\(\int \frac {\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [68]

Optimal result

Integrand size = 23, antiderivative size = 204 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{9/2} f}-\frac {\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 (a-b)^4 f}+\frac {(10 a-3 b) \cos ^3(e+f x)}{15 (a-b)^3 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )} \]

[Out]

-1/5*(5*a^2+10*a*b-b^2)*cos(f*x+e)/(a-b)^4/f+1/15*(10*a-3*b)*cos(f*x+e)^3/(a-b)^3/f-1/5*cos(f*x+e)^5/(a-b)/f/(
a-b+b*sec(f*x+e)^2)-1/10*b*(5*a^2+2*b^2)*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)-1/2*a*(3*a+4*b)*arctan(sec(
f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(9/2)/f

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3745, 473, 467, 1275, 211} \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 f (a-b)^4}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 f (a-b)^{9/2}}+\frac {(10 a-3 b) \cos ^3(e+f x)}{15 f (a-b)^3}-\frac {\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )} \]

[In]

Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-1/2*(a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/((a - b)^(9/2)*f) - ((5*a^2 + 10*a*b -
b^2)*Cos[e + f*x])/(5*(a - b)^4*f) + ((10*a - 3*b)*Cos[e + f*x]^3)/(15*(a - b)^3*f) - Cos[e + f*x]^5/(5*(a -
b)*f*(a - b + b*Sec[e + f*x]^2)) - (b*(5*a^2 + 2*b^2)*Sec[e + f*x])/(10*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2)
)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 467

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] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[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])

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[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]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {-10 a+3 b+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f} \\ & = -\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \frac {\frac {2 (10 a-3 b)}{(a-b) b}-\frac {2 \left (5 a^2+2 b^2\right ) x^2}{(a-b)^2 b}+\frac {\left (5 a^2+2 b^2\right ) x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{10 (a-b) f} \\ & = -\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \left (\frac {2 (10 a-3 b)}{(a-b)^2 b x^4}-\frac {2 \left (5 a^2+10 a b-b^2\right )}{(a-b)^3 b x^2}+\frac {5 a (3 a+4 b)}{(a-b)^3 \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{10 (a-b) f} \\ & = -\frac {\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 (a-b)^4 f}+\frac {(10 a-3 b) \cos ^3(e+f x)}{15 (a-b)^3 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {(a b (3 a+4 b)) \text {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 (a-b)^4 f} \\ & = -\frac {a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{9/2} f}-\frac {\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 (a-b)^4 f}+\frac {(10 a-3 b) \cos ^3(e+f x)}{15 (a-b)^3 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.48 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {120 a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {120 a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {-30 \cos (e+f x) \left (18 a b+b^2+a^2 \left (5+\frac {8 b}{a+b+(a-b) \cos (2 (e+f x))}\right )\right )+(a-b) (5 (5 a+3 b) \cos (3 (e+f x))+3 (-a+b) \cos (5 (e+f x)))}{(a-b)^4}}{240 f} \]

[In]

Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

((120*a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (120*a*S
qrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (-30*Cos[e + f*x]
*(18*a*b + b^2 + a^2*(5 + (8*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 3*b)*Cos[3*(e + f*x)]
+ 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^4)/(240*f)

Maple [A] (verified)

Time = 30.60 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {2 a b \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )+2 a b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {a b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\)	\(197\)
default	\(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {2 a b \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )+2 a b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {a b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\)	\(197\)
risch	\(-\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )} a}{96 \left (-a +b \right )^{3} f}-\frac {{\mathrm e}^{3 i \left (f x +e \right )} b}{32 \left (-a +b \right )^{3} f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a^{2}}{16 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {9 \,{\mathrm e}^{i \left (f x +e \right )} a b}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {{\mathrm e}^{i \left (f x +e \right )} b^{2}}{16 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{16 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {9 \,{\mathrm e}^{-i \left (f x +e \right )} a b}{8 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{16 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )} a}{96 \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 )} b}{32 \left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) f}-\frac {b \,a^{2} \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (a^{2}-2 a b +b^{2}\right )^{2} f \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 )}\, 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 )}{4 \left (a -b \right )^{5} 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 ) b}{\left (a -b \right )^{5} f}+\frac {3 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 )}{4 \left (a -b \right )^{5} 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 ) b}{\left (a -b \right )^{5} f}-\frac {\cos \left (5 f x +5 e \right )}{80 f \left (a -b \right )^{2}}\)	\(743\)

