3.2.29 \(\int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx\) [129]

Optimal. Leaf size=345 \[ \frac {\log (\cos (e+f x)) \tan (e+f x)}{a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {5 \log (1-\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {11 \log (1+\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a^2 c f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

ln(cos(f*x+e))*tan(f*x+e)/a^2/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+5/16*ln(1-sec(f*x+e))*tan(f*x+
e)/a^2/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+11/16*ln(1+sec(f*x+e))*tan(f*x+e)/a^2/c/f/(a+a*sec(f*
x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/a^2/c/f/(1-sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e
))^(1/2)-1/8*tan(f*x+e)/a^2/c/f/(1+sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/2*tan(f*x+e)/
a^2/c/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3997, 90} \begin {gather*} -\frac {\tan (e+f x)}{8 a^2 c f (1-\sec (e+f x)) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a^2 c f (\sec (e+f x)+1) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (\sec (e+f x)+1)^2 \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {5 \tan (e+f x) \log (1-\sec (e+f x))}{16 a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {11 \tan (e+f x) \log (\sec (e+f x)+1)}{16 a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \log (\cos (e+f x))}{a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(Log[Cos[e + f*x]]*Tan[e + f*x])/(a^2*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (5*Log[1 - Sec[
e + f*x]]*Tan[e + f*x])/(16*a^2*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (11*Log[1 + Sec[e + f
*x]]*Tan[e + f*x])/(16*a^2*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(8*a^2*c*f*(1
- Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(8*a^2*c*f*(1 + Sec[e + f*x
])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(2*a^2*c*f*(1 + Sec[e + f*x])*Sqrt[a +
a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3997

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

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx &=-\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \frac {1}{x (a+a x)^3 (c-c x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \left (\frac {1}{8 a^3 c^2 (-1+x)^2}-\frac {5}{16 a^3 c^2 (-1+x)}+\frac {1}{a^3 c^2 x}-\frac {1}{4 a^3 c^2 (1+x)^3}-\frac {1}{2 a^3 c^2 (1+x)^2}-\frac {11}{16 a^3 c^2 (1+x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {\log (\cos (e+f x)) \tan (e+f x)}{a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {5 \log (1-\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {11 \log (1+\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a^2 c f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.44, size = 275, normalized size = 0.80 \begin {gather*} \frac {\left (-14+16 i f x-8 i f x \cos (3 (e+f x))-10 \log \left (1-e^{i (e+f x)}\right )+5 \cos (3 (e+f x)) \log \left (1-e^{i (e+f x)}\right )+\cos (e+f x) \left (-12+8 i f x-5 \log \left (1-e^{i (e+f x)}\right )-11 \log \left (1+e^{i (e+f x)}\right )\right )-22 \log \left (1+e^{i (e+f x)}\right )+11 \cos (3 (e+f x)) \log \left (1+e^{i (e+f x)}\right )+2 \cos (2 (e+f x)) \left (5-8 i f x+5 \log \left (1-e^{i (e+f x)}\right )+11 \log \left (1+e^{i (e+f x)}\right )\right )\right ) \tan (e+f x)}{32 a^2 c f (-1+\cos (e+f x)) (1+\cos (e+f x))^2 \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

((-14 + (16*I)*f*x - (8*I)*f*x*Cos[3*(e + f*x)] - 10*Log[1 - E^(I*(e + f*x))] + 5*Cos[3*(e + f*x)]*Log[1 - E^(
I*(e + f*x))] + Cos[e + f*x]*(-12 + (8*I)*f*x - 5*Log[1 - E^(I*(e + f*x))] - 11*Log[1 + E^(I*(e + f*x))]) - 22
*Log[1 + E^(I*(e + f*x))] + 11*Cos[3*(e + f*x)]*Log[1 + E^(I*(e + f*x))] + 2*Cos[2*(e + f*x)]*(5 - (8*I)*f*x +
5*Log[1 - E^(I*(e + f*x))] + 11*Log[1 + E^(I*(e + f*x))]))*Tan[e + f*x])/(32*a^2*c*f*(-1 + Cos[e + f*x])*(1 +
Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]
time = 0.27, size = 284, normalized size = 0.82

