3.289 \(\int \frac{\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{a (3 a+5 b) \tan (e+f x)}{3 b^2 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{b^{5/2} f}-\frac{a \tan (e+f x) \sec ^2(e+f x)}{3 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(b^(5/2)*f) - (a*Sec[e + f*x]^2*Tan[e + f*x])/(
3*b*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (a*(3*a + 5*b)*Tan[e + f*x])/(3*b^2*(a + b)^2*f*Sqrt[a + b +
b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.139465, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4146, 413, 385, 217, 206} \[ -\frac{a (3 a+5 b) \tan (e+f x)}{3 b^2 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{b^{5/2} f}-\frac{a \tan (e+f x) \sec ^2(e+f x)}{3 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(b^(5/2)*f) - (a*Sec[e + f*x]^2*Tan[e + f*x])/(
3*b*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (a*(3*a + 5*b)*Tan[e + f*x])/(3*b^2*(a + b)^2*f*Sqrt[a + b +
b*Tan[e + f*x]^2])

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 413

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{\left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a+3 b+3 (a+b) x^2}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 b (a+b) f}\\ &=-\frac{a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{b^2 f}\\ &=-\frac{a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{b^2 f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac{a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 10.0698, size = 607, normalized size = 4.56 \[ \frac{\tan (e+f x) \sec ^6(e+f x) \sqrt{1-\frac{2 a \sin ^2(e+f x)}{2 a+2 b}} (a \cos (2 e+2 f x)+a+2 b)^{5/2} \left (-\frac{24 b \sin ^2(e+f x) \cos ^2(e+f x) \left (-\frac{b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{(a+b)^2}\right )^{5/2} \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{9}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{a+b}-\frac{\sec ^6(e+f x) \left (24 b^3 \sin ^6(e+f x) \left (a^2 \left (3 \sin ^4(e+f x)-7 \sin ^2(e+f x)+4\right )+a b \left (8-7 \sin ^2(e+f x)\right )+4 b^2\right ) \text{Hypergeometric2F1}\left (2,2,\frac{9}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sqrt{-\frac{b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{(a+b)^2}}+35 (a+b) \cos ^2(e+f x) \left (a^2 \left (8 \sin ^4(e+f x)-20 \sin ^2(e+f x)+15\right )+10 a b \left (3-2 \sin ^2(e+f x)\right )+15 b^2\right ) \left (\cos ^2(e+f x) \left (-\left (-3 a^2 \cos ^2(e+f x)+2 a b \left (\sin ^2(e+f x)-3\right )+b^2 \left (-\left (\sin ^2(e+f x)+3\right )\right )\right )\right ) \sqrt{-\frac{b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{(a+b)^2}}-3 \left (-a \sin ^2(e+f x)+a+b\right )^2 \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right )\right )\right )}{(a+b)^5}\right )}{315 f (2 a+2 b)^2 \sqrt{-2 a \sin ^2(e+f x)+2 a+2 b} \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right )^{3/2} \left (-\frac{b \tan ^2(e+f x)}{a+b}\right )^{5/2} \left (a+b \sec ^2(e+f x)\right )^{5/2} \sqrt{\frac{a+b \sec ^2(e+f x)}{a+b}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^6*Sqrt[1 - (2*a*Sin[e + f*x]^2)/(2*a + 2*b)]*Tan[e + f*x]*(
(-24*b*Cos[e + f*x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*(-(
(b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(5/2))/(a + b) - (Sec[e + f*x]^6*(24*
b^3*Hypergeometric2F1[2, 2, 9/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^6*(4*b^2 + a*b*(8 - 7*Sin[e + f*x
]^2) + a^2*(4 - 7*Sin[e + f*x]^2 + 3*Sin[e + f*x]^4))*Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[
e + f*x]^2)/(a + b)^2)] + 35*(a + b)*Cos[e + f*x]^2*(15*b^2 + 10*a*b*(3 - 2*Sin[e + f*x]^2) + a^2*(15 - 20*Sin
[e + f*x]^2 + 8*Sin[e + f*x]^4))*(-3*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*(a + b - a*Sin[e + f*x]^2)^2
- Cos[e + f*x]^2*(-3*a^2*Cos[e + f*x]^2 + 2*a*b*(-3 + Sin[e + f*x]^2) - b^2*(3 + Sin[e + f*x]^2))*Sqrt[-((b*Se
c[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)])))/(a + b)^5))/(315*(2*a + 2*b)^2*f*(a + b
*Sec[e + f*x]^2)^(5/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[2*a + 2*b - 2*a*Sin[e + f*x]^2]*(1 - (a*Sin[e
+ f*x]^2)/(a + b))^(3/2)*(-((b*Tan[e + f*x]^2)/(a + b)))^(5/2))

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Maple [C]  time = 0.469, size = 3010, normalized size = 22.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

1/3/f/(a+b)^2/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/b^2*sin(f*x+e)*(b+a*cos(f*x+e)^2)*(6*sin(f*x+e)*cos(f*x+
e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-
2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+c
os(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)
*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^3+12*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/
(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f
*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a
+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b+6*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*
x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b
^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2
)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^2-3*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^
(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2
)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1
/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^3-6*sin(f*x+e)*cos(
f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2
)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-
1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^
2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b-3*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*
a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2
)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(
f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^2+6*sin(f*x+e)*2^(1/2)*(1
/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(
a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b+12*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2
)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^
(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b
))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2))*a*b^2+6*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+
a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b
)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a
^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^
3-3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2
/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos
(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a
*b-b^2)/(a+b)^2)^(1/2))*a^2*b*sin(f*x+e)-6*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*
cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/
(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3
/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^2*sin(f*x+e)-3*2^(1/2)*(1/(a+b)*(I*cos(f*x+
e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(
1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a
-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^3*sin(
f*x+e)-3*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3-5*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)*a^2*b+3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+5*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)*a^2*b-4*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-6*cos(f*x+e)*((2*I*a^(1/
2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+6*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)*a*b^2)/(-1+cos(f*x+e))/cos(f*x+e)^5/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.46035, size = 1586, normalized size = 11.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*c
os(f*x + e)^2)*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos
(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(
f*x + e)^4) - 4*((3*a^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^4 + 2*a
^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f), 1/6*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e
)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sqrt(-b)*arctan(-1/2*((a - b)*co
s(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b
^2)*sin(f*x + e))) - 2*((3*a^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^
4 + 2*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)