3.1.79 \(\int \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^{5/2}} \, dx\) [79]

Optimal. Leaf size=260 \[ \frac {2 c^5 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {23 \sqrt {2} c^5 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}+\frac {21 c^5 \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a+a \sec (e+f x))^{3/2}}+\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}} \]

[Out]

2*c^5*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f-23*c^5*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)
/(a+a*sec(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/f+21*c^5*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)-19/6*c^5*tan(f*x+e)^
3/a/f/(a+a*sec(f*x+e))^(3/2)+3/4*c^5*sec(1/2*f*x+1/2*e)^2*sin(f*x+e)*tan(f*x+e)^4/f/(a+a*sec(f*x+e))^(5/2)+1/4
*a*c^5*sec(1/2*f*x+1/2*e)^4*sin(f*x+e)^2*tan(f*x+e)^5/f/(a+a*sec(f*x+e))^(7/2)

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Rubi [A]
time = 0.23, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3989, 3972, 481, 592, 596, 536, 209} \begin {gather*} \frac {2 c^5 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac {23 \sqrt {2} c^5 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{a^{5/2} f}+\frac {21 c^5 \tan (e+f x)}{a^2 f \sqrt {a \sec (e+f x)+a}}-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a \sec (e+f x)+a)^{3/2}}+\frac {a c^5 \sin ^2(e+f x) \tan ^5(e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right )}{4 f (a \sec (e+f x)+a)^{7/2}}+\frac {3 c^5 \sin (e+f x) \tan ^4(e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{4 f (a \sec (e+f x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^5*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (23*Sqrt[2]*c^5*ArcTan[(Sqrt[a]*
Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(a^(5/2)*f) + (21*c^5*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e
+ f*x]]) - (19*c^5*Tan[e + f*x]^3)/(6*a*f*(a + a*Sec[e + f*x])^(3/2)) + (3*c^5*Sec[(e + f*x)/2]^2*Sin[e + f*x
]*Tan[e + f*x]^4)/(4*f*(a + a*Sec[e + f*x])^(5/2)) + (a*c^5*Sec[(e + f*x)/2]^4*Sin[e + f*x]^2*Tan[e + f*x]^5)/
(4*f*(a + a*Sec[e + f*x])^(7/2))

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 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 592

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c -
a*d)*(p + 1))), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 596

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q +
1) + 1))), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

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 \frac {(c-c \sec (e+f x))^5}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\left (\left (a^5 c^5\right ) \int \frac {\tan ^{10}(e+f x)}{(a+a \sec (e+f x))^{15/2}} \, dx\right )\\ &=\frac {\left (2 a^3 c^5\right ) \text {Subst}\left (\int \frac {x^{10}}{\left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}+\frac {\left (a c^5\right ) \text {Subst}\left (\int \frac {x^6 \left (14+10 a x^2\right )}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{2 f}\\ &=\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}-\frac {c^5 \text {Subst}\left (\int \frac {x^4 \left (-30 a-38 a^2 x^2\right )}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{4 a f}\\ &=-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a+a \sec (e+f x))^{3/2}}+\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}+\frac {c^5 \text {Subst}\left (\int \frac {x^2 \left (-228 a^2-252 a^3 x^2\right )}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{12 a^3 f}\\ &=\frac {21 c^5 \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a+a \sec (e+f x))^{3/2}}+\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}-\frac {c^5 \text {Subst}\left (\int \frac {-504 a^3-528 a^4 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{12 a^5 f}\\ &=\frac {21 c^5 \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a+a \sec (e+f x))^{3/2}}+\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}-\frac {\left (2 c^5\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 )}{a^2 f}+\frac {\left (46 c^5\right ) \text {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^2 f}\\ &=\frac {2 c^5 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {23 \sqrt {2} c^5 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}+\frac {21 c^5 \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {19 c^5 \tan ^3(e+f x)}{6 a f (a+a \sec (e+f x))^{3/2}}+\frac {3 c^5 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {a c^5 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^5(e+f x)}{4 f (a+a \sec (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 3.70, size = 180, normalized size = 0.69 \begin {gather*} \frac {c^5 \cot \left (\frac {1}{2} (e+f x)\right ) \left ((81-30 \cos (e+f x)+52 \cos (2 (e+f x))-66 \cos (3 (e+f x))-37 \cos (4 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )+96 \text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right ) \cos ^2(e+f x) \sqrt {-1+\sec (e+f x)}-1104 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {-1+\sec (e+f x)}}{\sqrt {2}}\right ) \cos ^2(e+f x) \sqrt {-1+\sec (e+f x)}\right ) \sec ^2(e+f x)}{48 a^2 f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^5*Cot[(e + f*x)/2]*((81 - 30*Cos[e + f*x] + 52*Cos[2*(e + f*x)] - 66*Cos[3*(e + f*x)] - 37*Cos[4*(e + f*x)]
)*Sec[(e + f*x)/2]^4 + 96*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]^2*Sqrt[-1 + Sec[e + f*x]] - 1104*Sqrt[2
]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*Cos[e + f*x]^2*Sqrt[-1 + Sec[e + f*x]])*Sec[e + f*x]^2)/(48*a^2*f*Sq
rt[a*(1 + Sec[e + f*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(725\) vs. \(2(229)=458\).
time = 0.27, size = 726, normalized size = 2.79

