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

Optimal. Leaf size=95 \[ \frac{c^2 \tan (e+f x) \log (\sec (e+f x)+1)}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.278487, antiderivative size = 95, normalized size of antiderivative = 1., 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 = {3954, 3952} \[ \frac{c^2 \tan (e+f x) \log (\sec (e+f x)+1)}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3954

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(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] && IGtQ[n - 1/2, 0
] && LtQ[m, -2^(-1)]

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]

Rubi steps

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

Mathematica [C]  time = 1.05326, size = 132, normalized size = 1.39 \[ -\frac{c \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )+\left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)-2\right )}{a f (\cos (e+f x)+1) \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.253, size = 193, normalized size = 2. \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( \cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -\cos \left ( fx+e \right ) +1 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53183, size = 134, normalized size = 1.41 \begin{align*} \frac{\frac{c^{\frac{3}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt{-a} a} + \frac{c^{\frac{3}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt{-a} a} + \frac{c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{\sqrt{-a} a{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 3.56969, size = 123, normalized size = 1.29 \begin{align*} -\frac{{\left (c^{4} \log \left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right ) +{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{3}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{\sqrt{-a c} a c f{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^4*log(c*tan(1/2*f*x + 1/2*e)^2 - c) + (c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan
(1/2*f*x + 1/2*e))/(sqrt(-a*c)*a*c*f*abs(c))