3.1.80 \(\int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^{5/2}} \, dx\) [80]

Optimal. Leaf size=229 \[ \frac {2 c^4 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {11 c^4 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {2} a^{5/2} f}+\frac {7 c^4 \tan (e+f x)}{2 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^2(e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}} \]

[Out]

2*c^4*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f-11/2*c^4*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/
2)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)+7/2*c^4*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)-1/4*c^4*sec(1/2*f
*x+1/2*e)^2*sin(f*x+e)*tan(f*x+e)^2/a/f/(a+a*sec(f*x+e))^(3/2)-1/4*c^4*sec(1/2*f*x+1/2*e)^4*sin(f*x+e)^2*tan(f
*x+e)^3/f/(a+a*sec(f*x+e))^(5/2)

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Rubi [A]
time = 0.20, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, 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^4 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac {11 c^4 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {2} a^{5/2} f}+\frac {7 c^4 \tan (e+f x)}{2 a^2 f \sqrt {a \sec (e+f x)+a}}-\frac {c^4 \sin ^2(e+f x) \tan ^3(e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right )}{4 f (a \sec (e+f x)+a)^{5/2}}-\frac {c^4 \sin (e+f x) \tan ^2(e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{4 a f (a \sec (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(2*c^4*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (11*c^4*ArcTan[(Sqrt[a]*Tan[e +
f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) + (7*c^4*Tan[e + f*x])/(2*a^2*f*Sqrt[a + a*Sec[
e + f*x]]) - (c^4*Sec[(e + f*x)/2]^2*Sin[e + f*x]*Tan[e + f*x]^2)/(4*a*f*(a + a*Sec[e + f*x])^(3/2)) - (c^4*Se
c[(e + f*x)/2]^4*Sin[e + f*x]^2*Tan[e + f*x]^3)/(4*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 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))^4}{(a+a \sec (e+f x))^{5/2}} \, dx &=\left (a^4 c^4\right ) \int \frac {\tan ^8(e+f x)}{(a+a \sec (e+f x))^{13/2}} \, dx\\ &=-\frac {\left (2 a^2 c^4\right ) \text {Subst}\left (\int \frac {x^8}{\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 {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {c^4 \text {Subst}\left (\int \frac {x^4 \left (10+6 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 {c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^2(e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}+\frac {c^4 \text {Subst}\left (\int \frac {x^2 \left (-6 a-14 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^2 f}\\ &=\frac {7 c^4 \tan (e+f x)}{2 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^2(e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {c^4 \text {Subst}\left (\int \frac {-28 a^2-36 a^3 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 )}{4 a^4 f}\\ &=\frac {7 c^4 \tan (e+f x)}{2 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^2(e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {\left (2 c^4\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 (11 c^4\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^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {11 c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {2} a^{5/2} f}+\frac {7 c^4 \tan (e+f x)}{2 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^2(e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac {c^4 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan ^3(e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 2.72, size = 164, normalized size = 0.72 \begin {gather*} \frac {c^4 \cot \left (\frac {1}{2} (e+f x)\right ) \left ((-4+19 \cos (e+f x)-12 \cos (2 (e+f x))-3 \cos (3 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )+32 \text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right ) \cos (e+f x) \sqrt {-1+\sec (e+f x)}-88 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {-1+\sec (e+f x)}}{\sqrt {2}}\right ) \cos (e+f x) \sqrt {-1+\sec (e+f x)}\right ) \sec (e+f x)}{16 a^2 f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(c^4*Cot[(e + f*x)/2]*((-4 + 19*Cos[e + f*x] - 12*Cos[2*(e + f*x)] - 3*Cos[3*(e + f*x)])*Sec[(e + f*x)/2]^4 +
32*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]*Sqrt[-1 + Sec[e + f*x]] - 88*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e +
f*x]]/Sqrt[2]]*Cos[e + f*x]*Sqrt[-1 + Sec[e + f*x]])*Sec[e + f*x])/(16*a^2*f*Sqrt[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. \(549\) vs. \(2(200)=400\).
time = 0.24, size = 550, normalized size = 2.40

method	result	size
default	\(-\frac {c^{4} \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (2 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \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}+11 \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \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 ^{4}\left (f x +e \right )\right )-4 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \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 (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}-22 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \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 )+6 \left (\cos ^{5}\left (f x +e \right )\right )+2 \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 ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+11 \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \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 )-32 \left (\cos ^{3}\left (f x +e \right )\right )+36 \left (\cos ^{2}\left (f x +e \right )\right )-6 \cos \left (f x +e \right )-4\right )}{2 f \sin \left (f x +e \right )^{5} a^{3}}\)	\(550\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*c^4/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(2*cos(f*x+e)^4*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(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^(1/2)+11*sin(f*x+e)*(-2*cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)*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))*co
s(f*x+e)^4-4*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(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))*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)-22*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*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)+6*cos(f*x+e)^5+2*2^(1/2)*a
rctanh(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))^
(1/2)*sin(f*x+e)+11*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*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))-32*cos(f*x+e)^3+36*cos(f*x+e)^2-6*cos(f*x+e)-4)/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))^4/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 4.73, size = 706, normalized size = 3.08 \begin {gather*} \left [-\frac {11 \, \sqrt {2} {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {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 ) - 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 4 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\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 (3 \, c^{4} \cos \left (f x + e\right )^{2} + 9 \, c^{4} \cos \left (f x + e\right ) + 2 \, c^{4}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{4 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}, \frac {11 \, \sqrt {2} {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \sqrt {a} \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 ) - 4 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\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 (3 \, c^{4} \cos \left (f x + e\right )^{2} + 9 \, c^{4} \cos \left (f x + e\right ) + 2 \, c^{4}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/4*(11*sqrt(2)*(c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + 3*c^4*cos(f*x + e) + c^4)*sqrt(-a)*log(-(2*sqrt
(2)*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 3*a*cos(f*x + e)^2 - 2*a*cos(
f*x + e) + a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 4*(c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + 3*c^4*co
s(f*x + e) + c^4)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(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*(3*c^4*cos(f*x + e)^2 + 9*c^4*cos(f*x + e) +
2*c^4)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 +
3*a^3*f*cos(f*x + e) + a^3*f), 1/2*(11*sqrt(2)*(c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + 3*c^4*cos(f*x + e
) + c^4)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) -
4*(c^4*cos(f*x + e)^3 + 3*c^4*cos(f*x + e)^2 + 3*c^4*cos(f*x + e) + c^4)*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*(3*c^4*cos(f*x + e)^2 + 9*c^4*cos(f*x + e) + 2*c^4
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3
*f*cos(f*x + e) + a^3*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{4} \left (\int \left (- \frac {4 \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}}\right )\, dx + \int \frac {6 \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}}\, dx + \int \left (- \frac {4 \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}}\right )\, dx + \int \frac {\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}}\, dx + \int \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}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**4/(a+a*sec(f*x+e))**(5/2),x)

[Out]

c**4*(Integral(-4*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(6*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) + Integr
al(-4*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(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(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))^4/(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 )}^4}{{\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))^4/(a + a/cos(e + f*x))^(5/2),x)

[Out]

int((c - c/cos(e + f*x))^4/(a + a/cos(e + f*x))^(5/2), x)

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