3.1.59 \(\int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x)) \, dx\) [59]

Optimal. Leaf size=132 \[ \frac {2 a^{5/2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}-\frac {2 a^3 c \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {2 a^4 c \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}-\frac {2 a^5 c \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}} \]

[Out]

2*a^(5/2)*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/f-2*a^3*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2*
a^4*c*tan(f*x+e)^3/f/(a+a*sec(f*x+e))^(3/2)-2/5*a^5*c*tan(f*x+e)^5/f/(a+a*sec(f*x+e))^(5/2)

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Rubi [A]
time = 0.10, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3989, 3972, 472, 209} \begin {gather*} \frac {2 a^{5/2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}-\frac {2 a^5 c \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}-\frac {2 a^4 c \tan ^3(e+f x)}{f (a \sec (e+f x)+a)^{3/2}}-\frac {2 a^3 c \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^3*c*Tan[e + f*x])/(f*Sqrt[a + a
*Sec[e + f*x]]) - (2*a^4*c*Tan[e + f*x]^3)/(f*(a + a*Sec[e + f*x])^(3/2)) - (2*a^5*c*Tan[e + f*x]^5)/(5*f*(a +
a*Sec[e + f*x])^(5/2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3972

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[-2*(a^(m/2 +
n + 1/2)/d), Subst[Int[x^m*((2 + a*x^2)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 3989

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

Rubi steps

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

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Mathematica [A]
time = 0.84, size = 110, normalized size = 0.83 \begin {gather*} -\frac {a^2 c \left (-10 \text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right ) \cos ^2(e+f x)+(3+6 \cos (e+f x)+\cos (2 (e+f x))) \sqrt {-1+\sec (e+f x)}\right ) \sec ^2(e+f x) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{5 f \sqrt {-1+\sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*c*(-10*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]^2 + (3 + 6*Cos[e + f*x] + Cos[2*(e + f*x)])*Sqrt
[-1 + Sec[e + f*x]])*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(f*Sqrt[-1 + Sec[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(118)=236\).
time = 0.18, size = 303, normalized size = 2.30

method	result	size
default	\(-\frac {c \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (5 \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+10 \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {2}+5 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \sin \left (f x +e \right )-8 \left (\cos ^{3}\left (f x +e \right )\right )-16 \left (\cos ^{2}\left (f x +e \right )\right )+16 \cos \left (f x +e \right )+8\right ) a^{2}}{20 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2}}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*c/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(5*sin(f*x+e)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*s
in(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*cos(f*x+e)^2*2^(1/2)+10*sin(f*x+e)*arctanh(
1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*c
os(f*x+e)*2^(1/2)+5*2^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-
2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*sin(f*x+e)-8*cos(f*x+e)^3-16*cos(f*x+e)^2+16*cos(f*x+e)+8)/sin(f*x+e)/cos(f
*x+e)^2*a^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1501 vs. \(2 (126) = 252\).
time = 0.61, size = 1501, normalized size = 11.37 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(30*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/
4)*((12*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(2*f*x + 2*e) - 3*a^2*sin(2*f*x + 2*e) - 4
*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(3/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (12*a^2*sin(2*f*x + 2*e)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) + 3*a^2*cos(2*f*x + 2*e) - a^2 + 4*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*cos(3/2*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) + 1)))*sqrt(a) + 3*((a^2*cos(2*f*x + 2*e)^
2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*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) + 1))*sin(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(c
os(1/2*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))) +
sin(1/2*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)))
) + 1) - (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x +
2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e)))*sin(1/2*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) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2
*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*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))) + sin(1/2*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)))) - 1) - (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2
*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/
4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) + (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2
*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) +
1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 1))*sqrt(a))*c/((cos(
2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*f)

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Fricas [A]
time = 3.35, size = 382, normalized size = 2.89 \begin {gather*} \left [\frac {5 \, {\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, {\left (a^{2} c \cos \left (f x + e\right )^{2} + 3 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{5 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac {2 \, {\left (5 \, {\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + {\left (a^{2} c \cos \left (f x + e\right )^{2} + 3 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5*(5*(a^2*c*cos(f*x + e)^3 + a^2*c*cos(f*x + e)^2)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*co
s(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 2*(a^2*c*c
os(f*x + e)^2 + 3*a^2*c*cos(f*x + e) + a^2*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x
+ e)^3 + f*cos(f*x + e)^2), -2/5*(5*(a^2*c*cos(f*x + e)^3 + a^2*c*cos(f*x + e)^2)*sqrt(a)*arctan(sqrt((a*cos(
f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (a^2*c*cos(f*x + e)^2 + 3*a^2*c*cos(f*x + e
) + a^2*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - c \left (\int \left (- a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}\right )\, dx + \int \left (- a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}\right )\, dx + \int a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )}\, dx + \int a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{3}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(Integral(-a**2*sqrt(a*sec(e + f*x) + a), x) + Integral(-a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x), x) + I
ntegral(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2, x) + Integral(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x
)**3, x))

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Giac [A]
time = 1.43, size = 227, normalized size = 1.72 \begin {gather*} -\frac {\frac {5 \, \sqrt {-a} a^{3} c \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (\sqrt {2} a^{5} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, \sqrt {2} a^{5} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}}}{5 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(5*sqrt(-a)*a^3*c*log(abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 4*s
qrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 4*sqrt(2)
*abs(a) - 6*a))*sgn(cos(f*x + e))/abs(a) - 2*(sqrt(2)*a^5*c*sgn(cos(f*x + e))*tan(1/2*f*x + 1/2*e)^4 - 5*sqrt(
2)*a^5*c*sgn(cos(f*x + e)))*tan(1/2*f*x + 1/2*e)/((a*tan(1/2*f*x + 1/2*e)^2 - a)^2*sqrt(-a*tan(1/2*f*x + 1/2*e
)^2 + a)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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