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(-1/(a^2-2*a*b+b^2)/(a-b)^2*(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+2/3*a*b*cos(f*x+e)^3+a^2*cos(f*x+e)+2*a*b*cos(f*x+e))+a*b/(a-b)^4*(-1/2*a*cos(f*x+e)/(a*cos(f*x+e)^2-b*
cos(f*x+e)^2+b)+1/2*(3*a+4*b)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.91 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [-\frac {12 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2} + {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{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 ) + 30 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{60 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac {6 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2} + {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{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 ) + 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{30 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/60*(12*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 4*(10*a^3 - 23*a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x +
e)^5 + 20*(3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*
x + e)^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)) + 30*(3*a^2*b + 4*a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b
^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f), -1/30*(6*(a^3 - 3*a^2*b + 3*a
*b^2 - b^3)*cos(f*x + e)^7 - 2*(10*a^3 - 23*a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x + e)^5 + 10*(3*a^3 + a^2*b - 4*a
*b^2)*cos(f*x + e)^3 + 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt(b/(a - b))*arcta
n(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(3*a^2*b + 4*a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2
- 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (186) = 372\).

Time = 0.68 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.60 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {15 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \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^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b - b^{2}}} + \frac {30 \, {\left (a^{2} b + \frac {a^{2} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}} - \frac {4 \, {\left (8 \, a^{2} + 34 \, a b + 3 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {140 \, a b {\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 {160 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {180 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {30 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {15 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{30 \, f} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*(3*a^2*b + 4*a*b^2)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sq
rt(a*b - b^2)))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b - b^2)) + 30*(a^2*b + a^2*b*(cos(f*x + e
) - 1)/(cos(f*x + e) + 1) - 2*a*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b
^3 + b^4)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(
f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) - 4*(8*a^2 + 34*a*b + 3*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) +
1) - 140*a*b*(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 + 160*a
*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 30*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 180*a*b*(cos
(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*a*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 15*b^2*(cos(f*x + e
) - 1)^4/(cos(f*x + e) + 1)^4)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*((cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) - 1)^5))/f

Mupad [B] (verification not implemented)

Time = 14.67 (sec) , antiderivative size = 1049, normalized size of antiderivative = 5.14 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {16\,a^3+83\,a^2\,b+6\,a\,b^2}{15\,\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (32\,a^3-83\,a^2\,b+366\,a\,b^2\right )}{3\,\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (16\,a^3+223\,a^2\,b+1336\,a\,b^2\right )}{15\,\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,\left (6\,a^2\,b+11\,a\,b^2+4\,b^3\right )}{\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (-12\,a^3+32\,a^2\,b+73\,a\,b^2+12\,b^3\right )}{3\,\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (24\,a^3+134\,a^2\,b+145\,a\,b^2+12\,b^3\right )}{15\,\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}\,\left (4\,b^2+3\,a\,b\right )}{\left (a-b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}+\left (3\,a+4\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+\left (a+20\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+\left (40\,b-5\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+\left (40\,b-5\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (a+20\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (3\,a+4\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\right )}-\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {{\left (a-b\right )}^9\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (3\,a+4\,b\right )\,\left (24\,a^{12}\,b-160\,a^{11}\,b^2+416\,a^{10}\,b^3-448\,a^9\,b^4-112\,a^8\,b^5+896\,a^7\,b^6-1120\,a^6\,b^7+704\,a^5\,b^8-232\,a^4\,b^9+32\,a^3\,b^{10}\right )}{4\,{\left (a-b\right )}^{17/2}}-\frac {a\,\sqrt {b}\,\left (a-2\,b\right )\,{\left (3\,a+4\,b\right )}^2\,\left (-16\,a^{15}+224\,a^{14}\,b-1440\,a^{13}\,b^2+5632\,a^{12}\,b^3-14960\,a^{11}\,b^4+28512\,a^{10}\,b^5-40128\,a^9\,b^6+42240\,a^8\,b^7-33264\,a^7\,b^8+19360\,a^6\,b^9-8096\,a^5\,b^{10}+2304\,a^4\,b^{11}-400\,a^3\,b^{12}+32\,a^2\,b^{13}\right )}{32\,{\left (a-b\right )}^{27/2}}\right )-\frac {a\,\sqrt {b}\,\left (a-2\,b\right )\,{\left (3\,a+4\,b\right )}^2\,\left (16\,a^{15}-192\,a^{14}\,b+1056\,a^{13}\,b^2-3520\,a^{12}\,b^3+7920\,a^{11}\,b^4-12672\,a^{10}\,b^5+14784\,a^9\,b^6-12672\,a^8\,b^7+7920\,a^7\,b^8-3520\,a^6\,b^9+1056\,a^5\,b^{10}-192\,a^4\,b^{11}+16\,a^3\,b^{12}\right )}{32\,{\left (a-b\right )}^{27/2}}\right )}{9\,a^{14}\,b-48\,a^{13}\,b^2+76\,a^{12}\,b^3+40\,a^{11}\,b^4-266\,a^{10}\,b^5+280\,a^9\,b^6+28\,a^8\,b^7-296\,a^7\,b^8+265\,a^6\,b^9-104\,a^5\,b^{10}+16\,a^4\,b^{11}}\right )\,\left (3\,a+4\,b\right )}{2\,f\,{\left (a-b\right )}^{9/2}} \]

