Integrand size = 23, antiderivative size = 133 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{7/2} f}-\frac {(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )} \] Output:
-1/2*b^(1/2)*(3*a+2*b)*arctan(b^(1/2)*sec(f*x+e)/(a-b)^(1/2))/(a-b)^(7/2)/ f-(a+b)*cos(f*x+e)/(a-b)^3/f+1/3*cos(f*x+e)^3/(a-b)^2/f-1/2*a*b*sec(f*x+e) /(a-b)^3/f/(a-b+b*sec(f*x+e)^2)
Time = 2.55 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.37 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {6 \sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}+\frac {6 \sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}-\frac {\cos (e+f x) \left (9 a+15 b+\frac {12 a b}{a+b+(a-b) \cos (2 (e+f x))}\right )+(-a+b) \cos (3 (e+f x))}{(a-b)^3}}{12 f} \] Input:
Integrate[Sin[e + f*x]^3/(a + b*Tan[e + f*x]^2)^2,x]
Output:
((6*Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sq rt[b]])/(a - b)^(7/2) + (6*Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[a - b] + Sqrt[ a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) - (Cos[e + f*x]*(9*a + 15*b + (12*a*b)/(a + b + (a - b)*Cos[2*(e + f*x)])) + (-a + b)*Cos[3*(e + f*x)]) /(a - b)^3)/(12*f)
Time = 0.65 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4147, 25, 361, 25, 1584, 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.
| \(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {\frac {1}{2} b \int -\frac {\cos ^4(e+f x) \left (\frac {a \sec ^4(e+f x)}{(a-b)^3}-\frac {2 a \sec ^2(e+f x)}{(a-b)^2 b}+\frac {2}{(a-b) b}\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\frac {a b \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {1}{2} b \int \frac {\cos ^4(e+f x) \left (\frac {a \sec ^4(e+f x)}{(a-b)^3}-\frac {2 a \sec ^2(e+f x)}{(a-b)^2 b}+\frac {2}{(a-b) b}\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\frac {a b \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {-\frac {1}{2} b \int \left (\frac {2 \cos ^4(e+f x)}{(a-b)^2 b}+\frac {2 (a+b) \cos ^2(e+f x)}{b (b-a)^3}+\frac {3 a+2 b}{(a-b)^3 \left (b \sec ^2(e+f x)+a-b\right )}\right )d\sec (e+f x)-\frac {a b \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{2} b \left (\frac {(3 a+2 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{\sqrt {b} (a-b)^{7/2}}-\frac {2 \cos ^3(e+f x)}{3 b (a-b)^2}+\frac {2 (a+b) \cos (e+f x)}{b (a-b)^3}\right )-\frac {a b \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
Input:
Int[Sin[e + f*x]^3/(a + b*Tan[e + f*x]^2)^2,x]
Output:
(-1/2*(b*(((3*a + 2*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/((a - b )^(7/2)*Sqrt[b]) + (2*(a + b)*Cos[e + f*x])/((a - b)^3*b) - (2*Cos[e + f*x ]^3)/(3*(a - b)^2*b))) - (a*b*Sec[e + f*x])/(2*(a - b)^3*(a - b + b*Sec[e + f*x]^2)))/f
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[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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 = 15.54 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )-\cos \left (f x +e \right ) b}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\) | \(152\) |
| default | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )-\cos \left (f x +e \right ) b}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\) | \(152\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (f x +e \right )}}{24 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} a}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} a}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )}}{24 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {a b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (a -b \right )^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{4 \left (a -b \right )^{4} f}+\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 \left (a -b \right )^{4} f}-\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{4 \left (a -b \right )^{4} f}-\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 \left (a -b \right )^{4} f}\) | \(541\) |
Input:
int(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(1/(a^2-2*a*b+b^2)/(a-b)*(1/3*a*cos(f*x+e)^3-1/3*b*cos(f*x+e)^3-a*cos( f*x+e)-cos(f*x+e)*b)+b/(a-b)^3*(-1/2*a*cos(f*x+e)/(a*cos(f*x+e)^2-b*cos(f* x+e)^2+b)+1/2*(3*a+2*b)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^ (1/2))))
Time = 0.13 (sec) , antiderivative size = 456, normalized size of antiderivative = 3.43 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, {\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac {2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 3 \, {\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\right ] \] Input:
integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
Output:
[1/12*(4*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(3*a^2 - a*b - 2*b^2)*cos( f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2)*cos(f*x + e)^2 + 3*a*b + 2*b^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)) - 6*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3* b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f), 1/6*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e) ^5 - 2*(3*a^2 - a*b - 2*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2)*cos (f*x + e)^2 + 3*a*b + 2*b^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b ))*cos(f*x + e)/b) - 3*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b + 6*a ^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**3/(a+b*tan(f*x+e)**2)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
Output:
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. 354 vs. \(2 (119) = 238\).
