Integrand size = 21, antiderivative size = 101 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} f}-\frac {3 \cos (e+f x)}{2 (a-b)^2 f}+\frac {\cos (e+f x)}{2 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )} \]
-3/2*cos(f*x+e)/(a-b)^2/f+1/2*cos(f*x+e)/(a-b)/f/(a-b+b*sec(f*x+e)^2)-3/2* arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(5/2)/f
Time = 1.38 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}+\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}+\frac {2 \cos (e+f x) \left (-1-\frac {b}{a+b+(a-b) \cos (2 (e+f x))}\right )}{(a-b)^2}}{2 f} \]
((3*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(5/2) + (3*Sqrt[b]*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqr t[b]])/(a - b)^(5/2) + (2*Cos[e + f*x]*(-1 - b/(a + b + (a - b)*Cos[2*(e + f*x)])))/(a - b)^2)/(2*f)
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4147, 253, 264, 218}
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.
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {3 \int \frac {\cos ^2(e+f x)}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {3 \left (-\frac {b \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a-b}-\frac {\cos (e+f x)}{a-b}\right )}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}-\frac {\cos (e+f x)}{a-b}\right )}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
((3*(-((Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a - b)^(3/2)) - Cos[e + f*x]/(a - b)))/(2*(a - b)) + Cos[e + f*x]/(2*(a - b)*(a - b + b *Sec[e + f*x]^2)))/f
3.1.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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[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 = 1.54 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {-\frac {\cos \left (f x +e \right )}{a^{2}-2 a b +b^{2}}+\frac {b \left (-\frac {\cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {3 \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{2}}}{f}\) | \(103\) |
| default | \(\frac {-\frac {\cos \left (f x +e \right )}{a^{2}-2 a b +b^{2}}+\frac {b \left (-\frac {\cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {3 \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{2}}}{f}\) | \(103\) |
| risch | \(-\frac {{\mathrm e}^{i \left (f x +e \right )}}{2 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {{\mathrm e}^{-i \left (f x +e \right )}}{2 \left (a^{2}-2 a b +b^{2}\right ) f}+\frac {b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (-a +b \right )^{2} \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )}-\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right )^{3} f}+\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right )^{3} f}\) | \(265\) |
1/f*(-1/(a^2-2*a*b+b^2)*cos(f*x+e)+b/(a-b)^2*(-1/2*cos(f*x+e)/(a*cos(f*x+e )^2-b*cos(f*x+e)^2+b)+3/2/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b) )^(1/2))))
Time = 0.31 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.04 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [-\frac {4 \, {\left (a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 6 \, b \cos \left (f x + e\right )}{4 \, {\left ({\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f\right )}}, -\frac {2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 3 \, b \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f\right )}}\right ] \]
[-1/4*(4*(a - b)*cos(f*x + e)^3 - 3*((a - b)*cos(f*x + e)^2 + b)*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)) + 6*b*cos(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f*cos(f*x + e)^2 + (a^2*b - 2*a*b^2 + b^3)*f), -1/2*(2*( a - b)*cos(f*x + e)^3 + 3*((a - b)*cos(f*x + e)^2 + b)*sqrt(b/(a - b))*arc tan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 3*b*cos(f*x + e))/((a^3 - 3 *a^2*b + 3*a*b^2 - b^3)*f*cos(f*x + e)^2 + (a^2*b - 2*a*b^2 + b^3)*f)]
\[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\int \frac {\sin {\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.62 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.46 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {f^{3} \cos \left (f x + e\right )}{a^{2} f^{4} - 2 \, a b f^{4} + b^{2} f^{4}} + \frac {3 \, b \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b - b^{2}} f} - \frac {b \cos \left (f x + e\right )}{2 \, {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )} f} \]
-f^3*cos(f*x + e)/(a^2*f^4 - 2*a*b*f^4 + b^2*f^4) + 3/2*b*arctan((a*cos(f* x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^2 - 2*a*b + b^2)*sqrt(a*b - b^2)*f) - 1/2*b*cos(f*x + e)/((a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)*(a ^2 - 2*a*b + b^2)*f)
Time = 12.98 (sec) , antiderivative size = 436, normalized size of antiderivative = 4.32 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {2\,a+b}{{\left (a-b\right )}^2}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,a^2-a\,b+2\,b^2\right )}{a\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-2\,a^2+4\,a\,b+b^2\right )}{a\,{\left (a-b\right )}^2}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (4\,b-a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (4\,b-a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {{\left (a-b\right )}^5\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (18\,a^6\,b-72\,a^5\,b^2+108\,a^4\,b^3-72\,a^3\,b^4+18\,a^2\,b^5\right )}{a\,{\left (a-b\right )}^{9/2}}-\frac {9\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-16\,a^9+128\,a^8\,b-432\,a^7\,b^2+800\,a^6\,b^3-880\,a^5\,b^4+576\,a^4\,b^5-208\,a^3\,b^6+32\,a^2\,b^7\right )}{32\,a\,{\left (a-b\right )}^{15/2}}\right )-\frac {9\,\sqrt {b}\,\left (a-2\,b\right )\,\left (16\,a^9-96\,a^8\,b+240\,a^7\,b^2-320\,a^6\,b^3+240\,a^5\,b^4-96\,a^4\,b^5+16\,a^3\,b^6\right )}{32\,a\,{\left (a-b\right )}^{15/2}}\right )}{9\,a^6\,b-36\,a^5\,b^2+54\,a^4\,b^3-36\,a^3\,b^4+9\,a^2\,b^5}\right )}{2\,f\,{\left (a-b\right )}^{5/2}} \]
- ((2*a + b)/(a - b)^2 + (tan(e/2 + (f*x)/2)^4*(2*a^2 - a*b + 2*b^2))/(a*( a^2 - 2*a*b + b^2)) + (2*tan(e/2 + (f*x)/2)^2*(4*a*b - 2*a^2 + b^2))/(a*(a - b)^2))/(f*(a - tan(e/2 + (f*x)/2)^2*(a - 4*b) - tan(e/2 + (f*x)/2)^4*(a - 4*b) + a*tan(e/2 + (f*x)/2)^6)) - (3*b^(1/2)*atan(((a - b)^5*(tan(e/2 + (f*x)/2)^2*((b^(1/2)*(18*a^6*b + 18*a^2*b^5 - 72*a^3*b^4 + 108*a^4*b^3 - 72*a^5*b^2))/(a*(a - b)^(9/2)) - (9*b^(1/2)*(a - 2*b)*(128*a^8*b - 16*a^9 + 32*a^2*b^7 - 208*a^3*b^6 + 576*a^4*b^5 - 880*a^5*b^4 + 800*a^6*b^3 - 432 *a^7*b^2))/(32*a*(a - b)^(15/2))) - (9*b^(1/2)*(a - 2*b)*(16*a^9 - 96*a^8* b + 16*a^3*b^6 - 96*a^4*b^5 + 240*a^5*b^4 - 320*a^6*b^3 + 240*a^7*b^2))/(3 2*a*(a - b)^(15/2))))/(9*a^6*b + 9*a^2*b^5 - 36*a^3*b^4 + 54*a^4*b^3 - 36* a^5*b^2)))/(2*f*(a - b)^(5/2))