∫ sec(e+fx)(a+a sec(e+fx))3∕2 ___________________________________________

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

Optimal. Leaf size=95 \[ \frac {2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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

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Rubi [A] time = 0.26, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {3955, 3952} \[ \frac {2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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]

Rule 3955

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)

)^(n_), x_Symbol] :> -Simp[(d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(f*(m + n)), x

] + Dist[(c*(2*n - 1))/(m + n), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1), x], x] /

; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && !LtQ[m, -2

^(-1)] && !(IGtQ[m - 1/2, 0] && LtQ[m, n])

Rubi steps

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

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Mathematica [C] time = 1.42, size = 174, normalized size = 1.83 \[ \frac {\sqrt {2} a \sin \left (\frac {1}{2} (e+f x)\right ) \sec ^{\frac {3}{2}}(e+f x) \sqrt {a (\sec (e+f x)+1)} \left (1+\left (4 \log \left (1-e^{i (e+f x)}\right )-2 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (1+e^{i (e+f x)}\right ) \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

*Sqrt[a*(1 + Sec[e + f*x])]*(Cos[(e + f*x)/2] + I*Sin[(e + f*x)/2])*Sin[(e + f*x)/2])/((1 + E^(I*(e + f*x)))*S

qrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*f*Sqrt[c - c*Sec[e + f*x]])

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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sec \left (f x + e\right )^{2} + a \sec \left (f x + e\right )\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{c \sec \left (f x + e\right ) - c}, x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

e) - c), 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(sec(f*x+e)*(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2),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)Warning, integration of abs or sign assumes cons

tant sign by intervals (correct if the argument is real):Check [abs(cos(f*x+exp(1)))]Unable 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*pi/x/2)>(-2*pi/x/2)Una

ble 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*

pi/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)Una

ble 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*

pi/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)Una

ble 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)Warning, 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 variableWarning, assuming -2*a+a is p

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

ume to make assumptions on a variableUnable 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)Una

ble 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_n

ostep/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 sig

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

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