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

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

[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))

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Rubi [A]  time = 0.295997, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3904, 3887, 470, 578, 582, 522, 203} \[ \frac{2 c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{11 c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{2} a^{5/2} f}+\frac{7 c^4 \tan (e+f x)}{2 a^2 f \sqrt{a \sec (e+f x)+a}}-\frac{c^4 \sin ^2(e+f x) \tan ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right )}{4 f (a \sec (e+f x)+a)^{5/2}}-\frac{c^4 \sin (e+f x) \tan ^2(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{4 a f (a \sec (e+f x)+a)^{3/2}} \]

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 3904

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])

Rule 3887

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 470

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 578

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 582

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 522

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 203

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

Mathematica [A]  time = 2.59924, size = 164, normalized size = 0.72 \[ \frac{c^4 \cot \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left ((19 \cos (e+f x)-12 \cos (2 (e+f x))-3 \cos (3 (e+f x))-4) \sec ^4\left (\frac{1}{2} (e+f x)\right )+32 \cos (e+f x) \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )-88 \sqrt{2} \cos (e+f x) \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)-1}}{\sqrt{2}}\right )\right )}{16 a^2 f \sqrt{a (\sec (e+f x)+1)}} \]

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]  time = 0.272, size = 550, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

-1/2*c^4/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^4*sin(
f*x+e)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))+11*cos(f*x+e)^4
*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e
)+1)/sin(f*x+e))-4*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-
2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))+6*cos(f*x+e)^5-22*sin(f*x+e)*cos(f*x+e)^2*(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))+2*2^(
1/2)*sin(f*x+e)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)
/(1+cos(f*x+e)))^(1/2)-32*cos(f*x+e)^3+11*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(
1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))+36*cos(f*x+e)^2-6*cos(f*x+e)-4)/sin(f*x+e)^5

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

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 = 11.8767, size = 1673, normalized size = 7.31 \begin{align*} \text{result too large to display} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} c^{4} \left (\int - \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}}\, 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 - \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}}\, 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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out