Integrand size = 21, antiderivative size = 79 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=-\frac {3 (a+4 b) \text {arctanh}(\cos (e+f x))}{8 f}-\frac {(5 a+4 b) \cot (e+f x) \csc (e+f x)}{8 f}-\frac {a \cot ^3(e+f x) \csc (e+f x)}{4 f}+\frac {b \sec (e+f x)}{f} \] Output:
-3/8*(a+4*b)*arctanh(cos(f*x+e))/f-1/8*(5*a+4*b)*cot(f*x+e)*csc(f*x+e)/f-1 /4*a*cot(f*x+e)^3*csc(f*x+e)/f+b*sec(f*x+e)/f
Time = 5.81 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {-2 (3 a+4 b) \csc ^2\left (\frac {1}{2} (e+f x)\right )-a \csc ^4\left (\frac {1}{2} (e+f x)\right )-24 (a+4 b) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+(6 a+8 b) \sec ^2\left (\frac {1}{2} (e+f x)\right )+a \sec ^4\left (\frac {1}{2} (e+f x)\right )+128 b \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )}{64 f} \] Input:
Integrate[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2),x]
Output:
(-2*(3*a + 4*b)*Csc[(e + f*x)/2]^2 - a*Csc[(e + f*x)/2]^4 - 24*(a + 4*b)*( Log[Cos[(e + f*x)/2]] - Log[Sin[(e + f*x)/2]]) + (6*a + 8*b)*Sec[(e + f*x) /2]^2 + a*Sec[(e + f*x)/2]^4 + 128*b*Sec[e + f*x]*Sin[(e + f*x)/2]^2)/(64* f)
Time = 0.45 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4147, 25, 360, 25, 1471, 299, 219}
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.
| \(\displaystyle \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \tan (e+f x)^2}{\sin (e+f x)^5}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x) \left (b \sec ^2(e+f x)+a-b\right )}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x) \left (b \sec ^2(e+f x)+a-b\right )}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {-\frac {1}{4} \int -\frac {4 b \sec ^4(e+f x)+4 a \sec ^2(e+f x)+a}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)-\frac {a \sec (e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {4 b \sec ^4(e+f x)+4 a \sec ^2(e+f x)+a}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)-\frac {a \sec (e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {(5 a+4 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {8 b \sec ^2(e+f x)+3 a+4 b}{1-\sec ^2(e+f x)}d\sec (e+f x)\right )-\frac {a \sec (e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 b \sec (e+f x)-3 (a+4 b) \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)\right )+\frac {(5 a+4 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {a \sec (e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} (8 b \sec (e+f x)-3 (a+4 b) \text {arctanh}(\sec (e+f x)))+\frac {(5 a+4 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {a \sec (e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
Input:
Int[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2),x]
Output:
(-1/4*(a*Sec[e + f*x])/(1 - Sec[e + f*x]^2)^2 + ((-3*(a + 4*b)*ArcTanh[Sec [e + f*x]] + 8*b*Sec[e + f*x])/2 + ((5*a + 4*b)*Sec[e + f*x])/(2*(1 - Sec[ e + f*x]^2)))/4)/f
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 3.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29
| method | result | size |
| derivativedivides | \(\frac {b \left (-\frac {1}{2 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {3}{2 \cos \left (f x +e \right )}+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )+a \left (\left (-\frac {\csc \left (f x +e \right )^{3}}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{8}\right )}{f}\) | \(102\) |
| default | \(\frac {b \left (-\frac {1}{2 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {3}{2 \cos \left (f x +e \right )}+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )+a \left (\left (-\frac {\csc \left (f x +e \right )^{3}}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{8}\right )}{f}\) | \(102\) |
| risch | \(\frac {{\mathrm e}^{i \left (f x +e \right )} \left (3 a \,{\mathrm e}^{8 i \left (f x +e \right )}+12 b \,{\mathrm e}^{8 i \left (f x +e \right )}-8 a \,{\mathrm e}^{6 i \left (f x +e \right )}-32 b \,{\mathrm e}^{6 i \left (f x +e \right )}-22 a \,{\mathrm e}^{4 i \left (f x +e \right )}+40 b \,{\mathrm e}^{4 i \left (f x +e \right )}-8 a \,{\mathrm e}^{2 i \left (f x +e \right )}-32 b \,{\mathrm e}^{2 i \left (f x +e \right )}+3 a +12 b \right )}{4 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 f}-\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 f}\) | \(237\) |
Input:
int(csc(f*x+e)^5*(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(b*(-1/2/sin(f*x+e)^2/cos(f*x+e)+3/2/cos(f*x+e)+3/2*ln(csc(f*x+e)-cot( f*x+e)))+a*((-1/4*csc(f*x+e)^3-3/8*csc(f*x+e))*cot(f*x+e)+3/8*ln(csc(f*x+e )-cot(f*x+e))))
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (73) = 146\).
