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

Optimal. Leaf size=142 \[ -\frac {a \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[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]
time = 0.18, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3992, 3996, 31} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {\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 [C] Result contains complex when optimal does not.
time = 1.36, size = 152, normalized size = 1.07 \begin {gather*} \frac {\left (3-3 i f x+\cos (e+f x) \left (-4+4 i f x-8 \log \left (1-e^{i (e+f x)}\right )\right )+6 \log \left (1-e^{i (e+f x)}\right )+\cos (2 (e+f x)) \left (-i f x+2 \log \left (1-e^{i (e+f x)}\right )\right )\right ) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{2 c^2 f (-1+\cos (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 - (3*I)*f*x + Cos[e + f*x]*(-4 + (4*I)*f*x - 8*Log[1 - E^(I*(e + f*x))]) + 6*Log[1 - E^(I*(e + f*x))] + Co
s[2*(e + f*x)]*((-I)*f*x + 2*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(2*c^2*f*
(-1 + Cos[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [A]
time = 0.27, size = 226, normalized size = 1.59

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (8 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-16 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+7 \left (\cos ^{2}\left (f x +e \right )\right )-16 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+32 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \cos \left (f x +e \right )+8 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-16 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-5\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{8 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(226\)
risch	\(\frac {\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}-\frac {2 \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {2 i \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (2 \,{\mathrm e}^{3 i \left (f x +e \right )}-3 \,{\mathrm e}^{2 i \left (f x +e \right )}+2 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {2 i \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(422\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*(-1+cos(f*x+e))*(8*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-16*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))+7*c
os(f*x+e)^2-16*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+32*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+2*cos(f*x+e)+8*ln
(2/(cos(f*x+e)+1))-16*ln(-(-1+cos(f*x+e))/sin(f*x+e))-5)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)/sin(f*x+e)/cos(f*
x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1288 vs. \(2 (138) = 276\).
time = 0.75, size = 1288, normalized size = 9.07 \begin {gather*} -\frac {{\left ({\left (f x + e\right )} \cos \left (4 \, f x + 4 \, e\right )^{2} + 16 \, {\left (f x + e\right )} \cos \left (3 \, f x + 3 \, e\right )^{2} + 36 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 16 \, {\left (f x + e\right )} \cos \left (f x + e\right )^{2} + {\left (f x + e\right )} \sin \left (4 \, f x + 4 \, e\right )^{2} + 16 \, {\left (f x + e\right )} \sin \left (3 \, f x + 3 \, e\right )^{2} + 36 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + 16 \, {\left (f x + e\right )} \sin \left (f x + e\right )^{2} + f x + 2 \, {\left (2 \, {\left (4 \, \cos \left (3 \, f x + 3 \, e\right ) - 6 \, \cos \left (2 \, f x + 2 \, e\right ) + 4 \, \cos \left (f x + e\right ) - 1\right )} \cos \left (4 \, f x + 4 \, e\right ) - \cos \left (4 \, f x + 4 \, e\right )^{2} + 8 \, {\left (6 \, \cos \left (2 \, f x + 2 \, e\right ) - 4 \, \cos \left (f x + e\right ) + 1\right )} \cos \left (3 \, f x + 3 \, e\right ) - 16 \, \cos \left (3 \, f x + 3 \, e\right )^{2} + 12 \, {\left (4 \, \cos \left (f x + e\right ) - 1\right )} \cos \left (2 \, f x + 2 \, e\right ) - 36 \, \cos \left (2 \, f x + 2 \, e\right )^{2} - 16 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (2 \, \sin \left (3 \, f x + 3 \, e\right ) - 3 \, \sin \left (2 \, f x + 2 \, e\right ) + 2 \, \sin \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) - \sin \left (4 \, f x + 4 \, e\right )^{2} + 16 \, {\left (3 \, \sin \left (2 \, f x + 2 \, e\right ) - 2 \, \sin \left (f x + e\right )\right )} \sin \left (3 \, f x + 3 \, e\right ) - 16 \, \sin \left (3 \, f x + 3 \, e\right )^{2} - 36 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 48 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 16 \, \sin \left (f x + e\right )^{2} + 8 \, \cos \left (f x + e\right ) - 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + 2 \, {\left (f x - 4 \, {\left (f x + e\right )} \cos \left (3 \, f x + 3 \, e\right ) + 6 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e + 2 \, \sin \left (3 \, f x + 3 \, e\right ) - 3 \, \sin \left (2 \, f x + 2 \, e\right ) + 2 \, \sin \left (f x + e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) - 8 \, {\left (f x + 6 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e\right )} \cos \left (3 \, f x + 3 \, e\right ) + 12 \, {\left (f x - 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 8 \, {\left (f x + e\right )} \cos \left (f x + e\right ) - 2 \, {\left (4 \, {\left (f x + e\right )} \sin \left (3 \, f x + 3 \, e\right ) - 6 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) + 4 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + 2 \, \cos \left (3 \, f x + 3 \, e\right ) - 3 \, \cos \left (2 \, f x + 2 \, e\right ) + 2 \, \cos \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) - 4 \, {\left (12 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) - 8 \, {\left (f x + e\right )} \sin \left (f x + e\right ) - 1\right )} \sin \left (3 \, f x + 3 \, e\right ) - 6 \, {\left (8 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + 1\right )} \sin \left (2 \, f x + 2 \, e\right ) + e + 4 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (c^{3} \cos \left (4 \, f x + 4 \, e\right )^{2} + 16 \, c^{3} \cos \left (3 \, f x + 3 \, e\right )^{2} + 36 \, c^{3} \cos \left (2 \, f x + 2 \, e\right )^{2} + 16 \, c^{3} \cos \left (f x + e\right )^{2} + c^{3} \sin \left (4 \, f x + 4 \, e\right )^{2} + 16 \, c^{3} \sin \left (3 \, f x + 3 \, e\right )^{2} + 36 \, c^{3} \sin \left (2 \, f x + 2 \, e\right )^{2} - 48 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 16 \, c^{3} \sin \left (f x + e\right )^{2} - 8 \, c^{3} \cos \left (f x + e\right ) + c^{3} - 2 \, {\left (4 \, c^{3} \cos \left (3 \, f x + 3 \, e\right ) - 6 \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) + 4 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \cos \left (4 \, f x + 4 \, e\right ) - 8 \, {\left (6 \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \cos \left (3 \, f x + 3 \, e\right ) - 12 \, {\left (4 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (2 \, c^{3} \sin \left (3 \, f x + 3 \, e\right ) - 3 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) + 2 \, c^{3} \sin \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) - 16 \, {\left (3 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) - 2 \, c^{3} \sin \left (f x + e\right )\right )} \sin \left (3 \, f x + 3 \, e\right )\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]
time = 1.85, size = 178, normalized size = 1.25 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {8 \, \sqrt {2} \sqrt {-a c} a \log \left (2 \, {\left | a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \right |}\right )}{c^{3} {\left | a \right |}} - \frac {8 \, \sqrt {2} \sqrt {-a c} a \log \left ({\left | -a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \right |}\right )}{c^{3} {\left | a \right |}} - \frac {\sqrt {2} {\left (12 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )}^{2} \sqrt {-a c} a + 18 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )} \sqrt {-a c} a^{2} + 7 \, \sqrt {-a c} a^{3}\right )}}{a^{2} c^{3} {\left | a \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )}}{16 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*(8*sqrt(2)*sqrt(-a*c)*a*log(2*abs(a*tan(1/2*f*x + 1/2*e)^2))/(c^3*abs(a)) - 8*sqrt(2)*sqrt(-a*c)
*a*log(abs(-a*tan(1/2*f*x + 1/2*e)^2 - a))/(c^3*abs(a)) - sqrt(2)*(12*(a*tan(1/2*f*x + 1/2*e)^2 - a)^2*sqrt(-a
*c)*a + 18*(a*tan(1/2*f*x + 1/2*e)^2 - a)*sqrt(-a*c)*a^2 + 7*sqrt(-a*c)*a^3)/(a^2*c^3*abs(a)*tan(1/2*f*x + 1/2
*e)^4))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________