Integrand size = 25, antiderivative size = 131 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a-3 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 a^{5/2} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b (a+3 b) \tan (e+f x)}{2 a^2 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:
1/2*(a-3*b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/ f+1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/2*b*(a+3*b)*t an(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 14.09 (sec) , antiderivative size = 2059, normalized size of antiderivative = 15.72 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(3/2),x]
Output:
(3*(a + b)*AppellF1[1/2, -2, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^6*Sin[e + f*x])/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x )]]*(a + b*Sec[e + f*x]^2)^(3/2)*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*App ellF1[1/2, -2, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3* a*AppellF1[3/2, -2, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2) /(a + b)])*Sin[e + f*x]^2)*((3*a*(a + b)*AppellF1[1/2, -2, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(Sqr t[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)^2*(3*(a + b)*Ap pellF1[1/2, -2, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3 *a*AppellF1[3/2, -2, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2 )/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b)*AppellF1[1/2, -2, 3/2, 3/2, Sin[ e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5)/(2*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*AppellF1[1/2, -2 , 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/ 2, -2, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*A ppellF1[3/2, -1, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Si n[e + f*x]^2)) - (6*(a + b)*AppellF1[1/2, -2, 3/2, 3/2, Sin[e + f*x]^2, (a *Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^3*Sin[e + f*x]^2)/(Sqrt[a + 2*b ...
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4634, 316, 25, 402, 27, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int -\frac {2 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {2 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\frac {\int \frac {(a-3 b) (a+b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {b (a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {(a-3 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {b (a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\frac {(a-3 b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {b (a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {(a-3 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {b (a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
Input:
Int[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(3/2),x]
Output:
(Tan[e + f*x]/(2*a*(1 + Tan[e + f*x]^2)*Sqrt[a + b + b*Tan[e + f*x]^2]) + (((a - 3*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]) /a^(3/2) + (b*(a + 3*b)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f* x]^2]))/(2*a))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(665\) vs. \(2(115)=230\).
Time = 7.39 (sec) , antiderivative size = 666, normalized size of antiderivative = 5.08
method | result | size |
default | \(\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-2-2 \sec \left (f x +e \right )+\sec \left (f x +e \right )^{2}+\sec \left (f x +e \right )^{3}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-3-3 \sec \left (f x +e \right )-2 \sec \left (f x +e \right )^{2}-2 \sec \left (f x +e \right )^{3}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (-3 \sec \left (f x +e \right )^{2}-3 \sec \left (f x +e \right )^{3}\right )+\sqrt {-a}\, a^{3} \cos \left (f x +e \right ) \sin \left (f x +e \right )+\left (2+\cos \left (f x +e \right )^{2}\right ) \sqrt {-a}\, a^{2} b \tan \left (f x +e \right )+\sqrt {-a}\, a \,b^{2} \left (4 \tan \left (f x +e \right )+\tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )+3 \sqrt {-a}\, b^{3} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{2 f \left (a +b \right ) a^{2} \sqrt {-a}\, \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(666\) |
Input:
int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/f/(a+b)/a^2/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(3/2)*(((b+a*cos(f*x+e)^2)/( 1+cos(f*x+e))^2)^(1/2)*a^3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) ^(1/2)-4*sin(f*x+e)*a)*(sec(f*x+e)+1)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 )^(1/2)*a^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)* cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin( f*x+e)*a)*(-2-2*sec(f*x+e)+sec(f*x+e)^2+sec(f*x+e)^3)+((b+a*cos(f*x+e)^2)/ (1+cos(f*x+e))^2)^(1/2)*a*b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f *x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) ^2)^(1/2)-4*sin(f*x+e)*a)*(-3-3*sec(f*x+e)-2*sec(f*x+e)^2-2*sec(f*x+e)^3)+ ((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*ln(4*(-a)^(1/2)*((b+a*cos( f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e) ^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(-3*sec(f*x+e)^2-3*sec(f*x+e)^ 3)+(-a)^(1/2)*a^3*cos(f*x+e)*sin(f*x+e)+(2+cos(f*x+e)^2)*(-a)^(1/2)*a^2*b* tan(f*x+e)+(-a)^(1/2)*a*b^2*(4*tan(f*x+e)+tan(f*x+e)*sec(f*x+e)^2)+3*(-a)^ (1/2)*b^3*tan(f*x+e)*sec(f*x+e)^2)
Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (115) = 230\).
Time = 0.45 (sec) , antiderivative size = 699, normalized size of antiderivative = 5.34 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
[1/16*((a^2*b - 2*a*b^2 - 3*b^3 + (a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^2 )*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^ 2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a ^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)* cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*((a^3 + a^2*b)*cos(f*x + e)^3 + (a^2*b + 3*a*b^2)*cos(f*x + e)) *sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)* f*cos(f*x + e)^2 + (a^4*b + a^3*b^2)*f), -1/8*((a^2*b - 2*a*b^2 - 3*b^3 + (a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f* x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e) )*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e) ^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(( a^3 + a^2*b)*cos(f*x + e)^3 + (a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt((a*cos( f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)*f*cos(f*x + e )^2 + (a^4*b + a^3*b^2)*f)]
\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cos(f*x+e)**2/(a+b*sec(f*x+e)**2)**(3/2),x)
Output:
Integral(cos(e + f*x)**2/(a + b*sec(e + f*x)**2)**(3/2), x)
\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(3/2), x)
\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
integrate(cos(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(3/2), x)
Timed out. \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:
int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^(3/2),x)
Output:
int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^(3/2), x)
\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:
int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*cos(e + f*x)**2)/(sec(e + f*x)**4*b**2 + 2*sec(e + f*x)**2*a*b + a**2),x)