\(\int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [92]

Optimal result

Integrand size = 30, antiderivative size = 142 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

-1/2*a*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2)-a*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)/(a+a
*sec(f*x+e))^(1/2)+a*ln(1-cos(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 = 0.37 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3992, 3996, 31} \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x)}{2 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

-1/2*(a*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) - (a*Tan[e + f*x])/(c*f*Sqrt[a +
a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) + (a*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 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 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 {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c} \\ & = -\frac {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2} \\ & = -\frac {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {(a \tan (e+f x)) \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 {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \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 = 0.82 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a \left (-2 \log (\cos (e+f x))-2 \log (1-\sec (e+f x))+\frac {3-2 \sec (e+f x)}{(-1+\sec (e+f x))^2}\right ) \tan (e+f x)}{2 c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

-1/2*(a*(-2*Log[Cos[e + f*x]] - 2*Log[1 - Sec[e + f*x]] + (3 - 2*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 [A] (verified)

Time = 2.38 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (8 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-16 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-6 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \csc \left (f x +e \right )}{16 f \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\)	\(237\)
risch	\(-\frac {\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (-4 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )} f x -8 i {\mathrm e}^{3 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+2 \,{\mathrm e}^{4 i \left (f x +e \right )} e -4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +12 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-8 \,{\mathrm e}^{3 i \left (f x +e \right )} e +6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x +2 i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+12 \,{\mathrm e}^{2 i \left (f x +e \right )} e -4 \,{\mathrm e}^{i \left (f x +e \right )} f x +6 i {\mathrm e}^{2 i \left (f x +e \right )}-8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-4 i {\mathrm e}^{3 i \left (f x +e \right )}+2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-8 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\)	\(348\)

[In]

int((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/16/f*2^(1/2)*(-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))*(8*ln((1-cos(f*x+e))^2*csc(f*x+e
)^2+1)*(1-cos(f*x+e))^4*csc(f*x+e)^4-16*ln(-cot(f*x+e)+csc(f*x+e))*(1-cos(f*x+e))^4*csc(f*x+e)^4-6*(1-cos(f*x+
e))^2*csc(f*x+e)^2+1)*csc(f*x+e)

Fricas [F]

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

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-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]

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

[In]

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

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))/(-c*(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. 1173 vs. \(2 (128) = 256\).

Time = 0.48 (sec) , antiderivative size = 1173, normalized size of antiderivative = 8.26 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-((f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x + e)*cos(3*f*x + 3*e)^2 + 36*(f*x + e)*cos(2*f*x + 2*e)^2 + 16*(f*x +
e)*cos(f*x + e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 16*(f*x + e)*sin(3*f*x + 3*e)^2 + 36*(f*x + e)*sin(2*f*x +
2*e)^2 + 16*(f*x + e)*sin(f*x + e)^2 + f*x + 2*(2*(4*cos(3*f*x + 3*e) - 6*cos(2*f*x + 2*e) + 4*cos(f*x + e) -
1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 + 8*(6*cos(2*f*x + 2*e) - 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) - 16*
cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) - 1)*cos(2*f*x + 2*e) - 36*cos(2*f*x + 2*e)^2 - 16*cos(f*x + e)^2 + 4*
(2*sin(3*f*x + 3*e) - 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) - sin(4*f*x + 4*e)^2 + 16*(3*sin(2
*f*x + 2*e) - 2*sin(f*x + e))*sin(3*f*x + 3*e) - 16*sin(3*f*x + 3*e)^2 - 36*sin(2*f*x + 2*e)^2 + 48*sin(2*f*x
+ 2*e)*sin(f*x + e) - 16*sin(f*x + e)^2 + 8*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) + 2*(f*x
- 4*(f*x + e)*cos(3*f*x + 3*e) + 6*(f*x + e)*cos(2*f*x + 2*e) - 4*(f*x + e)*cos(f*x + e) + e + 2*sin(3*f*x +
3*e) - 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*cos(4*f*x + 4*e) - 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) - 4*(f*x
+ e)*cos(f*x + e) + e)*cos(3*f*x + 3*e) + 12*(f*x - 4*(f*x + e)*cos(f*x + e) + e)*cos(2*f*x + 2*e) - 8*(f*x +
e)*cos(f*x + e) - 2*(4*(f*x + e)*sin(3*f*x + 3*e) - 6*(f*x + e)*sin(2*f*x + 2*e) + 4*(f*x + e)*sin(f*x + e) +
2*cos(3*f*x + 3*e) - 3*cos(2*f*x + 2*e) + 2*cos(f*x + e))*sin(4*f*x + 4*e) - 4*(12*(f*x + e)*sin(2*f*x + 2*e)
- 8*(f*x + e)*sin(f*x + e) - 1)*sin(3*f*x + 3*e) - 6*(8*(f*x + e)*sin(f*x + e) + 1)*sin(2*f*x + 2*e) + e + 4*s
in(f*x + e))*sqrt(a)*sqrt(c)/((c^3*cos(4*f*x + 4*e)^2 + 16*c^3*cos(3*f*x + 3*e)^2 + 36*c^3*cos(2*f*x + 2*e)^2
+ 16*c^3*cos(f*x + e)^2 + c^3*sin(4*f*x + 4*e)^2 + 16*c^3*sin(3*f*x + 3*e)^2 + 36*c^3*sin(2*f*x + 2*e)^2 - 48*
c^3*sin(2*f*x + 2*e)*sin(f*x + e) + 16*c^3*sin(f*x + e)^2 - 8*c^3*cos(f*x + e) + c^3 - 2*(4*c^3*cos(3*f*x + 3*
e) - 6*c^3*cos(2*f*x + 2*e) + 4*c^3*cos(f*x + e) - c^3)*cos(4*f*x + 4*e) - 8*(6*c^3*cos(2*f*x + 2*e) - 4*c^3*c
os(f*x + e) + c^3)*cos(3*f*x + 3*e) - 12*(4*c^3*cos(f*x + e) - c^3)*cos(2*f*x + 2*e) - 4*(2*c^3*sin(3*f*x + 3*
e) - 3*c^3*sin(2*f*x + 2*e) + 2*c^3*sin(f*x + e))*sin(4*f*x + 4*e) - 16*(3*c^3*sin(2*f*x + 2*e) - 2*c^3*sin(f*
x + e))*sin(3*f*x + 3*e))*f)

Giac [F]

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

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(5/2),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(5/2), x)