\(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx\) [98]

Optimal result

Integrand size = 30, antiderivative size = 100 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {2 a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \log (1-\cos (e+f x)) \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3993, 3996, 31} \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx=\frac {a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 a^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \]

[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 31

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

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 {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 ) \text {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 [A] (verified)

Time = 0.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx=\frac {a^2 \left (\log (\cos (e+f x))+\log (1-\sec (e+f x))+\frac {2}{-1+\sec (e+f x)}\right ) \tan (e+f x)}{c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.45

method	result	size
default	\(\frac {a \left (\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+\cos \left (f x +e \right )+1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right )}{f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \left (\sec \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )}\)	\(145\)
risch	\(-\frac {a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (2 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+{\mathrm e}^{2 i \left (f x +e \right )} f x -4 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+2 \,{\mathrm e}^{2 i \left (f x +e \right )} e -2 \,{\mathrm e}^{i \left (f x +e \right )} f x -4 i {\mathrm e}^{i \left (f x +e \right )}+2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-4 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\)	\(227\)

[In]

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

[Out]

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

Fricas [F]

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

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx=\frac {\frac {2 \, \sqrt {-a} a \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {-a} a \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {-a} a {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}}{f} \]

[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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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