Integrand size = 21, antiderivative size = 51 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-a x-\frac {a \cot (e+f x)}{f}+\frac {a \cot ^3(e+f x)}{3 f}-\frac {(a+b) \cot ^5(e+f x)}{5 f} \] Output:
-a*x-a*cot(f*x+e)/f+1/3*a*cot(f*x+e)^3/f-1/5*(a+b)*cot(f*x+e)^5/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \cot ^5(e+f x)}{5 f}-\frac {a \cot ^5(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right )}{5 f} \] Input:
Integrate[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
-1/5*(b*Cot[e + f*x]^5)/f - (a*Cot[e + f*x]^5*Hypergeometric2F1[-5/2, 1, - 3/2, -Tan[e + f*x]^2])/(5*f)
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4629, 2075, 359, 264, 264, 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.
\(\displaystyle \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^2}{\tan (e+f x)^6}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {-a \int \frac {\cot ^4(e+f x)}{\tan ^2(e+f x)+1}d\tan (e+f x)-\frac {1}{5} (a+b) \cot ^5(e+f x)}{f}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-a \left (-\int \frac {\cot ^2(e+f x)}{\tan ^2(e+f x)+1}d\tan (e+f x)-\frac {1}{3} \cot ^3(e+f x)\right )-\frac {1}{5} (a+b) \cot ^5(e+f x)}{f}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-a \left (\int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)-\frac {1}{3} \cot ^3(e+f x)+\cot (e+f x)\right )-\frac {1}{5} (a+b) \cot ^5(e+f x)}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {-a \left (\arctan (\tan (e+f x))-\frac {1}{3} \cot ^3(e+f x)+\cot (e+f x)\right )-\frac {1}{5} (a+b) \cot ^5(e+f x)}{f}\) |
Input:
Int[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
(-1/5*((a + b)*Cot[e + f*x]^5) - a*(ArcTan[Tan[e + f*x]] + Cot[e + f*x] - Cot[e + f*x]^3/3))/f
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -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])
Time = 1.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}}{f}\) | \(63\) |
default | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}}{f}\) | \(63\) |
risch | \(-a x -\frac {2 i \left (45 a \,{\mathrm e}^{8 i \left (f x +e \right )}+15 b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a \,{\mathrm e}^{6 i \left (f x +e \right )}+140 a \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b \,{\mathrm e}^{4 i \left (f x +e \right )}-70 a \,{\mathrm e}^{2 i \left (f x +e \right )}+23 a +3 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}\) | \(104\) |
Input:
int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-1/5*cot(f*x+e)^5+1/3*cot(f*x+e)^3-cot(f*x+e)-f*x-e)-1/5*b/sin(f*x +e)^5*cos(f*x+e)^5)
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (47) = 94\).
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.16 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {{\left (23 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{5} - 35 \, a \cos \left (f x + e\right )^{3} + 15 \, a \cos \left (f x + e\right ) + 15 \, {\left (a f x \cos \left (f x + e\right )^{4} - 2 \, a f x \cos \left (f x + e\right )^{2} + a f x\right )} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )} \sin \left (f x + e\right )} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
-1/15*((23*a + 3*b)*cos(f*x + e)^5 - 35*a*cos(f*x + e)^3 + 15*a*cos(f*x + e) + 15*(a*f*x*cos(f*x + e)^4 - 2*a*f*x*cos(f*x + e)^2 + a*f*x)*sin(f*x + e))/((f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)*sin(f*x + e))
\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cot ^{6}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*cot(e + f*x)**6, x)
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {15 \, {\left (f x + e\right )} a + \frac {15 \, a \tan \left (f x + e\right )^{4} - 5 \, a \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
-1/15*(15*(f*x + e)*a + (15*a*tan(f*x + e)^4 - 5*a*tan(f*x + e)^2 + 3*a + 3*b)/tan(f*x + e)^5)/f
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (47) = 94\).
Time = 0.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.33 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 480 \, {\left (f x + e\right )} a + 330 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \frac {330 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{480 \, f} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/480*(3*a*tan(1/2*f*x + 1/2*e)^5 + 3*b*tan(1/2*f*x + 1/2*e)^5 - 35*a*tan( 1/2*f*x + 1/2*e)^3 - 15*b*tan(1/2*f*x + 1/2*e)^3 - 480*(f*x + e)*a + 330*a *tan(1/2*f*x + 1/2*e) + 30*b*tan(1/2*f*x + 1/2*e) - (330*a*tan(1/2*f*x + 1 /2*e)^4 + 30*b*tan(1/2*f*x + 1/2*e)^4 - 35*a*tan(1/2*f*x + 1/2*e)^2 - 15*b *tan(1/2*f*x + 1/2*e)^2 + 3*a + 3*b)/tan(1/2*f*x + 1/2*e)^5)/f
Time = 15.96 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-a\,x-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^4-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{3}+\frac {a}{5}+\frac {b}{5}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5} \] Input:
int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2),x)
Output:
- a*x - (a/5 + b/5 - (a*tan(e + f*x)^2)/3 + a*tan(e + f*x)^4)/(f*tan(e + f *x)^5)
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.22 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {-23 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +11 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b -3 \cos \left (f x +e \right ) a -3 \cos \left (f x +e \right ) b -15 \sin \left (f x +e \right )^{5} a f x}{15 \sin \left (f x +e \right )^{5} f} \] Input:
int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x)
Output:
( - 23*cos(e + f*x)*sin(e + f*x)**4*a - 3*cos(e + f*x)*sin(e + f*x)**4*b + 11*cos(e + f*x)*sin(e + f*x)**2*a + 6*cos(e + f*x)*sin(e + f*x)**2*b - 3* cos(e + f*x)*a - 3*cos(e + f*x)*b - 15*sin(e + f*x)**5*a*f*x)/(15*sin(e + f*x)**5*f)