\(\int \frac {\sec ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [357]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b \sin (e+f x)}{2 a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec (e+f x) \tan (e+f x)}{2 (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

1/2*(a+4*b)*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b)^(5/2)/f-1/2*(a-2*b)*b*sin(f*x+e)/a/
(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/2*sec(f*x+e)*tan(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3269, 425, 541, 12, 385, 212} \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f (a+b)^{5/2}}-\frac {b (a-2 b) \sin (e+f x)}{2 a f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x) \sec (e+f x)}{2 f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \]

[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 12

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

Rule 212

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

Rule 385

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 425

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]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

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 3269

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\text {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 {\text {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) \text {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) \text {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) \text {arctanh}\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] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 7.37 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.67 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sec ^5(e+f x) \left (16 (a+b) \, _3F_2\left (2,2,3;1,\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2+16 (a+b) \operatorname {Hypergeometric2F1}\left (2,3,\frac {9}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (4 a^2+7 a b \sin ^2(e+f x)+3 b^2 \sin ^4(e+f x)\right )-7 a \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (1,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \left (15 a^2+20 a b \sin ^2(e+f x)+8 b^2 \sin ^4(e+f x)\right )\right ) \tan (e+f x)}{105 a^4 f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

-1/105*(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)]*Si
n[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])/(
a^4*f*Sqrt[a + b*Sin[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3218\) vs. \(2(118)=236\).

Time = 3.54 (sec) , antiderivative size = 3219, normalized size of antiderivative = 24.02

method	result	size
default	\(\text {Expression too large to display}\)	\(3219\)

[In]

int(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[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*b^(19/2)*(a+b)^
(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))+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*b^(17/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b
^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*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*b^(15/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+
b^2*sin(f*x+e))/b^(3/2))*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*b^(13/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin
(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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)^(1/2)*(a+b-b*cos(f*x+e)^2)^(3
/2)*a*b^5*(a^3+3*a^2*b+3*a*b^2+b^3)-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)-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
*b^(19/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))+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*b^(17/2)*(a+b)^(1/2)*ln(((-b*cos
(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*a+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*co
s(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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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*b^(19/2)*(a+b)^(1/2)*ln(
((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))+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*b^(17/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b
^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*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*b^(15/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x
+e))/b^(3/2))*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/(sin(f*x+e)-1)*((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/(sin(f*x+e
)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^10)*c
os(f*x+e)^4-2*cos(f*x+e)^4*sin(f*x+e)*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b^7*(cos(f*x+e)^2*a*b+cos(f*x+e
)^2*b^2+a^2+2*a*b+b^2)+2*cos(f*x+e)^6*sin(f*x+e)*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b^8)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (118) = 236\).

Time = 0.60 (sec) , antiderivative size = 625, normalized size of antiderivative = 4.66 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\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 ] \]

[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)]

Sympy [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2)),x)

[Out]

int(1/(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2)), x)