Integrand size = 23, antiderivative size = 84 \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}-\frac {a \cos (e+f x)}{(a-b)^2 f}+\frac {\cos ^3(e+f x)}{3 (a-b) f} \]
-a*cos(f*x+e)/(a-b)^2/f+1/3*cos(f*x+e)^3/(a-b)/f-a*arctan(sec(f*x+e)*b^(1/ 2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(5/2)/f
Time = 1.18 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.77 \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {6 a \sqrt {a-b} \sqrt {b} \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+6 a \sqrt {a-b} \sqrt {b} \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+(a-b) \cos (e+f x) (-5 a-b+(a-b) \cos (2 (e+f x)))}{6 (a-b)^3 f} \]
(6*a*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/S qrt[b]] + 6*a*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f *x)/2])/Sqrt[b]] + (a - b)*Cos[e + f*x]*(-5*a - b + (a - b)*Cos[2*(e + f*x )]))/(6*(a - b)^3*f)
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4147, 25, 359, 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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{a+b \tan (e+f x)^2}dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \frac {\int -\frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {a \int \frac {\cos ^2(e+f x)}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a-b}+\frac {\cos ^3(e+f x)}{3 (a-b)}}{f}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\frac {a \left (-\frac {b \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a-b}-\frac {\cos (e+f x)}{a-b}\right )}{a-b}+\frac {\cos ^3(e+f x)}{3 (a-b)}}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {a \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}-\frac {\cos (e+f x)}{a-b}\right )}{a-b}+\frac {\cos ^3(e+f x)}{3 (a-b)}}{f}\) |
(Cos[e + f*x]^3/(3*(a - b)) + (a*(-((Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x]) /Sqrt[a - b]])/(a - b)^(3/2)) - Cos[e + f*x]/(a - b)))/(a - b))/f
3.1.56.3.1 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)/(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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -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 = 2.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a}{\left (a -b \right )^{2}}+\frac {a b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \sqrt {b \left (a -b \right )}}}{f}\) | \(87\) |
default | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a}{\left (a -b \right )^{2}}+\frac {a b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \sqrt {b \left (a -b \right )}}}{f}\) | \(87\) |
risch | \(-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} a}{8 \left (-a +b \right )^{2} f}-\frac {{\mathrm e}^{i \left (f x +e \right )} b}{8 \left (-a +b \right )^{2} f}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} a}{8 \left (-a +b \right )^{2} f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b}{8 \left (-a +b \right )^{2} f}+\frac {i \sqrt {b \left (a -b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right )^{3} f}-\frac {i \sqrt {b \left (a -b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right )^{3} f}+\frac {\cos \left (3 f x +3 e \right )}{12 f \left (a -b \right )}\) | \(237\) |
1/f*(1/(a-b)^2*(1/3*a*cos(f*x+e)^3-1/3*b*cos(f*x+e)^3-cos(f*x+e)*a)+a*b/(a -b)^2/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2)))
Time = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.45 \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{3} + 3 \, a \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, a \cos \left (f x + e\right )}{6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}, \frac {{\left (a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, a \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 3 \, a \cos \left (f x + e\right )}{3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \]
[1/6*(2*(a - b)*cos(f*x + e)^3 + 3*a*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*a*cos(f*x + e))/((a^2 - 2*a*b + b^2)*f), 1/3*((a - b)*cos(f *x + e)^3 - 3*a*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) - 3*a*cos(f*x + e))/((a^2 - 2*a*b + b^2)*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: ValueError} \]
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. 173 vs. \(2 (74) = 148\).
Time = 0.50 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.06 \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {a b \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b - b^{2}} f} + \frac {a^{2} f^{5} \cos \left (f x + e\right )^{3} - 2 \, a b f^{5} \cos \left (f x + e\right )^{3} + b^{2} f^{5} \cos \left (f x + e\right )^{3} - 3 \, a^{2} f^{5} \cos \left (f x + e\right ) + 3 \, a b f^{5} \cos \left (f x + e\right )}{3 \, {\left (a^{3} f^{6} - 3 \, a^{2} b f^{6} + 3 \, a b^{2} f^{6} - b^{3} f^{6}\right )}} \]
a*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) + 1/3*(a^2*f^5*cos(f*x + e)^3 - 2*a*b*f^5*cos( f*x + e)^3 + b^2*f^5*cos(f*x + e)^3 - 3*a^2*f^5*cos(f*x + e) + 3*a*b*f^5*c os(f*x + e))/(a^3*f^6 - 3*a^2*b*f^6 + 3*a*b^2*f^6 - b^3*f^6)
Time = 12.29 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.55 \[ \int \frac {\sin ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {2\,\left (2\,a+b\right )}{3\,{\left (a-b\right )}^2}+\frac {4\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{{\left (a-b\right )}^2}+\frac {2\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{{\left (a-b\right )}^2}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (8\,a^7\,b-32\,a^6\,b^2+48\,a^5\,b^3-32\,a^4\,b^4+8\,a^3\,b^5\right )}{{\left (a-b\right )}^{9/2}}-\frac {a\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-16\,a^9+128\,a^8\,b-432\,a^7\,b^2+800\,a^6\,b^3-880\,a^5\,b^4+576\,a^4\,b^5-208\,a^3\,b^6+32\,a^2\,b^7\right )}{8\,{\left (a-b\right )}^{15/2}}\right )-\frac {a\,\sqrt {b}\,\left (a-2\,b\right )\,\left (16\,a^9-96\,a^8\,b+240\,a^7\,b^2-320\,a^6\,b^3+240\,a^5\,b^4-96\,a^4\,b^5+16\,a^3\,b^6\right )}{8\,{\left (a-b\right )}^{15/2}}\right )\,{\left (a-b\right )}^5}{4\,a^8\,b-16\,a^7\,b^2+24\,a^6\,b^3-16\,a^5\,b^4+4\,a^4\,b^5}\right )}{f\,{\left (a-b\right )}^{5/2}} \]
- ((2*(2*a + b))/(3*(a - b)^2) + (4*a*tan(e/2 + (f*x)/2)^2)/(a - b)^2 + (2 *b*tan(e/2 + (f*x)/2)^4)/(a - b)^2)/(f*(3*tan(e/2 + (f*x)/2)^2 + 3*tan(e/2 + (f*x)/2)^4 + tan(e/2 + (f*x)/2)^6 + 1)) - (a*b^(1/2)*atan(((tan(e/2 + ( f*x)/2)^2*((b^(1/2)*(8*a^7*b + 8*a^3*b^5 - 32*a^4*b^4 + 48*a^5*b^3 - 32*a^ 6*b^2))/(a - b)^(9/2) - (a*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) )/(8*(a - b)^(15/2))) - (a*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))/(8*(a - b)^(15 /2)))*(a - b)^5)/(4*a^8*b + 4*a^4*b^5 - 16*a^5*b^4 + 24*a^6*b^3 - 16*a^7*b ^2)))/(f*(a - b)^(5/2))