\(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [106]

Optimal result

Integrand size = 30, antiderivative size = 100 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

-2*a^3*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2)+a^3*ln(1-cos(f*x+e))*tan(f*x+e)/c^2/f/(a+a*s
ec(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 = {3995, 3996, 31} \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(-2*a^3*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) + (a^3*Log[1 - Cos[e + f*x]]*Tan
[e + f*x])/(c^2*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 3995

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 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^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2} \\ & = -\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {\left (a^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = -\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^3 \left (\log (\cos (e+f x))+\log (1-\sec (e+f x))-\frac {2}{(-1+\sec (e+f x))^2}\right ) \tan (e+f x)}{c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(239\) vs. \(2(92)=184\).

Time = 2.01 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.40

method	result	size
default	\(\frac {\sqrt {2}\, a^{2} \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+2 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \csc \left (f x +e \right )}{4 f \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\)	\(240\)
risch	\(\frac {a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (8 i {\mathrm e}^{i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )} f x -8 i {\mathrm e}^{2 i \left (f x +e \right )}-2 \,{\mathrm e}^{4 i \left (f x +e \right )} e +4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +8 i {\mathrm e}^{3 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 \,{\mathrm e}^{3 i \left (f x +e \right )} e -6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x -2 i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +4 \,{\mathrm e}^{i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 i {\mathrm e}^{3 i \left (f x +e \right )}-12 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e -f x -2 e \right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\)	\(352\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.39 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {\frac {4 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {5}{2}}} - \frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}} - \frac {{\left (\sqrt {-a} a^{2} \sqrt {c} - \frac {2 \, \sqrt {-a} a^{2} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{4}}{c^{3} \sin \left (f x + e\right )^{4}}}{2 \, f} \]

[In]

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

[Out]

1/2*(4*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1))/c^(5/2) - 2*sqrt(-a)*a^2*log(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1)/c^(5/2) - (sqrt(-a)*a^2*sqrt(c) - 2*sqrt(-a)*a^2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)
*(cos(f*x + e) + 1)^4/(c^3*sin(f*x + e)^4))/f

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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