Time = 0.59 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.66 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {a^{4} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a^{3} b f^{11} \cos \left (f x + e\right )^{3} + 6 \, a^{2} b^{2} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a b^{3} f^{11} \cos \left (f x + e\right )^{3} + b^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) + 6 \, a^{3} b f^{11} \cos \left (f x + e\right ) - 6 \, a b^{3} f^{11} \cos \left (f x + e\right ) + 3 \, b^{4} f^{11} \cos \left (f x + e\right )}{3 \, {\left (a^{6} f^{12} - 6 \, a^{5} b f^{12} + 15 \, a^{4} b^{2} f^{12} - 20 \, a^{3} b^{3} f^{12} + 15 \, a^{2} b^{4} f^{12} - 6 \, a b^{5} f^{12} + b^{6} f^{12}\right )}} - \frac {a b \cos \left (f x + e\right )}{2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} f} + \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}} f} \] Input:
integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/3*(a^4*f^11*cos(f*x + e)^3 - 4*a^3*b*f^11*cos(f*x + e)^3 + 6*a^2*b^2*f^1 1*cos(f*x + e)^3 - 4*a*b^3*f^11*cos(f*x + e)^3 + b^4*f^11*cos(f*x + e)^3 - 3*a^4*f^11*cos(f*x + e) + 6*a^3*b*f^11*cos(f*x + e) - 6*a*b^3*f^11*cos(f* x + e) + 3*b^4*f^11*cos(f*x + e))/(a^6*f^12 - 6*a^5*b*f^12 + 15*a^4*b^2*f^ 12 - 20*a^3*b^3*f^12 + 15*a^2*b^4*f^12 - 6*a*b^5*f^12 + b^6*f^12) - 1/2*a* b*cos(f*x + e)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a*cos(f*x + e)^2 - b*cos( f*x + e)^2 + b)*f) + 1/2*(3*a*b + 2*b^2)*arctan((a*cos(f*x + e) - b*cos(f* x + e))/sqrt(a*b - b^2))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)* f)
Time = 11.31 (sec) , antiderivative size = 737, normalized size of antiderivative = 5.54 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(sin(e + f*x)^3/(a + b*tan(e + f*x)^2)^2,x)
Output:
(b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*((b^(1/2)*(3*a + 2*b)*(24*a^9*b + 16* a^2*b^8 - 72*a^3*b^7 + 96*a^4*b^6 + 40*a^5*b^5 - 240*a^6*b^4 + 264*a^7*b^3 - 128*a^8*b^2))/(4*a*(a - b)^(13/2)) + (b^(1/2)*(a - 2*b)*(3*a + 2*b)^2*( 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^1 0*b^2))/(32*a*(a - b)^(21/2))) + (b^(1/2)*(a - 2*b)*(3*a + 2*b)^2*(144*a^1 1*b - 16*a^12 + 16*a^3*b^9 - 144*a^4*b^8 + 576*a^5*b^7 - 1344*a^6*b^6 + 20 16*a^7*b^5 - 2016*a^8*b^4 + 1344*a^9*b^3 - 576*a^10*b^2))/(32*a*(a - b)^(2 1/2)))*(a - b)^7)/(12*a^3*b^8 - 4*a^2*b^9 - 9*a^10*b + 3*a^4*b^7 - 46*a^5* b^6 + 45*a^6*b^5 + 24*a^7*b^4 - 67*a^8*b^3 + 42*a^9*b^2))*(3*a + 2*b))/(2* f*(a - b)^(7/2)) - ((11*a*b + 4*a^2)/(3*(a - b)*(a^2 - 2*a*b + b^2)) + (ta n(e/2 + (f*x)/2)^8*(3*a*b + 2*b^2))/((a - b)*(a^2 - 2*a*b + b^2)) + (2*tan (e/2 + (f*x)/2)^6*(2*a^2 - 3*a*b + 11*b^2))/((a - b)*(a^2 - 2*a*b + b^2)) + (2*tan(e/2 + (f*x)/2)^2*(9*a*b + 2*a^2 + 19*b^2))/(3*(a - b)*(a^2 - 2*a* b + b^2)) + (2*tan(e/2 + (f*x)/2)^4*(22*a*b - 10*a^2 + 33*b^2))/(3*(a - b) *(a^2 - 2*a*b + b^2)))/(f*(a + tan(e/2 + (f*x)/2)^2*(a + 4*b) + tan(e/2 + (f*x)/2)^8*(a + 4*b) - tan(e/2 + (f*x)/2)^4*(2*a - 12*b) - tan(e/2 + (f*x) /2)^6*(2*a - 12*b) + a*tan(e/2 + (f*x)/2)^10))
Time = 0.22 (sec) , antiderivative size = 818, normalized size of antiderivative = 6.15 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,x)
Output:
(9*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f*x)/2))/sqrt( b))*sin(e + f*x)**2*a**2 - 3*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt( a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a*b - 6*sqrt(b)*sqrt(a - b)* atan((sqrt(a - b) - sqrt(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*b** 2 - 9*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f*x)/2))/sq rt(b))*a**2 - 6*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f *x)/2))/sqrt(b))*a*b + 9*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) + sqrt(a)*t an((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a**2 - 3*sqrt(b)*sqrt(a - b)*ata n((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a*b - 6*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b ))*sin(e + f*x)**2*b**2 - 9*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) + sqrt(a )*tan((e + f*x)/2))/sqrt(b))*a**2 - 6*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b ) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*a*b - 2*cos(e + f*x)*sin(e + f*x)** 4*a**3 + 6*cos(e + f*x)*sin(e + f*x)**4*a**2*b - 6*cos(e + f*x)*sin(e + f* x)**4*a*b**2 + 2*cos(e + f*x)*sin(e + f*x)**4*b**3 - 2*cos(e + f*x)*sin(e + f*x)**2*a**3 - 4*cos(e + f*x)*sin(e + f*x)**2*a**2*b + 14*cos(e + f*x)*s in(e + f*x)**2*a*b**2 - 8*cos(e + f*x)*sin(e + f*x)**2*b**3 + 4*cos(e + f* x)*a**3 + 7*cos(e + f*x)*a**2*b - 11*cos(e + f*x)*a*b**2 + 4*sin(e + f*x)* *2*a**3 + 3*sin(e + f*x)**2*a**2*b - 18*sin(e + f*x)**2*a*b**2 + 11*sin(e + f*x)**2*b**3 - 4*a**3 - 7*a**2*b + 11*a*b**2)/(6*f*(sin(e + f*x)**2*a...