3.357 \(\int \frac{\sec ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=134 \[ -\frac{b (a-2 b) \sin (e+f x)}{2 a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{(a+4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f (a+b)^{5/2}}+\frac{\tan (e+f x) \sec (e+f x)}{2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

((a + 4*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*(a + b)^(5/2)*f) - ((a - 2*b)*b*
Sin[e + f*x])/(2*a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + (Sec[e + f*x]*Tan[e + f*x])/(2*(a + b)*f*Sqrt[a +
b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.175953, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3190, 414, 527, 12, 377, 206} \[ -\frac{b (a-2 b) \sin (e+f x)}{2 a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{(a+4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f (a+b)^{5/2}}+\frac{\tan (e+f x) \sec (e+f x)}{2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 4*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*(a + b)^(5/2)*f) - ((a - 2*b)*b*
Sin[e + f*x])/(2*a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + (Sec[e + f*x]*Tan[e + f*x])/(2*(a + b)*f*Sqrt[a +
b*Sin[e + f*x]^2])

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 414

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

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

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

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

Mathematica [C]  time = 7.46951, size = 224, normalized size = 1.67 \[ -\frac{\tan (e+f x) \sec ^5(e+f x) \left (16 (a+b) \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, _3F_2\left (2,2,3;1,\frac{9}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right )+16 (a+b) \sin ^2(e+f x) \left (4 a^2+7 a b \sin ^2(e+f x)+3 b^2 \sin ^4(e+f x)\right ) \, _2F_1\left (2,3;\frac{9}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right )-7 a \cos ^2(e+f x) \left (15 a^2+20 a b \sin ^2(e+f x)+8 b^2 \sin ^4(e+f x)\right ) \, _2F_1\left (1,2;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right )\right )}{105 a^4 f \sqrt{a+b \sin ^2(e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sec[e + f*x]^5*(16*(a + b)*HypergeometricPFQ[{2, 2, 3}, {1, 9/2}, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x
]^2*(a + b*Sin[e + f*x]^2)^2 + 16*(a + b)*Hypergeometric2F1[2, 3, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e +
f*x]^2*(4*a^2 + 7*a*b*Sin[e + f*x]^2 + 3*b^2*Sin[e + f*x]^4) - 7*a*Cos[e + f*x]^2*Hypergeometric2F1[1, 2, 7/2,
-(((a + b)*Tan[e + f*x]^2)/a)]*(15*a^2 + 20*a*b*Sin[e + f*x]^2 + 8*b^2*Sin[e + f*x]^4))*Tan[e + f*x])/(105*a^
4*f*Sqrt[a + b*Sin[e + f*x]^2])

