Integrand size = 23, antiderivative size = 129 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\left (3 a^2+6 a b-b^2\right ) x}{8 (a-b)^3}-\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^3 f}-\frac {(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f} \]
1/8*(3*a^2+6*a*b-b^2)*x/(a-b)^3-1/8*(5*a-b)*cos(f*x+e)*sin(f*x+e)/(a-b)^2/ f+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f-a^(3/2)*arctan(b^(1/2)*tan(f*x+e)/a^ (1/2))*b^(1/2)/(a-b)^3/f
Time = 0.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.77 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {4 \left (3 a^2+6 a b-b^2\right ) (e+f x)-32 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-8 a (a-b) \sin (2 (e+f x))+(a-b)^2 \sin (4 (e+f x))}{32 (a-b)^3 f} \]
(4*(3*a^2 + 6*a*b - b^2)*(e + f*x) - 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Ta n[e + f*x])/Sqrt[a]] - 8*a*(a - b)*Sin[2*(e + f*x)] + (a - b)^2*Sin[4*(e + f*x)])/(32*(a - b)^3*f)
Time = 0.34 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4146, 372, 402, 397, 216, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{a+b \tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int \frac {a-(4 a-b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {a (3 a+b)-(5 a-b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+6 a b-b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a-b}-\frac {8 a^2 b \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+6 a b-b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {8 a^2 b \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+6 a b-b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {8 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a-b}}{2 (a-b)}}{4 (a-b)}}{f}\) |
(Tan[e + f*x]/(4*(a - b)*(1 + Tan[e + f*x]^2)^2) - (-1/2*(((3*a^2 + 6*a*b - b^2)*ArcTan[Tan[e + f*x]])/(a - b) - (8*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]* Tan[e + f*x])/Sqrt[a]])/(a - b))/(a - b) + ((5*a - b)*Tan[e + f*x])/(2*(a - b)*(1 + Tan[e + f*x]^2)))/(4*(a - b)))/f
3.1.62.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 5.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {3}{4} a b -\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {1}{4} a b +\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+6 a b -b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{3}}}{f}\) | \(131\) |
| default | \(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {3}{4} a b -\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {1}{4} a b +\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+6 a b -b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{3}}}{f}\) | \(131\) |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a -b \right )^{3}}+\frac {3 x a b}{4 \left (a -b \right )^{3}}-\frac {x \,b^{2}}{8 \left (a -b \right )^{3}}+\frac {i a \,{\mathrm e}^{2 i \left (f x +e \right )}}{8 \left (a -b \right )^{2} f}-\frac {i a \,{\mathrm e}^{-2 i \left (f x +e \right )}}{8 \left (a^{2}-2 a b +b^{2}\right ) f}+\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 \left (a -b \right )^{3} f}-\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 \left (a -b \right )^{3} f}+\frac {\sin \left (4 f x +4 e \right )}{32 \left (a -b \right ) f}\) | \(218\) |
1/f*(-a^2*b/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+1/(a-b)^3 *(((-5/8*a^2+3/4*a*b-1/8*b^2)*tan(f*x+e)^3+(-3/8*a^2+1/4*a*b+1/8*b^2)*tan( f*x+e))/(1+tan(f*x+e)^2)^2+1/8*(3*a^2+6*a*b-b^2)*arctan(tan(f*x+e))))
Time = 0.32 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.97 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x - 2 \, \sqrt {-a b} a \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x + 4 \, \sqrt {a b} a \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \]
[1/8*((3*a^2 + 6*a*b - b^2)*f*x - 2*sqrt(-a*b)*a*log(((a^2 + 6*a*b + b^2)* cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 - 4*((a + b)*cos(f*x + e)^ 3 - b*cos(f*x + e))*sqrt(-a*b)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*co s(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)) + (2*(a^2 - 2*a*b + b^ 2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b + b^2)*cos(f*x + e))*sin(f*x + e))/((a^ 3 - 3*a^2*b + 3*a*b^2 - b^3)*f), 1/8*((3*a^2 + 6*a*b - b^2)*f*x + 4*sqrt(a *b)*a*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b)/(a*b*cos(f*x + e)* sin(f*x + e))) + (2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b + b^2)*cos(f*x + e))*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)]
Timed out. \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.42 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {8 \, a^{2} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a - b\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a + b\right )} \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}}}{8 \, f} \]
-1/8*(8*a^2*b*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (3*a^2 + 6*a*b - b^2)*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^ 2 - b^3) + ((5*a - b)*tan(f*x + e)^3 + (3*a + b)*tan(f*x + e))/((a^2 - 2*a *b + b^2)*tan(f*x + e)^4 + 2*(a^2 - 2*a*b + b^2)*tan(f*x + e)^2 + a^2 - 2* a*b + b^2))/f
Time = 0.50 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.41 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a^{2} b}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {5 \, a \tan \left (f x + e\right )^{3} - b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \]
-1/8*(8*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt( a*b)))*a^2*b/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (3*a^2 + 6*a*b - b^2)*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*tan(f*x + e)^3 - b *tan(f*x + e)^3 + 3*a*tan(f*x + e) + b*tan(f*x + e))/((a^2 - 2*a*b + b^2)* (tan(f*x + e)^2 + 1)^2))/f
Time = 13.78 (sec) , antiderivative size = 3588, normalized size of antiderivative = 27.81 \[ \int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]
(atan(((((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4* b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (((32*a*b^9 - 96* a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7 *b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^ 3*b^3 + 15*a^4*b^2)) - (tan(e + f*x)*(-a^3*b)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64*(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a* b^3 + b^4 + 6*a^2*b^2)))*(-a^3*b)^(1/2))/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3 )))*(-a^3*b)^(1/2)*1i)/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (((tan(e + f* x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a ^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b^2)/( 64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(e + f*x)*(-a^3*b)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 128 0*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64 *(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^ 2)))*(-a^3*b)^(1/2))/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))*(-a^3*b)^(1/2)*1 i)/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/((a^2*b^6 - 11*a^3*b^5 + 27*a^4*b^ 4 + 15*a^5*b^3)/(32*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b ^3 + 15*a^4*b^2)) + (((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a...