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

Optimal. Leaf size=96 \[ \frac{c^3 \tan (e+f x) \log (\cos (e+f x))}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.193182, antiderivative size = 96, 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 = {3910, 3905, 3475} \[ \frac{c^3 \tan (e+f x) \log (\cos (e+f x))}{a f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3910

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-8*a^3*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^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), 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 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 3475

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

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac{4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{c^2 \int \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)} \, dx}{a^2}\\ &=-\frac{4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{\left (c^3 \tan (e+f x)\right ) \int \tan (e+f x) \, dx}{a \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{c^3 \log (\cos (e+f x)) \tan (e+f x)}{a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.731135, size = 116, normalized size = 1.21 \[ \frac{i c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (i \log \left (1+e^{2 i (e+f x)}\right )+\left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+f x+4 i\right )}{a f (\cos (e+f x)+1) \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*c^2*Cot[(e + f*x)/2]*(4*I + f*x + Cos[e + f*x]*(f*x + I*Log[1 + E^((2*I)*(e + f*x))]) + I*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 = 0.27, size = 236, normalized size = 2.5 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( -1+\cos \left ( fx+e \right ) \right ) } \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 ) -\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \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 ) -2\,\cos \left ( fx+e \right ) -\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +2 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{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((c-c*sec(f*x+e))^(5/2)/(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))-cos(f*x+e)*ln(2/(1+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))/s
in(f*x+e))-2*cos(f*x+e)-ln(2/(1+cos(f*x+e)))+2)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*cos(f*x+e)^3*(1/cos(f*x+e
)*a*(1+cos(f*x+e)))^(1/2)/sin(f*x+e)^3/(-1+cos(f*x+e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.35289, size = 1098, normalized size = 11.44 \begin{align*} \left [-\frac{4 \, c^{2} \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 ) -{\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt{-\frac{c}{a}} \log \left (\frac{c \cos \left (f x + e\right )^{4} -{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt{-\frac{c}{a}} \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 ) + c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}, -\frac{2 \, c^{2} \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 ) -{\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt{\frac{c}{a}} \arctan \left (\frac{\sqrt{\frac{c}{a}} \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 )}{c \cos \left (f x + e\right )^{2} + c}\right )}{a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/2*(4*c^2*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^2*cos(f*x + e)^2 + 2*a*c^2*cos(f*x + e) + a*c^2)*sqrt(-c/a)*log(1/2*(c*cos(f*x + e)^4 - (cos(f
*x + e)^3 + cos(f*x + e))*sqrt(-c/a)*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) + c)/cos(f*x + e)^2))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f), -(2*c^2*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^2*
cos(f*x + e)^2 + 2*a*c^2*cos(f*x + e) + a*c^2)*sqrt(c/a)*arctan(sqrt(c/a)*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)/(c*cos(f*x + e)^2 + c)))/(a^2*f*cos(f*x
+ e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)]

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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((c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 3.05831, size = 185, normalized size = 1.93 \begin{align*} -\frac{c{\left (\frac{\sqrt{-a c} c^{2} \log \left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}{a^{2}{\left | c \right |}} - \frac{\sqrt{-a c} c^{2} \log \left ({\left | c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c \right |}\right )}{a^{2}{\left | c \right |}} + \frac{2 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} \sqrt{-a c} c}{a^{2}{\left | c \right |}}\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 )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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