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Maple [B]  time = 9.885, size = 3219, normalized size = 24. \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)^3/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/4/(a+b)^(1/2)/a/b^5/cos(f*x+e)^2/(a^4*b^2*cos(f*x+e)^4+4*a^3*b^3*cos(f*x+e)^4+6*a^2*b^4*cos(f*x+e)^4+4*a*b^5
*cos(f*x+e)^4+b^6*cos(f*x+e)^4-2*a^5*b*cos(f*x+e)^2-10*a^4*b^2*cos(f*x+e)^2-20*a^3*b^3*cos(f*x+e)^2-20*a^2*b^4
*cos(f*x+e)^2-10*a*b^5*cos(f*x+e)^2-2*b^6*cos(f*x+e)^2+a^6+6*a^5*b+15*a^4*b^2+20*a^3*b^3+15*a^2*b^4+6*a*b^5+b^
6)*(-a*(8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(19/2)*(a+b)^(1/2)-8*ln(((-b*cos(f*x
+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(19/2)*(a+b)^(1/2)+24*ln(((a+b-b*cos(f*x+e)^2)
^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(17/2)*(a+b)^(1/2)*a-24*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^
(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(17/2)*(a+b)^(1/2)*a+24*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/
b^(1/2))*b^(15/2)*(a+b)^(1/2)*a^2-24*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/
2))*b^(15/2)*(a+b)^(1/2)*a^2+8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(13/2)*(a+b)^(1
/2)*a^3-8*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(13/2)*(a+b)^(1/2)*a^
3+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^5*b^5+8*ln(2/(1+sin(f*x+e))*(
(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^4*b^6+22*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos
(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b^7+28*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin
(f*x+e)+a))*a^2*b^8+17*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^9+4*ln
(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^10-ln(2/(-1+sin(f*x+e))*((a+b)^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^5*b^5-8*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^
2)^(1/2)+b*sin(f*x+e)+a))*a^4*b^6-22*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)
+a))*a^3*b^7-28*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^8-17*ln(2/
(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^9-4*ln(2/(-1+sin(f*x+e))*((a+b)^(
1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^10)*cos(f*x+e)^2+2*sin(f*x+e)*(a+b-b*cos(f*x+e)^2)^(3/2)*(a
+b)^(1/2)*a*b^5*(a^3+3*a^2*b+3*a*b^2+b^3)-a*(8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b
^(19/2)*(a+b)^(1/2)-8*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(19/2)*(a
+b)^(1/2)+8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(17/2)*(a+b)^(1/2)*a-8*ln(((-b*cos
(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(17/2)*(a+b)^(1/2)*a+ln(2/(1+sin(f*x+e))*(
(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b^7+6*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(
f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^8+9*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f
*x+e)+a))*a*b^9+4*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^10-ln(2/(-1+s
in(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^7-6*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2
)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^8-9*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)
^(1/2)+b*sin(f*x+e)+a))*a*b^9-4*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*
b^10)*cos(f*x+e)^6+2*a*(8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(19/2)*(a+b)^(1/2)-8
*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(19/2)*(a+b)^(1/2)+16*ln(((a+b
-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*b^(17/2)*(a+b)^(1/2)*a-16*ln(((-b*cos(f*x+e)^2+(a*b^2+b^
3)/b^2)^(1/2)*b^(3/2)+sin(f*x+e)*b^2)/b^(3/2))*b^(17/2)*(a+b)^(1/2)*a+8*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)
+b*sin(f*x+e))/b^(1/2))*b^(15/2)*(a+b)^(1/2)*a^2-8*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+sin(f*x
+e)*b^2)/b^(3/2))*b^(15/2)*(a+b)^(1/2)*a^2+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f
*x+e)+a))*a^4*b^6+7*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b^7+15*ln
(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^8+13*ln(2/(1+sin(f*x+e))*((a+
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^9+4*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e
)^2)^(1/2)-b*sin(f*x+e)+a))*b^10-ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))
*a^4*b^6-7*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^7-15*ln(2/(-1+s
in(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^8-13*ln(2/(-1+sin(f*x+e))*((a+b)^(1/
2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^9-4*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^
(1/2)+b*sin(f*x+e)+a))*b^10)*cos(f*x+e)^4+2*sin(f*x+e)*cos(f*x+e)^6*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a*b
^8-2*cos(f*x+e)^2*sin(f*x+e)*(a+b)^(1/2)*b^6*(2*(a+b-b*cos(f*x+e)^2)^(3/2)*a^3+4*(a+b-b*cos(f*x+e)^2)^(3/2)*a^
2*b+2*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2-2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a^2*b-4*(-b*cos(f*x+e)^2+(a*b
^2+b^3)/b^2)^(3/2)*a*b^2-2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*b^3-(a+b-b*cos(f*x+e)^2)^(1/2)*a^4-3*(a+b-b
*cos(f*x+e)^2)^(1/2)*a^3*b-3*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2*b^2-(a+b-b*cos(f*x+e)^2)^(1/2)*a*b^3)-2*cos(f*x+e)
^4*sin(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*a*b^7*(a*b*cos(f*x+e)^2+b^2*cos(f*x+e)^2+a^2+2*a*b+b^2))/
f

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

[Out]

Timed out

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Fricas [B]  time = 6.26998, size = 1438, normalized size = 10.73 \begin{align*} \left [\frac{{\left ({\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a^{3} + 5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{a + b} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2} -{\left (a^{2} b - a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{8 \,{\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{4} -{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}, -\frac{{\left ({\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a^{3} + 5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a - b} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) + 2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2} -{\left (a^{2} b - a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{4 \,{\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{4} -{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(((a^2*b + 4*a*b^2)*cos(f*x + e)^4 - (a^3 + 5*a^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt(a + b)*log(((a^2 + 8*
a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b
)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) - 4*(a^3
+ 2*a^2*b + a*b^2 - (a^2*b - a*b^2 - 2*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^
4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^4 - (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(
f*x + e)^2), -1/4*(((a^2*b + 4*a*b^2)*cos(f*x + e)^4 - (a^3 + 5*a^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt(-a - b)*
arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*c
os(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e))) + 2*(a^3 + 2*a^2*b + a*b^2 - (a^2*b - a*b^2 - 2*b^3)*cos(f*x
+ e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)
^4 - (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(e + f*x)**3/(a + b*sin(e + f*x)**2)**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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