Integrand size = 21, antiderivative size = 46 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {(a+2 b) \cot (e+f x)}{f}-\frac {(a+b) \cot ^3(e+f x)}{3 f}+\frac {b \tan (e+f x)}{f} \] Output:
-(a+2*b)*cot(f*x+e)/f-1/3*(a+b)*cot(f*x+e)^3/f+b*tan(f*x+e)/f
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.83 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {2 a \cot (e+f x)}{3 f}-\frac {5 b \cot (e+f x)}{3 f}-\frac {a \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac {b \cot (e+f x) \csc ^2(e+f x)}{3 f}+\frac {b \tan (e+f x)}{f} \] Input:
Integrate[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(-2*a*Cot[e + f*x])/(3*f) - (5*b*Cot[e + f*x])/(3*f) - (a*Cot[e + f*x]*Csc [e + f*x]^2)/(3*f) - (b*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f) + (b*Tan[e + f* x])/f
Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^2}{\sin (e+f x)^4}dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {\int \left ((a+b) \cot ^4(e+f x)+(a+2 b) \cot ^2(e+f x)+b\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} (a+b) \cot ^3(e+f x)-(a+2 b) \cot (e+f x)+b \tan (e+f x)}{f}\) |
Input:
Int[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(-((a + 2*b)*Cot[e + f*x]) - ((a + b)*Cot[e + f*x]^3)/3 + 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 = 1.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.59
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {2}{3}-\frac {\csc \left (f x +e \right )^{2}}{3}\right ) \cot \left (f x +e \right )+b \left (-\frac {1}{3 \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )}+\frac {4}{3 \sin \left (f x +e \right ) \cos \left (f x +e \right )}-\frac {8 \cot \left (f x +e \right )}{3}\right )}{f}\) | \(73\) |
default | \(\frac {a \left (-\frac {2}{3}-\frac {\csc \left (f x +e \right )^{2}}{3}\right ) \cot \left (f x +e \right )+b \left (-\frac {1}{3 \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )}+\frac {4}{3 \sin \left (f x +e \right ) \cos \left (f x +e \right )}-\frac {8 \cot \left (f x +e \right )}{3}\right )}{f}\) | \(73\) |
risch | \(\frac {4 i \left (3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+8 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a -4 b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(76\) |
parallelrisch | \(\frac {\cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (\left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+\left (8 a +20 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (-18 a -90 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (8 a +20 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+a +b \right )}{24 f \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-24 f}\) | \(108\) |
norman | \(\frac {\frac {a +b}{24 f}+\frac {\left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{24 f}-\frac {3 \left (a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{4 f}+\frac {\left (2 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{6 f}+\frac {\left (2 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{6 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\) | \(123\) |
Input:
int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-2/3-1/3*csc(f*x+e)^2)*cot(f*x+e)+b*(-1/3/sin(f*x+e)^3/cos(f*x+e)+ 4/3/sin(f*x+e)/cos(f*x+e)-8/3*cot(f*x+e)))
Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {2 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b}{3 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:
integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
-1/3*(2*(a + 4*b)*cos(f*x + e)^4 - 3*(a + 4*b)*cos(f*x + e)^2 + 3*b)/((f*c os(f*x + e)^3 - f*cos(f*x + e))*sin(f*x + e))
\[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \csc ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)**4*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*csc(e + f*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right ) - \frac {3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{2} + a + b}{\tan \left (f x + e\right )^{3}}}{3 \, f} \] Input:
integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/3*(3*b*tan(f*x + e) - (3*(a + 2*b)*tan(f*x + e)^2 + a + b)/tan(f*x + e)^ 3)/f
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right ) - \frac {3 \, a \tan \left (f x + e\right )^{2} + 6 \, b \tan \left (f x + e\right )^{2} + a + b}{\tan \left (f x + e\right )^{3}}}{3 \, f} \] Input:
integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/3*(3*b*tan(f*x + e) - (3*a*tan(f*x + e)^2 + 6*b*tan(f*x + e)^2 + a + b)/ tan(f*x + e)^3)/f
Time = 12.60 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {b\,\mathrm {tan}\left (e+f\,x\right )}{f}-\frac {\left (a+2\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {a}{3}+\frac {b}{3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^3} \] Input:
int((a + b/cos(e + f*x)^2)/sin(e + f*x)^4,x)
Output:
(b*tan(e + f*x))/f - (a/3 + b/3 + tan(e + f*x)^2*(a + 2*b))/(f*tan(e + f*x )^3)
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.57 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {2 \sin \left (f x +e \right )^{4} a +8 \sin \left (f x +e \right )^{4} b -\sin \left (f x +e \right )^{2} a -4 \sin \left (f x +e \right )^{2} b -a -b}{3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} f} \] Input:
int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2),x)
Output:
(2*sin(e + f*x)**4*a + 8*sin(e + f*x)**4*b - sin(e + f*x)**2*a - 4*sin(e + f*x)**2*b - a - b)/(3*cos(e + f*x)*sin(e + f*x)**3*f)