method	result	size
default	\(-\frac {c^{5} \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (-1+\cos \left (f x +e \right )\right )^{2} \left (-3 \left (\cos ^{3}\left (f x +e \right )\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 {3}{2}} \sin \left (f x +e \right ) \sqrt {2}-69 \left (\cos ^{3}\left (f x +e \right )\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )-9 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{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 ) \sqrt {2}-207 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-9 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{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 ) \sqrt {2}-207 \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )-3 \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 {3}{2}} \sin \left (f x +e \right )-69 \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} \sin \left (f x +e \right )+148 \left (\cos ^{4}\left (f x +e \right )\right )+132 \left (\cos ^{3}\left (f x +e \right )\right )-200 \left (\cos ^{2}\left (f x +e \right )\right )-84 \cos \left (f x +e \right )+4\right )}{6 f \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} a^{3}}\)	\(726\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*c^5/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-1+cos(f*x+e))^2*(-3*cos(f*x+e)^3*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))^(3/2)*sin(f*x+e)*2^(1/2)-69
*cos(f*x+e)^3*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x
+e)+1)/sin(f*x+e))*sin(f*x+e)-9*cos(f*x+e)^2*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*arctanh(1/2*(-2*c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*2^(1/2)-207*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2
)*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)-9*cos(
f*x+e)*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*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^(1/2)-207*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*ln((sin(f*x+e)*(-2*cos(f*
x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*cos(f*x+e)-3*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))^(3/2)*sin(f*x+e)-69*ln((sin(f*x+e)*
(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*sin(f*x+e)
+148*cos(f*x+e)^4+132*cos(f*x+e)^3-200*cos(f*x+e)^2-84*cos(f*x+e)+4)/cos(f*x+e)/sin(f*x+e)^5/a^3

