3.126 \(\int \frac{(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=144 \[ \frac{c^2 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]
[Out]
-((c^2*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]])) - (c^2*Tan[e + f*x])/(a*f*(a + a
*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^2*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Se
c[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
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Rubi [A] time = 0.287751, antiderivative size = 144, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{3908, 3907, 3911, 31} \[ \frac{c^2 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sec[e + f*x])^(3/2)/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
-((c^2*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]])) - (c^2*Tan[e + f*x])/(a*f*(a + a
*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^2*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Se
c[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 3907
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[
(-2*a*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/c, Int[Sqrt[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{(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{c^2 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{c \int \frac{\sqrt{c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac{c^2 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{c \int \frac{\sqrt{c-c \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac{c^2 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{\left (c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a+a x} \, dx,x,\cos (e+f x)\right )}{a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{c^2 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{c^2 \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.838573, size = 152, normalized size = 1.06 \[ \frac{i c \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (6 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+\left (8 i \log \left (1+e^{i (e+f x)}\right )+4 f x+6 i\right ) \cos (e+f x)+3 f x+4 i\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sec[e + f*x])^(3/2)/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
((I/2)*c*Cot[(e + f*x)/2]*(4*I + 3*f*x + Cos[2*(e + f*x)]*(f*x + (2*I)*Log[1 + E^(I*(e + f*x))]) + Cos[e + f*x
]*(6*I + 4*f*x + (8*I)*Log[1 + E^(I*(e + f*x))]) + (6*I)*Log[1 + E^(I*(e + f*x))])*Sqrt[c - c*Sec[e + f*x]])/(
a^2*f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])
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Maple [A] time = 0.253, size = 152, normalized size = 1.1 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{4\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 4\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +4\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -3 \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((c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x)
[Out]
1/4/f/a^3*(-1+cos(f*x+e))*(4*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+5*cos(f*x+e)^2+8*cos(f*x+e)*ln(2/(1+cos(f*x+e))
)-2*cos(f*x+e)+4*ln(2/(1+cos(f*x+e)))-3)*(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)^5
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Maxima [B] time = 2.54643, size = 2411, normalized size = 16.74 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
-((f*x + e)*c*cos(4*f*x + 4*e)^2 + 36*(f*x + e)*c*cos(2*f*x + 2*e)^2 + 16*(f*x + e)*c*cos(3/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e)))^2 + 16*(f*x + e)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + (f*x
+ e)*c*sin(4*f*x + 4*e)^2 + 36*(f*x + e)*c*sin(2*f*x + 2*e)^2 + 16*(f*x + e)*c*sin(3/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e)))^2 + 16*(f*x + e)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*(f*x + e)
*c*cos(2*f*x + 2*e) + (f*x + e)*c - 2*(c*cos(4*f*x + 4*e)^2 + 36*c*cos(2*f*x + 2*e)^2 + 16*c*cos(3/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + c*sin(4*
f*x + 4*e)^2 + 12*c*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*c*sin(2*f*x + 2*e)^2 + 16*c*sin(3/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*(6*c*cos(2*f
*x + 2*e) + c)*cos(4*f*x + 4*e) + 12*c*cos(2*f*x + 2*e) + 8*(c*cos(4*f*x + 4*e) + 6*c*cos(2*f*x + 2*e) + 4*c*c
os(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 8*(c*cos(4*f*x + 4*e) + 6*c*cos(2*f*x + 2*e) + c)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 8*(
c*sin(4*f*x + 4*e) + 6*c*sin(2*f*x + 2*e) + 4*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 8*(c*sin(4*f*x + 4*e) + 6*c*sin(2*f*x + 2*e))*sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + c)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 2*(6*(f*x + e)*c*cos(2*f*x + 2*e) + (f*x + e)*c - 4*c*sin
(2*f*x + 2*e))*cos(4*f*x + 4*e) + 2*(4*(f*x + e)*c*cos(4*f*x + 4*e) + 24*(f*x + e)*c*cos(2*f*x + 2*e) + 16*(f*
x + e)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(f*x + e)*c + 3*c*sin(4*f*x + 4*e) + 2*c*sin
(2*f*x + 2*e))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(4*(f*x + e)*c*cos(4*f*x + 4*e) + 24*(
f*x + e)*c*cos(2*f*x + 2*e) + 4*(f*x + e)*c + 3*c*sin(4*f*x + 4*e) + 2*c*sin(2*f*x + 2*e))*cos(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(3*(f*x + e)*c*sin(2*f*x + 2*e) + 2*c*cos(2*f*x + 2*e))*sin(4*f*x + 4*e)
- 8*c*sin(2*f*x + 2*e) + 2*(4*(f*x + e)*c*sin(4*f*x + 4*e) + 24*(f*x + e)*c*sin(2*f*x + 2*e) + 16*(f*x + e)*c
*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 3*c*cos(4*f*x + 4*e) - 2*c*cos(2*f*x + 2*e) - 3*c)*sin
(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(4*(f*x + e)*c*sin(4*f*x + 4*e) + 24*(f*x + e)*c*sin(2*f
*x + 2*e) - 3*c*cos(4*f*x + 4*e) - 2*c*cos(2*f*x + 2*e) - 3*c)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))))*sqrt(a)*sqrt(c)/((a^3*cos(4*f*x + 4*e)^2 + 36*a^3*cos(2*f*x + 2*e)^2 + 16*a^3*cos(3/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e)))^2 + 16*a^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + a^3*sin(4*f*x
+ 4*e)^2 + 12*a^3*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*a^3*sin(2*f*x + 2*e)^2 + 16*a^3*sin(3/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*a^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*a^3*cos
(2*f*x + 2*e) + a^3 + 2*(6*a^3*cos(2*f*x + 2*e) + a^3)*cos(4*f*x + 4*e) + 8*(a^3*cos(4*f*x + 4*e) + 6*a^3*cos(
2*f*x + 2*e) + 4*a^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + a^3)*cos(3/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + 8*(a^3*cos(4*f*x + 4*e) + 6*a^3*cos(2*f*x + 2*e) + a^3)*cos(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + 8*(a^3*sin(4*f*x + 4*e) + 6*a^3*sin(2*f*x + 2*e) + 4*a^3*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 8*(a^3*sin(4*f*x + 4*e) + 6
*a^3*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sec \left (f x + e\right ) + a}{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(a*sec(f*x + e) + a)*(-c*sec(f*x + e) + c)^(3/2)/(a^3*sec(f*x + e)^3 + 3*a^3*sec(f*x + e)^2 + 3*a
^3*sec(f*x + e) + a^3), 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((c-c*sec(f*x+e))**(3/2)/(a+a*sec(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [A] time = 4.4283, size = 316, normalized size = 2.19 \begin{align*} -\frac{\frac{4 \, \sqrt{-a c} c^{2} \log \left (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 )}{a^{3}{\left | c \right |}} - \frac{4 \, \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 ) \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 )}{a^{3}{\left | c \right |}} - \frac{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt{-a c} a^{3}{\left | c \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 ) - 2 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} \sqrt{-a c} a^{3} c{\left | c \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 )}{a^{6} c^{2}}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
-1/4*(4*sqrt(-a*c)*c^2*log(2*abs(c))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/(a^3*abs(c)) - 4*sqrt(
-a*c)*c^2*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/(a^3*abs(c
)) - ((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^3*abs(c)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e)
) - 2*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^3*c*abs(c)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e)
))/(a^6*c^2))/f