Integrand size = 25, antiderivative size = 174 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f} \]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*(7* a+3*b)*cot(f*x+e)/a^2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*(a-3*b)*(3* a+b)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^3/f-1/3*b*cot(f*x+e)/ a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
Time = 7.77 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.42 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x)}{4 \sqrt {2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 \left (3 a^4+8 a^3 b+5 a^2 b^2-12 a b^3-4 b^4\right )+4 \left (3 a^4+6 a^3 b+8 a b^3+3 b^4\right ) \cos (2 (e+f x))+a \left (3 a^3+9 a b^2+4 b^3\right ) \cos (4 (e+f x))\right ) \csc (e+f x) \sec ^5(e+f x)}{48 a^2 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
-1/4*(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2 *b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5)/(Sqrt[2]*a^(5/2)*f*(a + b*S ec[e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])*(3*(3*a^4 + 8*a^3* b + 5*a^2*b^2 - 12*a*b^3 - 4*b^4) + 4*(3*a^4 + 6*a^3*b + 8*a*b^3 + 3*b^4)* Cos[2*(e + f*x)] + a*(3*a^3 + 9*a*b^2 + 4*b^3)*Cos[4*(e + f*x)])*Csc[e + f *x]*Sec[e + f*x]^5)/(48*a^2*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4629, 2075, 374, 441, 445, 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.
| \(\displaystyle \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\cot ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\cot ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^2(e+f x) \left (-4 b \tan ^2(e+f x)+3 a-b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\cot ^2(e+f x) \left ((a-3 b) (3 a+b)-2 b (7 a+3 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {3 (a+b)^3}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {-3 (a+b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {-3 (a+b)^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {-\frac {3 (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/3*(b*Cot[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (-(( b*(7*a + 3*b)*Cot[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) + ((-3*(a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2 ]])/Sqrt[a] - ((a - 3*b)*(3*a + b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x ]^2])/(a + b))/(a*(a + b)))/(3*a*(a + b)))/f
3.5.38.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(1861\) vs. \(2(156)=312\).
Time = 7.59 (sec) , antiderivative size = 1862, normalized size of antiderivative = 10.70
1/6/f/(a+b)^3/a^2/(-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x +e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2 *csc(f*x+e)^2+a+b)*(3*(-a)^(1/2)*a^4*(1-cos(f*x+e))^8*csc(f*x+e)^8+6*(-a)^ (1/2)*a^3*b*(1-cos(f*x+e))^8*csc(f*x+e)^8+3*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+ e))^8*csc(f*x+e)^8-12*a^4*(1-cos(f*x+e))^6*(-a)^(1/2)*csc(f*x+e)^6-24*(-a) ^(1/2)*a^2*b^2*(1-cos(f*x+e))^6*csc(f*x+e)^6-48*(-a)^(1/2)*a*b^3*(1-cos(f* x+e))^6*csc(f*x+e)^6-12*(-a)^(1/2)*b^4*(1-cos(f*x+e))^6*csc(f*x+e)^6+18*(- a)^(1/2)*a^4*(1-cos(f*x+e))^4*csc(f*x+e)^4-12*(-a)^(1/2)*a^3*b*(1-cos(f*x+ e))^4*csc(f*x+e)^4+90*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^4*csc(f*x+e)^4-32* (-a)^(1/2)*a*b^3*(1-cos(f*x+e))^4*csc(f*x+e)^4-24*(-a)^(1/2)*b^4*(1-cos(f* x+e))^4*csc(f*x+e)^4+6*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b *(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-co s(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(f*x+e)))/((1-cos(f *x+e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e) )^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*cs c(f*x+e)^2+a+b)^(3/2)*a^3*(csc(f*x+e)-cot(f*x+e))+18*ln(4*((-a)^(1/2)*(a*( 1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f* x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc (f*x+e)-cot(f*x+e)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^ 4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc...
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (156) = 312\).
Time = 2.15 (sec) , antiderivative size = 1097, normalized size of antiderivative = 6.30 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[-1/24*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^ 2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*co s(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))*sin(f*x + e) + 8*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)*cos(f *x + e)^5 + (6*a^4*b - 9*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4)*cos(f*x + e)^3 + ( 3*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b) /cos(f*x + e)^2))/(((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)*sin(f*x + e)), 1/12*(3*(a^3*b^2 + 3 *a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*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)))*sin(f*x + e) - 4*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)*cos...
\[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cot ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^2}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]