3.248 \(\int \sec ^4(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=165 \[ -\frac{(a-5 b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 b^{3/2} f}+\frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b f}-\frac{(a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 b f}-\frac{(a-5 b) (a+b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 b f} \]

[Out]

-((a - 5*b)*(a + b)^2*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*b^(3/2)*f) - ((a - 5
*b)*(a + b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*b*f) - ((a - 5*b)*Tan[e + f*x]*(a + b + b*Tan[e +
f*x]^2)^(3/2))/(24*b*f) + (Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*b*f)

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Rubi [A]  time = 0.130678, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4146, 388, 195, 217, 206} \[ -\frac{(a-5 b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 b^{3/2} f}+\frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b f}-\frac{(a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 b f}-\frac{(a-5 b) (a+b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 b f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((a - 5*b)*(a + b)^2*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*b^(3/2)*f) - ((a - 5
*b)*(a + b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*b*f) - ((a - 5*b)*Tan[e + f*x]*(a + b + b*Tan[e +
f*x]^2)^(3/2))/(24*b*f) + (Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*b*f)

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 388

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f}-\frac{(a-5 b) \operatorname{Subst}\left (\int \left (a+b+b x^2\right )^{3/2} \, dx,x,\tan (e+f x)\right )}{6 b f}\\ &=-\frac{(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f}-\frac{((a-5 b) (a+b)) \operatorname{Subst}\left (\int \sqrt{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 b f}\\ &=-\frac{(a-5 b) (a+b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 b f}-\frac{(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f}-\frac{\left ((a-5 b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=-\frac{(a-5 b) (a+b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 b f}-\frac{(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f}-\frac{\left ((a-5 b) (a+b)^2\right ) \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 )}{16 b f}\\ &=-\frac{(a-5 b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac{(a-5 b) (a+b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 b f}-\frac{(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f}\\ \end{align*}

Mathematica [C]  time = 9.38514, size = 400, normalized size = 2.42 \[ \frac{e^{i (e+f x)} \cos ^3(e+f x) \sqrt{4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\frac{3 (a-5 b) (a+b)^2 \log \left (\frac{4 i f \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}-4 \sqrt{b} f \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )}{\sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}-\frac{i \sqrt{b} \left (-1+e^{2 i (e+f x)}\right ) \left (3 a^2 \left (1+e^{2 i (e+f x)}\right )^4+2 a b \left (50 e^{2 i (e+f x)}+11 e^{4 i (e+f x)}+11\right ) \left (1+e^{2 i (e+f x)}\right )^2+b^2 \left (100 e^{2 i (e+f x)}+298 e^{4 i (e+f x)}+100 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}+15\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^6}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{12 \sqrt{2} b^{3/2} f (a \cos (2 e+2 f x)+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]^3*(((-I)*Sqrt[b]
*(-1 + E^((2*I)*(e + f*x)))*(3*a^2*(1 + E^((2*I)*(e + f*x)))^4 + 2*a*b*(1 + E^((2*I)*(e + f*x)))^2*(11 + 50*E^
((2*I)*(e + f*x)) + 11*E^((4*I)*(e + f*x))) + b^2*(15 + 100*E^((2*I)*(e + f*x)) + 298*E^((4*I)*(e + f*x)) + 10
0*E^((6*I)*(e + f*x)) + 15*E^((8*I)*(e + f*x)))))/(1 + E^((2*I)*(e + f*x)))^6 + (3*(a - 5*b)*(a + b)^2*Log[(-4
*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)
/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*(a + b*Sec[e + f*x
]^2)^(3/2))/(12*Sqrt[2]*b^(3/2)*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))

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Maple [C]  time = 0.502, size = 2519, normalized size = 15.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

1/48/f/b/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*sin(f*x+e)*(-9*sin(f*x+e)*cos(f*x+e)^6*2^(1/2)*(1/(a+b)*(I*co
s(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+32*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-32*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(
a+b))^(1/2)*a*b^2+22*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+15*cos(f*x+e)^7*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-22*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-15*cos(f*x+e)^6*
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+10*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-10*c
os(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+15*cos(f
*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-15*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^
3+3*sin(f*x+e)*cos(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^3-15*sin(f*x+e)*cos(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))*b^3-6*sin(f
*x+e)*cos(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)*El
lipticPi((-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^3+30*sin(f*x+e)*cos(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)*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*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)*a^3-3*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+8*cos(f*x+e)*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)*b^3+17*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-17*cos(f*x+e)^4*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+22*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-22*cos(f*
x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-27*sin(f*x+e)*cos(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+18*s
in(f*x+e)*cos(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
)*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+54*sin(f*x+e)*cos(
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)*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)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^
2)^(3/2)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)^2/cos(f*x+e)^3

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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)^4*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.4534, size = 1154, normalized size = 6.99 \begin{align*} \left [-\frac{3 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt{b} \cos \left (f x + e\right )^{5} \log \left (\frac{{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \,{\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \,{\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{2} f \cos \left (f x + e\right )^{5}}, -\frac{3 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt{-b} \arctan \left (-\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \,{\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \,{\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \,{\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{2} f \cos \left (f x + e\right )^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/192*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(b)*cos(f*x + e)^5*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^2*b + 22*a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8
*b^3 + 2*(7*a*b^2 + 5*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^2*f*co
s(f*x + e)^5), -1/96*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*c
os(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)))*c
os(f*x + e)^5 - 2*((3*a^2*b + 22*a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8*b^3 + 2*(7*a*b^2 + 5*b^3)*cos(f*x + e)^2)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^2*f*cos(f*x + e)^5)]

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Sympy [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(sec(f*x+e)**4*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^4, x)