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]
time = 7.11, size = 801, normalized size = 3.08 \begin {gather*} \left [\frac {69 \, \sqrt {2} {\left (a c^{5} \cos \left (f x + e\right )^{4} + 3 \, a c^{5} \cos \left (f x + e\right )^{3} + 3 \, a c^{5} \cos \left (f x + e\right )^{2} + a c^{5} \cos \left (f x + e\right )\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 6 \, {\left (c^{5} \cos \left (f x + e\right )^{4} + 3 \, c^{5} \cos \left (f x + e\right )^{3} + 3 \, c^{5} \cos \left (f x + e\right )^{2} + c^{5} \cos \left (f x + e\right )\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 ) + 4 \, {\left (37 \, c^{5} \cos \left (f x + e\right )^{3} + 70 \, c^{5} \cos \left (f x + e\right )^{2} + 20 \, c^{5} \cos \left (f x + e\right ) - c^{5}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{6 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f \cos \left (f x + e\right )\right )}}, -\frac {6 \, {\left (c^{5} \cos \left (f x + e\right )^{4} + 3 \, c^{5} \cos \left (f x + e\right )^{3} + 3 \, c^{5} \cos \left (f x + e\right )^{2} + c^{5} \cos \left (f x + e\right )\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 ) - 2 \, {\left (37 \, c^{5} \cos \left (f x + e\right )^{3} + 70 \, c^{5} \cos \left (f x + e\right )^{2} + 20 \, c^{5} \cos \left (f x + e\right ) - c^{5}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \frac {69 \, \sqrt {2} {\left (a c^{5} \cos \left (f x + e\right )^{4} + 3 \, a c^{5} \cos \left (f x + e\right )^{3} + 3 \, a c^{5} \cos \left (f x + e\right )^{2} + a c^{5} \cos \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {2} \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 )}{\sqrt {a}}}{3 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f \cos \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(69*sqrt(2)*(a*c^5*cos(f*x + e)^4 + 3*a*c^5*cos(f*x + e)^3 + 3*a*c^5*cos(f*x + e)^2 + a*c^5*cos(f*x + e))
*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) + 3*co
s(f*x + e)^2 + 2*cos(f*x + e) - 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 6*(c^5*cos(f*x + e)^4 + 3*c^5*cos(
f*x + e)^3 + 3*c^5*cos(f*x + e)^2 + c^5*cos(f*x + e))*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)) + 4*(37*c^5*
cos(f*x + e)^3 + 70*c^5*cos(f*x + e)^2 + 20*c^5*cos(f*x + e) - c^5)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*si
n(f*x + e))/(a^3*f*cos(f*x + e)^4 + 3*a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + a^3*f*cos(f*x + e)), -1/
3*(6*(c^5*cos(f*x + e)^4 + 3*c^5*cos(f*x + e)^3 + 3*c^5*cos(f*x + e)^2 + c^5*cos(f*x + e))*sqrt(a)*arctan(sqrt
((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 2*(37*c^5*cos(f*x + e)^3 + 70*c^5*c
os(f*x + e)^2 + 20*c^5*cos(f*x + e) - c^5)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 69*sqrt(2)*(
a*c^5*cos(f*x + e)^4 + 3*a*c^5*cos(f*x + e)^3 + 3*a*c^5*cos(f*x + e)^2 + a*c^5*cos(f*x + e))*arctan(sqrt(2)*sq
rt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/(a^3*f*cos(f*x + e)^4 + 3*
a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + a^3*f*cos(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - c^{5} \left (\int \frac {5 \sec {\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx + \int \left (- \frac {10 \sec ^{2}{\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\right )\, dx + \int \frac {10 \sec ^{3}{\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx + \int \left (- \frac {5 \sec ^{4}{\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\right )\, dx + \int \frac {\sec ^{5}{\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx + \int \left (- \frac {1}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**5*(Integral(5*sec(e + f*x)/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a
)*sec(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x) + Integral(-10*sec(e + f*x)**2/(a**2*sqrt(a*sec(e + f*x) +
a)*sec(e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x) + Inte
gral(10*sec(e + f*x)**3/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e
+ f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x) + Integral(-5*sec(e + f*x)**4/(a**2*sqrt(a*sec(e + f*x) + a)*sec(
e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x) + Integral(sec
(e + f*x)**5/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x) + a
**2*sqrt(a*sec(e + f*x) + a)), x) + Integral(-1/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 2*a**2*sqrt(a
*sec(e + f*x) + a)*sec(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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

[Out]

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

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