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

Optimal. Leaf size=100 \[ \frac{a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 a^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.185599, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3908, 3911, 31} \[ \frac{a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{2 a^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3908

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-4*a^2*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a/c, Int[Sqr
t[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] && LtQ[n, -2^(-1)]

Rule 3911

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 0.670817, size = 115, normalized size = 1.15 \[ \frac{a \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (-2 \log \left (1-e^{i (e+f x)}\right )+\left (2 \log \left (1-e^{i (e+f x)}\right )-i f x\right ) \cos (e+f x)+i f x-2\right )}{c f (\cos (e+f x)-1) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.268, size = 163, normalized size = 1.6 \begin{align*} -{\frac{a \left ( -1+\cos \left ( fx+e \right ) \right ) }{f\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) } \left ( 2\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -\cos \left ( fx+e \right ) -2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -1 \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{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*a*(-1+cos(f*x+e))*(2*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-cos(f*x+e)*ln(2/(1+cos(f*x+e)))-cos(f*x+e
)-2*ln(-(-1+cos(f*x+e))/sin(f*x+e))+ln(2/(1+cos(f*x+e)))-1)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/cos(f*x+e)/(
c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)

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Maxima [A]  time = 1.53536, size = 128, normalized size = 1.28 \begin{align*} \frac{\frac{2 \, \sqrt{-a} a \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac{3}{2}}} - \frac{\sqrt{-a} a \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac{3}{2}}} + \frac{\sqrt{-a} a{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*sqrt(-a)*a*log(sin(f*x + e)/(cos(f*x + e) + 1))/c^(3/2) - sqrt(-a)*a*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 + 1)/c^(3/2) + sqrt(-a)*a*(cos(f*x + e) + 1)^2/(c^(3/2)*sin(f*x + e)^2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(f*x + e) + a)^(3/2)*sqrt(-c*sec(f*x + e) + c)/(c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2),
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((a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(3/2),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((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out