Integrand size = 23, antiderivative size = 117 \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{7/2} f}-\frac {a^2 \cos (e+f x)}{(a-b)^3 f}+\frac {(2 a-b) \cos ^3(e+f x)}{3 (a-b)^2 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f} \]
-a^2*cos(f*x+e)/(a-b)^3/f+1/3*(2*a-b)*cos(f*x+e)^3/(a-b)^2/f-1/5*cos(f*x+e )^5/(a-b)/f-a^2*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(7/2) /f
Time = 3.78 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {120 a^2 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+120 a^2 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+\sqrt {a-b} \cos (e+f x) \left (-89 a^2-42 a b+11 b^2+4 \left (7 a^2-9 a b+2 b^2\right ) \cos (2 (e+f x))-3 (a-b)^2 \cos (4 (e+f x))\right )}{120 (a-b)^{7/2} f} \]
(120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + 120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + Sqrt[a - b]*Cos[e + f*x]*(-89*a^2 - 42*a*b + 11*b^2 + 4*(7*a^2 - 9*a*b + 2*b^2)*Cos[2*(e + f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)]))/(120*(a - b)^(7 /2)*f)
Time = 0.35 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4147, 364, 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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{a+b \tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {\int \left (\frac {\cos ^6(e+f x)}{a-b}+\frac {(b-2 a) \cos ^4(e+f x)}{(a-b)^2}+\frac {a^2 \cos ^2(e+f x)}{(a-b)^3}-\frac {a^2 b}{(a-b)^3 \left (b \sec ^2(e+f x)+a-b\right )}\right )d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{7/2}}-\frac {a^2 \cos (e+f x)}{(a-b)^3}-\frac {\cos ^5(e+f x)}{5 (a-b)}+\frac {(2 a-b) \cos ^3(e+f x)}{3 (a-b)^2}}{f}\) |
(-((a^2*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a - b)^(7/2)) - (a^2*Cos[e + f*x])/(a - b)^3 + ((2*a - b)*Cos[e + f*x]^3)/(3*(a - b)^2) - Cos[e + f*x]^5/(5*(a - b)))/f
3.1.55.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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 = 10.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.23
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+a b \cos \left (f x +e \right )^{3}-\frac {b^{2} \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )}{\left (a -b \right )^{3}}+\frac {a^{2} b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \sqrt {b \left (a -b \right )}}}{f}\) | \(144\) |
| default | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+a b \cos \left (f x +e \right )^{3}-\frac {b^{2} \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )}{\left (a -b \right )^{3}}+\frac {a^{2} b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \sqrt {b \left (a -b \right )}}}{f}\) | \(144\) |
| risch | \(-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a^{2}}{16 \left (a -b \right )^{3} f}-\frac {{\mathrm e}^{i \left (f x +e \right )} a b}{4 \left (a -b \right )^{3} f}+\frac {{\mathrm e}^{i \left (f x +e \right )} b^{2}}{16 \left (a -b \right )^{3} f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{16 \left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} a b}{4 \left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) f}+\frac {{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{16 \left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {i \sqrt {b \left (a -b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right )^{4} f}+\frac {i \sqrt {b \left (a -b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right )^{4} f}-\frac {\cos \left (5 f x +5 e \right )}{80 f \left (a -b \right )}+\frac {5 \cos \left (3 f x +3 e \right ) a}{48 f \left (a -b \right )^{2}}-\frac {\cos \left (3 f x +3 e \right ) b}{48 f \left (a -b \right )^{2}}\) | \(378\) |
1/f*(-1/(a-b)^3*(1/5*a^2*cos(f*x+e)^5-2/5*a*b*cos(f*x+e)^5+1/5*b^2*cos(f*x +e)^5-2/3*a^2*cos(f*x+e)^3+a*b*cos(f*x+e)^3-1/3*b^2*cos(f*x+e)^3+a^2*cos(f *x+e))+a^2*b/(a-b)^3/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/ 2)))
Time = 0.30 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.51 \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [-\frac {6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 15 \, a^{2} \cos \left (f x + e\right )}{15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \]
[-1/30*(6*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 10*(2*a^2 - 3*a*b + b^2)*co s(f*x + e)^3 + 15*a^2*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 3 0*a^2*cos(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), -1/15*(3*(a^2 - 2 *a*b + b^2)*cos(f*x + e)^5 - 5*(2*a^2 - 3*a*b + b^2)*cos(f*x + e)^3 + 15*a ^2*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*a^ 2*cos(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (105) = 210\).
