\(\int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\) [126]

Optimal result

Integrand size = 30, antiderivative size = 144 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^2 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^2 \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

-c^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2)-c^2*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(3/2)/(c-c
*sec(f*x+e))^(1/2)+c^2*ln(1+cos(f*x+e))*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3993, 3992, 3996, 31} \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\frac {c^2 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]

[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 31

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

Rule 3992

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 3993

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 3996

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

Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {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 [A] (verified)

Time = 0.59 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.60 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^2 \left (-\log (\cos (e+f x))-\log (1+\sec (e+f x))+\frac {2+\sec (e+f x)}{(1+\sec (e+f x))^2}\right ) \tan (e+f x)}{a^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (\sec \left (f x +e \right )-1\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (4 \cos \left (f x +e \right )^{2} \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+5 \cos \left (f x +e \right )^{2}+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \cos \left (f x +e \right )-3\right ) \cos \left (f x +e \right ) \cot \left (f x +e \right )}{4 f \,a^{3} \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )}\)	\(154\)
risch	\(-\frac {c \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (6 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )} f x +2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{4 i \left (f x +e \right )} e +4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +8 i {\mathrm e}^{2 i \left (f x +e \right )}+8 \,{\mathrm e}^{3 i \left (f x +e \right )} e +6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +4 \,{\mathrm e}^{i \left (f x +e \right )} f x +12 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+6 i {\mathrm e}^{3 i \left (f x +e \right )}+8 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}+2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(349\)

[In]

int((c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f/a^3*(sec(f*x+e)-1)*(-c*(sec(f*x+e)-1))^(1/2)*c*(a*(sec(f*x+e)+1))^(1/2)*(4*cos(f*x+e)^2*ln(2/(cos(f*x+e
)+1))+8*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+5*cos(f*x+e)^2+4*ln(2/(cos(f*x+e)+1))-2*cos(f*x+e)-3)/(cos(f*x+e)+1)^2
/(cos(f*x+e)-1)*cos(f*x+e)*cot(f*x+e)

Fricas [F]

\[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((-c*(sec(e + f*x) - 1))**(3/2)/(a*(sec(e + f*x) + 1))**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1786 vs. \(2 (132) = 264\).

Time = 0.51 (sec) , antiderivative size = 1786, normalized size of antiderivative = 12.40 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[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)

Giac [A] (verification not implemented)

none

Time = 1.61 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {4 \, \sqrt {-a c} c^{2} \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{a^{3} {\left | c \right |}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{3} {\left | c \right |} - 2 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{3} c {\left | c \right |}}{a^{6} c^{2}}}{4 \, f} \]

[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(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/(a^3*abs(c)) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^2
*sqrt(-a*c)*a^3*abs(c) - 2*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^3*c*abs(c))/(a^6*c^2))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((c - c/cos(e + f*x))^(3/2)/(a + a/cos(e + f*x))^(5/2),x)

[Out]

int((c - c/cos(e + f*x))^(3/2)/(a + a/cos(e + f*x))^(5/2), x)