[In]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^2,x)

[Out]

- ((6*a*b^2 + 83*a^2*b + 16*a^3)/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^8*(366*a*b
^2 - 83*a^2*b + 32*a^3))/(3*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^4*(1336*a*b^2 + 223
*a^2*b + 16*a^3))/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (2*tan(e/2 + (f*x)/2)^10*(11*a*b^2 + 6*a^2*b
+ 4*b^3))/((a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (4*tan(e/2 + (f*x)/2)^6*(73*a*b^2 + 32*a^2*b - 12*a^3 +
12*b^3))/(3*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (2*tan(e/2 + (f*x)/2)^2*(145*a*b^2 + 134*a^2*b + 24*a^3
+ 12*b^3))/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (a*tan(e/2 + (f*x)/2)^12*(3*a*b + 4*b^2))/((a - b)*
(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(f*(a + tan(e/2 + (f*x)/2)^4*(a + 20*b) + tan(e/2 + (f*x)/2)^10*(a + 20*b) +
tan(e/2 + (f*x)/2)^2*(3*a + 4*b) + tan(e/2 + (f*x)/2)^12*(3*a + 4*b) - tan(e/2 + (f*x)/2)^6*(5*a - 40*b) - ta
n(e/2 + (f*x)/2)^8*(5*a - 40*b) + a*tan(e/2 + (f*x)/2)^14)) - (a*b^(1/2)*atan(((a - b)^9*(tan(e/2 + (f*x)/2)^2
*((b^(1/2)*(3*a + 4*b)*(24*a^12*b + 32*a^3*b^10 - 232*a^4*b^9 + 704*a^5*b^8 - 1120*a^6*b^7 + 896*a^7*b^6 - 112
*a^8*b^5 - 448*a^9*b^4 + 416*a^10*b^3 - 160*a^11*b^2))/(4*(a - b)^(17/2)) - (a*b^(1/2)*(a - 2*b)*(3*a + 4*b)^2
*(224*a^14*b - 16*a^15 + 32*a^2*b^13 - 400*a^3*b^12 + 2304*a^4*b^11 - 8096*a^5*b^10 + 19360*a^6*b^9 - 33264*a^
7*b^8 + 42240*a^8*b^7 - 40128*a^9*b^6 + 28512*a^10*b^5 - 14960*a^11*b^4 + 5632*a^12*b^3 - 1440*a^13*b^2))/(32*
(a - b)^(27/2))) - (a*b^(1/2)*(a - 2*b)*(3*a + 4*b)^2*(16*a^15 - 192*a^14*b + 16*a^3*b^12 - 192*a^4*b^11 + 105
6*a^5*b^10 - 3520*a^6*b^9 + 7920*a^7*b^8 - 12672*a^8*b^7 + 14784*a^9*b^6 - 12672*a^10*b^5 + 7920*a^11*b^4 - 35
20*a^12*b^3 + 1056*a^13*b^2))/(32*(a - b)^(27/2))))/(9*a^14*b + 16*a^4*b^11 - 104*a^5*b^10 + 265*a^6*b^9 - 296
*a^7*b^8 + 28*a^8*b^7 + 280*a^9*b^6 - 266*a^10*b^5 + 40*a^11*b^4 + 76*a^12*b^3 - 48*a^13*b^2))*(3*a + 4*b))/(2
*f*(a - b)^(9/2))