method	result	size
default	\(-\frac {\left (20 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-32 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-13 \left (\cos ^{3}\left (f x +e \right )\right )+20 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-32 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+7 \left (\cos ^{2}\left (f x +e \right )\right )-20 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+32 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos \left (f x +e \right )-20 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+32 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-11\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{32 f \,c^{3} \sin \left (f x +e \right )^{5} a^{3}}\)	\(284\)
risch	\(\frac {\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{a^{2} c \left ({\mathrm e}^{2 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}}\, \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 \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{a^{2} c \left ({\mathrm e}^{2 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}}\, \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 {i \left (5 \,{\mathrm e}^{5 i \left (f x +e \right )}-6 \,{\mathrm e}^{4 i \left (f x +e \right )}-14 \,{\mathrm e}^{3 i \left (f x +e \right )}-6 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{2} c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \left ({\mathrm e}^{2 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 ) \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 {5 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 a^{2} c \left ({\mathrm e}^{2 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}}\, \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 {11 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 a^{2} c \left ({\mathrm e}^{2 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}}\, \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}\)	\(621\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/f*(20*cos(f*x+e)^3*ln(-(-1+cos(f*x+e))/sin(f*x+e))-32*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-13*cos(f*x+e)^3+
20*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))-32*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+7*cos(f*x+e)^2-20*cos(f*x
+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+32*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+cos(f*x+e)-20*ln(-(-1+cos(f*x+e))/sin(f
*x+e))+32*ln(2/(cos(f*x+e)+1))-11)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*cos(f*x+e)^2*(c*(-1+cos(f*x+e))/cos(f*x
+e))^(3/2)/c^3/sin(f*x+e)^5/a^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 4644 vs. \(2 (333) = 666\).
time = 3.36, size = 4644, normalized size = 13.46 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-1/8*(8*(f*x + e)*cos(6*f*x + 6*e)^2 + 8*(f*x + e)*cos(4*f*x + 4*e)^2 + 8*(f*x + e)*cos(2*f*x + 2*e)^2 + 32*(f
*x + e)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 128*(f*x + e)*cos(3/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e)))^2 + 32*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 8*(f*x + e)*si
n(6*f*x + 6*e)^2 + 8*(f*x + e)*sin(4*f*x + 4*e)^2 + 8*(f*x + e)*sin(2*f*x + 2*e)^2 + 32*(f*x + e)*sin(5/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 128*(f*x + e)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 32*(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 8*f*x + 11*(2*(cos(4*f*x + 4*e) +
cos(2*f*x + 2*e) - 1)*cos(6*f*x + 6*e) - cos(6*f*x + 6*e)^2 - 2*(cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(
4*f*x + 4*e)^2 - cos(2*f*x + 2*e)^2 - 4*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) - 4*cos(3/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)*cos(5
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 8
*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 2*cos(1/2*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))) - 16*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e)))^2 - 4*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) - 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*(sin(4*f*x + 4*e) + sin(2*f
*x + 2*e))*sin(6*f*x + 6*e) - sin(6*f*x + 6*e)^2 - sin(4*f*x + 4*e)^2 - 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) -
sin(2*f*x + 2*e)^2 - 4*(sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - sin(2*f*x + 2*e) - 4*sin(3/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(5/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - 4*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 8*(sin(6*f*x + 6*e) -
sin(4*f*x + 4*e) - sin(2*f*x + 2*e) + 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))) - 16*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 4*(sin(6*f*
x + 6*e) - sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*sin(1
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*cos(2*f*x + 2*e) - 1)*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) + 5*(2*(cos(4*f*x + 4*e)
+ cos(2*f*x + 2*e) - 1)*cos(6*f*x + 6*e) - cos(6*f*x + 6*e)^2 - 2*(cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - c
os(4*f*x + 4*e)^2 - cos(2*f*x + 2*e)^2 - 4*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) - 4*cos(3/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)*co
s(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2
+ 8*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 2*cos(1/2*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))) - 16*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e)))^2 - 4*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) - 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*(sin(4*f*x + 4*e) + sin(
2*f*x + 2*e))*sin(6*f*x + 6*e) - sin(6*f*x + 6*e)^2 - sin(4*f*x + 4*e)^2 - 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e)
- sin(2*f*x + 2*e)^2 - 4*(sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - sin(2*f*x + 2*e) - 4*sin(3/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))) + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(5/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) - 4*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 8*(sin(6*f*x + 6*e)
- sin(4*f*x + 4*e) - sin(2*f*x + 2*e) + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 16*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 4*(sin(6
*f*x + 6*e) - sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*si
n(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*cos(2*f*x + 2*e) - 1)*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) + 4*(4*f*x - 4*(f*x +
e)*cos(4*f*x + 4*e) - 4*(f*x + e)*cos(2*f*x + 2*e) + 4*e + 3*sin(4*f*x + 4*e) + 3*sin(2*f*x + 2*e))*cos(6*f*x
+ 6*e) - 16*(f*x - (f*x + e)*cos(2*f*x + 2*e) + e)*cos(4*f*x + 4*e) - 16*(f*x + e)*cos(2*f*x + 2*e) + 2*(16*f
*x + 16*(f*x + e)*cos(6*f*x + 6*e) - 16*(f*x + e)*cos(4*f*x + 4*e) - 16*(f*x + e)*cos(2*f*x + 2*e) - 64*(f*x +
e)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + 16*e + 5*sin(6*f*x + 6*e) + 7*sin(4*f*x + 4*e) + 7*sin(2*f*x + 2*e) + 8*sin(3/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))))*cos(5/2*arctan...

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

[Out]

integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^3*c^2*sec(f*x + e)^5 + a^3*c^2*sec(f*x + e)^4 -
2*a^3*c^2*sec(f*x + e)^3 - 2*a^3*c^2*sec(f*x + e)^2 + a^3*c^2*sec(f*x + e) + a^3*c^2), x)

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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Giac [A]
time = 2.02, size = 193, normalized size = 0.56 \begin {gather*} -\frac {\frac {10 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} a^{2} {\left | c \right |}} + \frac {32 \, \sqrt {-a c} \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{a^{3} c {\left | c \right |}} - \frac {2 \, {\left (5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}}{\sqrt {-a c} a^{2} c {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{3} c^{3} {\left | c \right |} - 8 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{3} c^{4} {\left | c \right |}}{a^{6} c^{8}}}{32 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/32*(10*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*a^2*abs(c)) + 32*sqrt(-a*c)*log(abs(c*tan(1/2*f*x + 1
/2*e)^2 + c))/(a^3*c*abs(c)) - 2*(5*c*tan(1/2*f*x + 1/2*e)^2 - c)/(sqrt(-a*c)*a^2*c*abs(c)*tan(1/2*f*x + 1/2*e
)^2) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^3*c^3*abs(c) - 8*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*
c)*a^3*c^4*abs(c))/(a^6*c^8))/f

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

[Out]

int(1/((a + a/cos(e + f*x))^(5/2)*(c - c/cos(e + f*x))^(3/2)), x)

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