Time = 0.52 (sec) , antiderivative size = 356, normalized size of antiderivative = 3.04 \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {15 \, a^{2} b \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}}} - \frac {2 \, {\left (8 \, a^{2} + 9 \, a b - 2 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {30 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {10 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {30 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \]
-1/15*(15*a^2*b*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2)))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqr t(a*b - b^2)) - 2*(8*a^2 + 9*a*b - 2*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos( f*x + e) + 1) - 30*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 10*b^2*(cos (f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 10*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 90*a*b*(co s(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*b^2*(cos(f*x + e) - 1)^3/(cos( f*x + e) + 1)^3 + 15*a*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5)) /f
Time = 13.62 (sec) , antiderivative size = 643, normalized size of antiderivative = 5.50 \[ \int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {2\,\left (8\,a^2+9\,a\,b-2\,b^2\right )}{15\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (8\,a^2+b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,a^2+3\,a\,b-b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (3\,a-b\right )}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a\,\sqrt {b}\,\left (16\,a^{10}\,b-96\,a^9\,b^2+240\,a^8\,b^3-320\,a^7\,b^4+240\,a^6\,b^5-96\,a^5\,b^6+16\,a^4\,b^7\right )}{2\,{\left (a-b\right )}^{13/2}}+\frac {a^3\,\sqrt {b}\,\left (a-2\,b\right )\,\left (16\,a^{12}-176\,a^{11}\,b+864\,a^{10}\,b^2-2496\,a^9\,b^3+4704\,a^8\,b^4-6048\,a^7\,b^5+5376\,a^6\,b^6-3264\,a^5\,b^7+1296\,a^4\,b^8-304\,a^3\,b^9+32\,a^2\,b^{10}\right )}{8\,{\left (a-b\right )}^{21/2}}\right )+\frac {a^3\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-16\,a^{12}+144\,a^{11}\,b-576\,a^{10}\,b^2+1344\,a^9\,b^3-2016\,a^8\,b^4+2016\,a^7\,b^5-1344\,a^6\,b^6+576\,a^5\,b^7-144\,a^4\,b^8+16\,a^3\,b^9\right )}{8\,{\left (a-b\right )}^{21/2}}\right )\,{\left (a-b\right )}^7}{4\,a^{12}\,b-24\,a^{11}\,b^2+60\,a^{10}\,b^3-80\,a^9\,b^4+60\,a^8\,b^5-24\,a^7\,b^6+4\,a^6\,b^7}\right )}{f\,{\left (a-b\right )}^{7/2}} \]
- ((2*(9*a*b + 8*a^2 - 2*b^2))/(15*(a - b)*(a^2 - 2*a*b + b^2)) + (4*tan(e /2 + (f*x)/2)^4*(8*a^2 + b^2))/(3*(a - b)*(a^2 - 2*a*b + b^2)) + (4*tan(e/ 2 + (f*x)/2)^2*(3*a*b + 4*a^2 - b^2))/(3*(a - b)*(a^2 - 2*a*b + b^2)) + (4 *b*tan(e/2 + (f*x)/2)^6*(3*a - b))/((a - b)*(a^2 - 2*a*b + b^2)) + (2*a*b* tan(e/2 + (f*x)/2)^8)/((a - b)*(a^2 - 2*a*b + b^2)))/(f*(5*tan(e/2 + (f*x) /2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 + 5*tan(e/2 + (f *x)/2)^8 + tan(e/2 + (f*x)/2)^10 + 1)) - (a^2*b^(1/2)*atan(((tan(e/2 + (f* x)/2)^2*((a*b^(1/2)*(16*a^10*b + 16*a^4*b^7 - 96*a^5*b^6 + 240*a^6*b^5 - 3 20*a^7*b^4 + 240*a^8*b^3 - 96*a^9*b^2))/(2*(a - b)^(13/2)) + (a^3*b^(1/2)* (a - 2*b)*(16*a^12 - 176*a^11*b + 32*a^2*b^10 - 304*a^3*b^9 + 1296*a^4*b^8 - 3264*a^5*b^7 + 5376*a^6*b^6 - 6048*a^7*b^5 + 4704*a^8*b^4 - 2496*a^9*b^ 3 + 864*a^10*b^2))/(8*(a - b)^(21/2))) + (a^3*b^(1/2)*(a - 2*b)*(144*a^11* b - 16*a^12 + 16*a^3*b^9 - 144*a^4*b^8 + 576*a^5*b^7 - 1344*a^6*b^6 + 2016 *a^7*b^5 - 2016*a^8*b^4 + 1344*a^9*b^3 - 576*a^10*b^2))/(8*(a - b)^(21/2)) )*(a - b)^7)/(4*a^12*b + 4*a^6*b^7 - 24*a^7*b^6 + 60*a^8*b^5 - 80*a^9*b^4 + 60*a^10*b^3 - 24*a^11*b^2)))/(f*(a - b)^(7/2))