Integrand size = 21, antiderivative size = 68 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {(a+3 b) \cot (e+f x)}{f}-\frac {(2 a+3 b) \cot ^3(e+f x)}{3 f}-\frac {(a+b) \cot ^5(e+f x)}{5 f}+\frac {b \tan (e+f x)}{f} \] Output:
-(a+3*b)*cot(f*x+e)/f-1/3*(2*a+3*b)*cot(f*x+e)^3/f-1/5*(a+b)*cot(f*x+e)^5/ f+b*tan(f*x+e)/f
Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.88 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {8 a \cot (e+f x)}{15 f}-\frac {11 b \cot (e+f x)}{5 f}-\frac {4 a \cot (e+f x) \csc ^2(e+f x)}{15 f}-\frac {3 b \cot (e+f x) \csc ^2(e+f x)}{5 f}-\frac {a \cot (e+f x) \csc ^4(e+f x)}{5 f}-\frac {b \cot (e+f x) \csc ^4(e+f x)}{5 f}+\frac {b \tan (e+f x)}{f} \] Input:
Integrate[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
(-8*a*Cot[e + f*x])/(15*f) - (11*b*Cot[e + f*x])/(5*f) - (4*a*Cot[e + f*x] *Csc[e + f*x]^2)/(15*f) - (3*b*Cot[e + f*x]*Csc[e + f*x]^2)/(5*f) - (a*Cot [e + f*x]*Csc[e + f*x]^4)/(5*f) - (b*Cot[e + f*x]*Csc[e + f*x]^4)/(5*f) + (b*Tan[e + f*x])/f
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4620, 355, 2009}
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.
\(\displaystyle \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^2}{\sin (e+f x)^6}dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {\int \left ((a+b) \cot ^6(e+f x)+(2 a+3 b) \cot ^4(e+f x)+(a+3 b) \cot ^2(e+f x)+b\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{5} (a+b) \cot ^5(e+f x)-\frac {1}{3} (2 a+3 b) \cot ^3(e+f x)-(a+3 b) \cot (e+f x)+b \tan (e+f x)}{f}\) |
Input:
Int[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
(-((a + 3*b)*Cot[e + f*x]) - ((2*a + 3*b)*Cot[e + f*x]^3)/3 - ((a + b)*Cot [e + f*x]^5)/5 + b*Tan[e + f*x])/f
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 1.50 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.31
method | result | size |
parallelrisch | \(-\frac {\sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \left (\left (a +6 b \right ) \cos \left (2 f x +2 e \right )+\frac {4 \left (-a -6 b \right ) \cos \left (4 f x +4 e \right )}{5}+\frac {\left (a +6 b \right ) \cos \left (6 f x +6 e \right )}{5}+2 a \right )}{384 f \cos \left (f x +e \right )}\) | \(89\) |
risch | \(-\frac {16 i \left (10 a \,{\mathrm e}^{6 i \left (f x +e \right )}+5 a \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b \,{\mathrm e}^{4 i \left (f x +e \right )}-4 a \,{\mathrm e}^{2 i \left (f x +e \right )}-24 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a +6 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(98\) |
derivativedivides | \(\frac {a \left (-\frac {8}{15}-\frac {\csc \left (f x +e \right )^{4}}{5}-\frac {4 \csc \left (f x +e \right )^{2}}{15}\right ) \cot \left (f x +e \right )+b \left (-\frac {1}{5 \sin \left (f x +e \right )^{5} \cos \left (f x +e \right )}-\frac {2}{5 \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )}+\frac {8}{5 \sin \left (f x +e \right ) \cos \left (f x +e \right )}-\frac {16 \cot \left (f x +e \right )}{5}\right )}{f}\) | \(101\) |
default | \(\frac {a \left (-\frac {8}{15}-\frac {\csc \left (f x +e \right )^{4}}{5}-\frac {4 \csc \left (f x +e \right )^{2}}{15}\right ) \cot \left (f x +e \right )+b \left (-\frac {1}{5 \sin \left (f x +e \right )^{5} \cos \left (f x +e \right )}-\frac {2}{5 \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )}+\frac {8}{5 \sin \left (f x +e \right ) \cos \left (f x +e \right )}-\frac {16 \cot \left (f x +e \right )}{5}\right )}{f}\) | \(101\) |
