Integrand size = 25, antiderivative size = 174 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f} \] Output:
arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f-b*cot(f*x+ e)^3/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/3*(3*a-b)*(a+3*b)*cot(f*x+e)*( a+b+b*tan(f*x+e)^2)^(1/2)/a/(a+b)^3/f-1/3*(a-3*b)*cot(f*x+e)^3*(a+b+b*tan( f*x+e)^2)^(1/2)/a/(a+b)^2/f
Time = 2.66 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.29 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{2 \sqrt {2} a^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 e+2 f x))^2 \sec ^3(e+f x) \left (\frac {(4 a+9 b) \csc (e+f x)}{12 (a+b)^3 f}-\frac {\csc ^3(e+f x)}{12 (a+b)^2 f}-\frac {b^3 \sin (e+f x)}{2 a (a+b)^3 f (a+2 b+a \cos (2 e+2 f x))}\right )}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:
Integrate[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
Output:
(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])^(3/2)*Sec[e + f*x]^3)/(2*Sqrt[2]*a^(3/2)*f*(a + b*Sec[ e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^3*(((4 *a + 9*b)*Csc[e + f*x])/(12*(a + b)^3*f) - Csc[e + f*x]^3/(12*(a + b)^2*f) - (b^3*Sin[e + f*x])/(2*a*(a + b)^3*f*(a + 2*b + a*Cos[2*e + 2*f*x]))))/( a + b*Sec[e + f*x]^2)^(3/2)
Time = 0.48 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.06, 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, 445, 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.
\(\displaystyle \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 374 |
\(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (-4 b \tan ^2(e+f x)+a-3 b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 445 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (2 (a-3 b) b \tan ^2(e+f x)+(3 a-b) (a+3 b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 445 |
\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {3 (a+b)^3}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {-\frac {-3 (a+b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {-\frac {-3 (a+b)^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {-\frac {-\frac {3 (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
Input:
Int[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
Output:
(-((b*Cot[e + f*x]^3)/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) + (-1/3* ((a - 3*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b) - ((-3*( a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sq rt[a] - ((3*a - b)*(a + 3*b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/ (a + b))/(3*(a + b)))/(a*(a + b)))/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[((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[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. \(917\) vs. \(2(158)=316\).
Time = 10.02 (sec) , antiderivative size = 918, normalized size of antiderivative = 5.28
Input:
int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*((2*sec(f*x+e)*csc(f*x+e)^3-3*cot(f*x+e)^3)*b*a^3*(-a)^(1/2)+(-6*cos(f *x+e)^4+4*cos(f*x+e)^2+1)*sec(f*x+e)^3*csc(f*x+e)^3*b^2*a^2*(-a)^(1/2)+(-c os(f*x+e)^6+2*cos(f*x+e)^4-4*cos(f*x+e)^2+8/3)*sec(f*x+e)^3*csc(f*x+e)^3*b ^3*a*(-a)^(1/2)+(-4/3*cot(f*x+e)^3+cot(f*x+e)*csc(f*x+e)^2)*a^4*(-a)^(1/2) -(-a)^(1/2)*b^4*tan(f*x+e)*sec(f*x+e)^2-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x +e))^2)^(1/2)*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)*a^4*(-3-3*sec(f*x+e))-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*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 )*a^3*b*(-9-9*sec(f*x+e)-3*sec(f*x+e)^2-3*sec(f*x+e)^3)-1/3*((b+a*cos(f*x+ e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*a^2*b^2*(-9-9*sec(f*x+e)-9*sec(f*x+e)^2-9*sec(f* x+e)^3)-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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*co s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^3*(-3-3*sec(f*x+e) -9*sec(f*x+e)^2-9*sec(f*x+e)^3)-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*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...
Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (158) = 316\).
Time = 1.81 (sec) , antiderivative size = 1061, normalized size of antiderivative = 6.10 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
[-1/24*(3*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3* a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*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 + 3 2*(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)*co s(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*((4*a^4 + 9*a^3*b + 3*a*b^3)*cos(f*x + e)^5 - (3*a^4 + 4*a^3*b - 9*a^2*b^2 + 6*a*b^3)*cos(f*x + e)^3 - (3*a^3*b + 8*a^2*b^2 - 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5*b - 2*a^ 3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b ^4)*f)*sin(f*x + e)), -1/12*(3*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f* x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - 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*((4*a^4 + 9*a^3*b + 3*a*b^3)*cos(f*x + e)^5 - (3*a^4 + 4*a^3*b - 9*a^2*b^2 + 6*a*...
\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cot(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
Output:
Integral(cot(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
integrate(cot(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:
int(cot(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2),x)
Output:
int(cot(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2), x)
\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:
int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**4)/(sec(e + f*x)**4*b**2 + 2*sec(e + f*x)**2*a*b + a**2),x)