(16*a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(8*a*c^2*f*(1

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

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

a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI

ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&

(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (e+f x))^{3/2} (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)^2 (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}{4 a^2 c^3 (-1+x)^3}+\frac {1}{2 a^2 c^3 (-1+x)^2}-\frac {11}{16 a^2 c^3 (-1+x)}+\frac {1}{a^2 c^3 x}-\frac {1}{8 a^2 c^3 (1+x)^2}-\frac {5}{16 a^2 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)}{a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {11 \log (1-\sec (e+f x)) \tan (e+f x)}{16 a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {5 \log (1+\sec (e+f x)) \tan (e+f x)}{16 a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1-\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a c^2 f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C] time = 2.62, size = 275, normalized size = 0.79 \[ \frac {\tan (e+f x) \left (22 \log \left (1-e^{i (e+f x)}\right )+10 \log \left (1+e^{i (e+f x)}\right )-8 i f x \cos (3 (e+f x))+11 \log \left (1-e^{i (e+f x)}\right ) \cos (3 (e+f x))+\left (-11 \log \left (1-e^{i (e+f x)}\right )-5 \log \left (1+e^{i (e+f x)}\right )+8 i f x-12\right ) \cos (e+f x)+2 \left (-11 \log \left (1-e^{i (e+f x)}\right )-5 \log \left (1+e^{i (e+f x)}\right )+8 i f x-5\right ) \cos (2 (e+f x))+5 \log \left (1+e^{i (e+f x)}\right ) \cos (3 (e+f x))-16 i f x+14\right )}{32 a c^2 f (\cos (e+f x)-1)^2 (\cos (e+f x)+1) \sqrt {a (\sec (e+f x)+1)} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

((14 - (16*I)*f*x - (8*I)*f*x*Cos[3*(e + f*x)] + 22*Log[1 - E^(I*(e + f*x))] + 11*Cos[3*(e + f*x)]*Log[1 - E^(

I*(e + f*x))] + Cos[e + f*x]*(-12 + (8*I)*f*x - 11*Log[1 - E^(I*(e + f*x))] - 5*Log[1 + E^(I*(e + f*x))]) + 2*

Cos[2*(e + f*x)]*(-5 + (8*I)*f*x - 11*Log[1 - E^(I*(e + f*x))] - 5*Log[1 + E^(I*(e + f*x))]) + 10*Log[1 + E^(I

*(e + f*x))] + 5*Cos[3*(e + f*x)]*Log[1 + E^(I*(e + f*x))])*Tan[e + f*x])/(32*a*c^2*f*(-1 + Cos[e + f*x])^2*(1

+ Cos[e + f*x])*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])

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fricas [F] time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{a^{2} c^{3} \sec \left (f x + e\right )^{5} - a^{2} c^{3} \sec \left (f x + e\right )^{4} - 2 \, a^{2} c^{3} \sec \left (f x + e\right )^{3} + 2 \, a^{2} c^{3} \sec \left (f x + e\right )^{2} + a^{2} c^{3} \sec \left (f x + e\right ) - a^{2} c^{3}}, x\right ) \]

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))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^2*c^3*sec(f*x + e)^5 - a^2*c^3*sec(f*x + e)^4

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

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch

eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable

to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)

Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant

sign by intervals (correct if the argument is real):Check [abs(cos(f*t_nostep+exp(1)))]Unable to check sign:

(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check

sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to

check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab

le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/

2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n

ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*

pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2

)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no

step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p

i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign

: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, assuming -2*a+a is positive. Hint: run assume to make assumptio

ns on a variableWarning, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarnin

g, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning, assuming -2*a+a is

positive. Hint: run assume to make assumptions on a variableUnable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n

ostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*

pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2

)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_no

step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*p

i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign

: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to chec

k sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable t

o check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Un

able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_noste

p/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t

_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-

4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep

/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_

nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4

*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check si

gn: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to ch

eck sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable

to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)

Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nos

tep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi

/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>

(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nost

ep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/

t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign:

(4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check

sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to

check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unab

le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/

2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n