\(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [129]

Optimal result

Integrand size = 36, antiderivative size = 145 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (1-\sec (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {4039, 4037} \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^2 \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

-1/2*(a*(a + a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(5/2)) + (a^2*Sqrt[a + a*Sec[e + f*x]
]*Tan[e + f*x])/(c*f*(c - c*Sec[e + f*x])^(3/2)) + (a^3*Log[1 - Sec[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 4037

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)])/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)], x_Symbol] :> Simp[a*c*Log[1 + (b/a)*Csc[e + f*x]]*(Cot[e + f*x]/(b*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c +
d*Csc[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4039

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m +
1))), x] - Dist[d*((2*n - 1)/(b*(2*m + 1))), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(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] && IGtQ[n - 1/2, 0
] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac {a \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c} \\ & = -\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2} \\ & = -\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (1-\sec (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 = 1.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a^3 \left (-\log (1-\sec (e+f x))+\frac {-2+4 \sec (e+f x)}{(-1+\sec (e+f x))^2}\right ) \tan (e+f x)}{c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

-((a^3*(-Log[1 - Sec[e + f*x]] + (-2 + 4*Sec[e + f*x])/(-1 + Sec[e + f*x])^2)*Tan[e + f*x])/(c^2*f*Sqrt[a*(1 +
Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]]))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(131)=262\).

Time = 3.60 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.88

method	result	size
default	\(-\frac {\sqrt {2}\, a^{2} \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+2 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \csc \left (f x +e \right )}{4 f \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\)	\(272\)
risch	\(\frac {8 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}-\frac {2 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}+\frac {i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\)	\(325\)

[In]

int(sec(f*x+e)*(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-co
s(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))*(2*ln(-cot(f*x+e)+csc(f*x+e)+
1)*(1-cos(f*x+e))^4*csc(f*x+e)^4+2*ln(-cot(f*x+e)+csc(f*x+e)-1)*(1-cos(f*x+e))^4*csc(f*x+e)^4-4*ln(-cot(f*x+e)
+csc(f*x+e))*(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]

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

[In]

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

[In]

integrate(sec(f*x+e)*(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 = 0.31 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{\frac {5}{2}}} + \frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} - \frac {4 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {5}{2}}} + \frac {{\left (\sqrt {-a} a^{2} \sqrt {c} + \frac {2 \, \sqrt {-a} a^{2} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{4}}{c^{3} \sin \left (f x + e\right )^{4}}}{2 \, f} \]

[In]

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

[Out]

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

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

[In]

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

[In]

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

[Out]

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