Time = 0.54 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.25 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {6 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{4} - 10 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left ({\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{3} + {\left (a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{3} + {\left (a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 16 \, b}{16 \, {\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )}} \] Input:
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2),x, algorithm="fricas")
Output:
1/16*(6*(a + 4*b)*cos(f*x + e)^4 - 10*(a + 4*b)*cos(f*x + e)^2 - 3*((a + 4 *b)*cos(f*x + e)^5 - 2*(a + 4*b)*cos(f*x + e)^3 + (a + 4*b)*cos(f*x + e))* log(1/2*cos(f*x + e) + 1/2) + 3*((a + 4*b)*cos(f*x + e)^5 - 2*(a + 4*b)*co s(f*x + e)^3 + (a + 4*b)*cos(f*x + e))*log(-1/2*cos(f*x + e) + 1/2) + 16*b )/(f*cos(f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*cos(f*x + e))
\[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right ) \csc ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)**5*(a+b*tan(f*x+e)**2),x)
Output:
Integral((a + b*tan(e + f*x)**2)*csc(e + f*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.28 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=-\frac {3 \, {\left (a + 4 \, b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (a + 4 \, b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{4} - 5 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, b\right )}}{\cos \left (f x + e\right )^{5} - 2 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )}}{16 \, f} \] Input:
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2),x, algorithm="maxima")
Output:
-1/16*(3*(a + 4*b)*log(cos(f*x + e) + 1) - 3*(a + 4*b)*log(cos(f*x + e) - 1) - 2*(3*(a + 4*b)*cos(f*x + e)^4 - 5*(a + 4*b)*cos(f*x + e)^2 + 8*b)/(co s(f*x + e)^5 - 2*cos(f*x + e)^3 + cos(f*x + e)))/f
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (73) = 146\).
Time = 0.55 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.03 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {12 \, {\left (a + 4 \, b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) - \frac {{\left (a - \frac {8 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {18 \, a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {72 \, b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) - 1\right )}^{2}} - \frac {8 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {128 \, b}{\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1}}{64 \, f} \] Input:
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2),x, algorithm="giac")
Output:
1/64*(12*(a + 4*b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) - (a - 8*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*b*(cos(f*x + e) - 1)/(cos( f*x + e) + 1) + 18*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 72*b*(cos (f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(cos(f*x + e) - 1)^2 - 8*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*b*(cos(f*x + e) - 1 )/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 128*b /((cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/f
Time = 7.42 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.75 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a}{8}+\frac {b}{8}\right )}{f}-\frac {\left (-2\,a-34\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (\frac {7\,a}{4}+2\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\frac {a}{4}}{f\,\left (16\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-16\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (\frac {3\,a}{8}+\frac {3\,b}{2}\right )}{f}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{64\,f} \] Input:
int((a + b*tan(e + f*x)^2)/sin(e + f*x)^5,x)
Output:
(tan(e/2 + (f*x)/2)^2*(a/8 + b/8))/f - (a/4 + tan(e/2 + (f*x)/2)^2*((7*a)/ 4 + 2*b) - tan(e/2 + (f*x)/2)^4*(2*a + 34*b))/(f*(16*tan(e/2 + (f*x)/2)^4 - 16*tan(e/2 + (f*x)/2)^6)) + (log(tan(e/2 + (f*x)/2))*((3*a)/8 + (3*b)/2) )/f + (a*tan(e/2 + (f*x)/2)^4)/(64*f)
Time = 0.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.99 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {3 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a +12 \cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} b -\cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a -9 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +3 \sin \left (f x +e \right )^{4} a +12 \sin \left (f x +e \right )^{4} b -\sin \left (f x +e \right )^{2} a -4 \sin \left (f x +e \right )^{2} b -2 a}{8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} f} \] Input:
int(csc(f*x+e)^5*(a+b*tan(f*x+e)^2),x)
Output:
(3*cos(e + f*x)*log(tan((e + f*x)/2))*sin(e + f*x)**4*a + 12*cos(e + f*x)* log(tan((e + f*x)/2))*sin(e + f*x)**4*b - cos(e + f*x)*sin(e + f*x)**4*a - 9*cos(e + f*x)*sin(e + f*x)**4*b + 3*sin(e + f*x)**4*a + 12*sin(e + f*x)* *4*b - sin(e + f*x)**2*a - 4*sin(e + f*x)**2*b - 2*a)/(8*cos(e + f*x)*sin( e + f*x)**4*f)