3.2.39 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx\) [139]

Optimal. Leaf size=142 \[ \frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2}} \]

[Out]

c*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(3/2)+4*c^3*ln(1+sec(f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*
x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+2*c^2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4039, 4040, 4037} \begin {gather*} \frac {4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{a f \sqrt {a \sec (e+f x)+a}}+\frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

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

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)]

Rule 4040

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx &=\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2}}-\frac {(2 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a}\\ &=\frac {2 c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2}}-\frac {\left (4 c^2\right ) \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a}\\ &=\frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 1.03, size = 183, normalized size = 1.29 \begin {gather*} -\frac {c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \left (-1+4 \log \left (1+e^{i (e+f x)}\right )+\cos (e+f x) \left (-5+8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right )+\cos (2 (e+f x)) \left (4 \log \left (1+e^{i (e+f x)}\right )-2 \log \left (1+e^{2 i (e+f x)}\right )\right )-2 \log \left (1+e^{2 i (e+f x)}\right )\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)}}{a f (1+\cos (e+f x)) \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

-((c^2*Cot[(e + f*x)/2]*(-1 + 4*Log[1 + E^(I*(e + f*x))] + Cos[e + f*x]*(-5 + 8*Log[1 + E^(I*(e + f*x))] - 4*L
og[1 + E^((2*I)*(e + f*x))]) + Cos[2*(e + f*x)]*(4*Log[1 + E^(I*(e + f*x))] - 2*Log[1 + E^((2*I)*(e + f*x))])
- 2*Log[1 + E^((2*I)*(e + f*x))])*Sec[e + f*x]*Sqrt[c - c*Sec[e + f*x]])/(a*f*(1 + Cos[e + f*x])*Sqrt[a*(1 + S
ec[e + f*x])]))

________________________________________________________________________________________

Maple [A]
time = 2.86, size = 235, normalized size = 1.65

method	result	size
default	\(-\frac {\left (4 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 \cos \left (f x +e \right )+1\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right )^{3} \left (-1+\cos \left (f x +e \right )\right ) a^{2}}\)	\(235\)
risch	\(\frac {2 i c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (5 \,{\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {8 i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}+\frac {4 i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*(4*cos(f*x+e)^2*ln(-(cos(f*x+e)-1+sin(f*x+e))/sin(f*x+e))+4*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*co
s(f*x+e)^2-cos(f*x+e)^2+4*cos(f*x+e)*ln(-(cos(f*x+e)-1+sin(f*x+e))/sin(f*x+e))+4*cos(f*x+e)*ln(-(-1+cos(f*x+e)
-sin(f*x+e))/sin(f*x+e))+4*cos(f*x+e)+1)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*cos(f*x+e)^2*(a*(cos(f*x+e)+1)/c
os(f*x+e))^(1/2)/sin(f*x+e)^3/(-1+cos(f*x+e))/a^2

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 2185 vs. \(2 (140) = 280\).
time = 0.78, size = 2185, normalized size = 15.39 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-2*(8*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
^2 + 8*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 8*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
))^2 + 8*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
)))^2 - 2*c^2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) + 2*c^2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e) + 2*c^2*sin(2*f*x +
2*e) + 2*(c^2*cos(4*f*x + 4*e)^2 + 4*c^2*cos(2*f*x + 2*e)^2 + c^2*sin(4*f*x + 4*e)^2 + 4*c^2*sin(4*f*x + 4*e)*
sin(2*f*x + 2*e) + 4*c^2*sin(2*f*x + 2*e)^2 + 4*c^2*cos(2*f*x + 2*e) + c^2 + 2*(2*c^2*cos(2*f*x + 2*e) + c^2)*
cos(4*f*x + 4*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*(c^2*cos(4*f*x + 4*e)^2 + 4*c^2*cos(2*f*
x + 2*e)^2 + 4*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*c^2*cos(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e)))^2 + c^2*sin(4*f*x + 4*e)^2 + 4*c^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*c^2*sin(2*f*
x + 2*e)^2 + 4*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*c^2*sin(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e)))^2 + 4*c^2*cos(2*f*x + 2*e) + c^2 + 2*(2*c^2*cos(2*f*x + 2*e) + c^2)*cos(4*f*x + 4*e)
+ 4*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2*e) + 2*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
) + c^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2*
e) + c^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2
*e) + 2*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 4*(c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))))*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 1) + (16*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) - 5*c^2*sin(4*f*x + 4*e) - 6*c^2*sin(2*f*x + 2*e) + 8*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x
+ 2*e) + c^2)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))) - (5*c^2*sin(4*f*x + 4*e) + 6*c^2*sin(2*f*x + 2*e) - 8*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2*e) +
c^2)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (
16*c^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
5*c^2*cos(4*f*x + 4*e) + 6*c^2*cos(2*f*x + 2*e) + 5*c^2 + 8*(c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2*e))*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (5*c^2*c
os(4*f*x + 4*e) + 6*c^2*cos(2*f*x + 2*e) + 5*c^2 + 8*(c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2*e))*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/
((a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x + 2*e)^2 + 4*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
))^2 + 4*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x
+ 4*e)*sin(2*f*x + 2*e) + 4*a^2*sin(2*f*x + 2*e)^2 + 4*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
))^2 + 4*a^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*a^2*cos(2*f*x + 2*e) + a^2 + 2*(2*a^2*
cos(2*f*x + 2*e) + a^2)*cos(4*f*x + 4*e) + 4*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + 2*a^2*cos(1/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + a^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(a^
2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(a
^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e) + 2*a^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin
(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*sin(1/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*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(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

________________________________________________________________________________________

Giac [A]
time = 1.61, size = 161, normalized size = 1.13 \begin {gather*} -\frac {2 \, c^{2} {\left (\frac {2 \, \sqrt {-a c} c \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}{a^{2} {\left | c \right |}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c}}{a^{2} {\left | c \right |}} - \frac {2 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} c + \sqrt {-a c} c^{2}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} a^{2} {\left | c \right |}}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-2*c^2*(2*sqrt(-a*c)*c*log(c*tan(1/2*f*x + 1/2*e)^2 - c)/(a^2*abs(c)) + (c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a
*c)/(a^2*abs(c)) - (2*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*c + sqrt(-a*c)*c^2)/((c*tan(1/2*f*x + 1/2*e)^2
- c)*a^2*abs(c)))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(5/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^(3/2)),x)

[Out]

int((c - c/cos(e + f*x))^(5/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^(3/2)), x)

________________________________________________________________________________________