∫ (a + a sec(e + fx))5∕2∘_________

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

Optimal. Leaf size=139 \[ \frac {a^3 c \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a^2 c \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}}-\frac {a c \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sec (e+f x)}} \]

[Out]

-1/2*a*c*(a+a*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)+a^3*c*ln(cos(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^2*c*(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 = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3906, 3905, 3475} \[ -\frac {a^2 c \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}}+\frac {a^3 c \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a c \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3905

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

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

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

Rule 3906

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp

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

rt[a + b*Csc[e + f*x]]*(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] && GtQ[n, 1/2]

Rubi steps

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

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Mathematica [C] time = 1.40, size = 164, normalized size = 1.18 \[ \frac {a^2 e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \left (\cot \left (\frac {1}{2} (e+f x)\right )+i\right ) \sec ^2(e+f x) \sqrt {a (\sec (e+f x)+1)} \sqrt {c-c \sec (e+f x)} \left (-\log \left (1+e^{2 i (e+f x)}\right )+4 \cos (e+f x)+\left (i f x-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))+i f x+1\right )}{4 f \left (1+e^{i (e+f x)}\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a^2*(1 + E^((2*I)*(e + f*x)))*(I + Cot[(e + f*x)/2])*(1 + I*f*x + 4*Cos[e + f*x] + Cos[2*(e + f*x)]*(I*f*x -

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

- c*Sec[e + f*x]])/(4*E^(I*(e + f*x))*(1 + E^(I*(e + f*x)))*f)

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fricas [A] time = 0.56, size = 420, normalized size = 3.02 \[ \left [\frac {{\left (5 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \sqrt {-a c} \log \left (\frac {a c \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + a c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}, \frac {{\left (5 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )^{2} + a c}\right )}{2 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \]

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))^(1/2),x, algorithm="fricas")

[Out]

[1/2*((5*a^2*cos(f*x + e) + a^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e

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

e)^3 + cos(f*x + e))*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)

)*sin(f*x + e) + a*c)/cos(f*x + e)^2))/(f*cos(f*x + e)^2 + f*cos(f*x + e)), 1/2*((5*a^2*cos(f*x + e) + a^2)*sq

rt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) + 2*(a^2*cos(f*x +

e)^2 + a^2*cos(f*x + e))*sqrt(a*c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x +

e) - c)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(a*c*cos(f*x + e)^2 + a*c)))/(f*cos(f*x + e)^2 + f*cos(f*x + e

))]

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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((a+a*sec(f*x+e))^(5/2)*(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)Warning, integration of ab

s or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(cos(f*x+exp(1)))]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