norman | \(\frac {\frac {a +b}{160 f}+\frac {\left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{160 f}-\frac {5 \left (a +7 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{8 f}+\frac {5 \left (5 a +21 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{96 f}+\frac {5 \left (5 a +21 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{96 f}+\frac {\left (11 a +21 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{240 f}+\frac {\left (11 a +21 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{240 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\) | \(169\) |
Input:
int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
-1/384*sec(1/2*f*x+1/2*e)^5*csc(1/2*f*x+1/2*e)^5*((a+6*b)*cos(2*f*x+2*e)+4 /5*(-a-6*b)*cos(4*f*x+4*e)+1/5*(a+6*b)*cos(6*f*x+6*e)+2*a)/f/cos(f*x+e)
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.34 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {8 \, {\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{6} - 20 \, {\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{4} + 15 \, {\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \, {\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:
integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
-1/15*(8*(a + 6*b)*cos(f*x + e)^6 - 20*(a + 6*b)*cos(f*x + e)^4 + 15*(a + 6*b)*cos(f*x + e)^2 - 15*b)/((f*cos(f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*co s(f*x + e))*sin(f*x + e))
Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\text {Timed out} \] Input:
integrate(csc(f*x+e)**6*(a+b*sec(f*x+e)**2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {15 \, b \tan \left (f x + e\right ) - \frac {15 \, {\left (a + 3 \, b\right )} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a + 3 \, b\right )} \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/15*(15*b*tan(f*x + e) - (15*(a + 3*b)*tan(f*x + e)^4 + 5*(2*a + 3*b)*tan (f*x + e)^2 + 3*a + 3*b)/tan(f*x + e)^5)/f
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {15 \, b \tan \left (f x + e\right ) - \frac {15 \, a \tan \left (f x + e\right )^{4} + 45 \, b \tan \left (f x + e\right )^{4} + 10 \, a \tan \left (f x + e\right )^{2} + 15 \, b \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/15*(15*b*tan(f*x + e) - (15*a*tan(f*x + e)^4 + 45*b*tan(f*x + e)^4 + 10* a*tan(f*x + e)^2 + 15*b*tan(f*x + e)^2 + 3*a + 3*b)/tan(f*x + e)^5)/f
Time = 12.72 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {b\,\mathrm {tan}\left (e+f\,x\right )}{f}-\frac {\left (a+3\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (\frac {2\,a}{3}+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {a}{5}+\frac {b}{5}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5} \] Input:
int((a + b/cos(e + f*x)^2)/sin(e + f*x)^6,x)
Output:
(b*tan(e + f*x))/f - (a/5 + b/5 + tan(e + f*x)^2*((2*a)/3 + b) + tan(e + f *x)^4*(a + 3*b))/(f*tan(e + f*x)^5)
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.38 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {8 \sin \left (f x +e \right )^{6} a +48 \sin \left (f x +e \right )^{6} b -4 \sin \left (f x +e \right )^{4} a -24 \sin \left (f x +e \right )^{4} b -\sin \left (f x +e \right )^{2} a -6 \sin \left (f x +e \right )^{2} b -3 a -3 b}{15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} f} \] Input:
int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x)
Output:
(8*sin(e + f*x)**6*a + 48*sin(e + f*x)**6*b - 4*sin(e + f*x)**4*a - 24*sin (e + f*x)**4*b - sin(e + f*x)**2*a - 6*sin(e + f*x)**2*b - 3*a - 3*b)/(15* cos(e + f*x)*sin(e + f*x)**5*f)