\(\int \frac {\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [128]

Optimal result

Integrand size = 25, antiderivative size = 199 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3745, 473, 464, 277, 197} \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 f (a-b)^3 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\cos ^5(e+f x)}{5 f (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 f (a-b)^2 \sqrt {a+b \sec ^2(e+f x)-b}} \]

[In]

Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 464

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

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

Mathematica [A] (verified)

Time = 3.63 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (150 a^3+1078 a^2 b+338 a b^2-30 b^3+\left (125 a^3+169 a^2 b-329 a b^2+35 b^3\right ) \cos (2 (e+f x))-2 (a-b)^2 (11 a+b) \cos (4 (e+f x))+3 a^3 \cos (6 (e+f x))-9 a^2 b \cos (6 (e+f x))+9 a b^2 \cos (6 (e+f x))-3 b^3 \cos (6 (e+f x))\right ) \sec (e+f x)}{240 \sqrt {2} (a-b)^4 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]

[In]

Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-1/240*((150*a^3 + 1078*a^2*b + 338*a*b^2 - 30*b^3 + (125*a^3 + 169*a^2*b - 329*a*b^2 + 35*b^3)*Cos[2*(e + f*x
)] - 2*(a - b)^2*(11*a + b)*Cos[4*(e + f*x)] + 3*a^3*Cos[6*(e + f*x)] - 9*a^2*b*Cos[6*(e + f*x)] + 9*a*b^2*Cos
[6*(e + f*x)] - 3*b^3*Cos[6*(e + f*x)])*Sec[e + f*x])/(Sqrt[2]*(a - b)^4*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*
x)])*Sec[e + f*x]^2])

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.37

method	result	size
default	\(\frac {a^{6} \left (a -b \right )^{2} \left (3 \sin \left (f x +e \right )^{4} \cos \left (f x +e \right )^{2} b^{3}+9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{4} a \,b^{2}-3 a^{3} \cos \left (f x +e \right )^{6}+9 a^{2} b \cos \left (f x +e \right )^{6}+2 \sin \left (f x +e \right )^{4} b^{3}-9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} a \,b^{2}+10 a^{3} \cos \left (f x +e \right )^{4}-24 a^{2} b \cos \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{2} a \,b^{2}-15 a^{3} \cos \left (f x +e \right )^{2}+5 a^{2} b \cos \left (f x +e \right )^{2}-30 a^{2} b \right ) \left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \sec \left (f x +e \right )^{3}}{15 f \left (\sqrt {-b \left (a -b \right )}+a -b \right )^{6} \left (\sqrt {-b \left (a -b \right )}-a +b \right )^{6} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(273\)

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/15/f*a^6*(a-b)^2/((-b*(a-b))^(1/2)+a-b)^6/((-b*(a-b))^(1/2)-a+b)^6*(3*sin(f*x+e)^4*cos(f*x+e)^2*b^3+9*sin(f*
x+e)^2*cos(f*x+e)^4*a*b^2-3*a^3*cos(f*x+e)^6+9*a^2*b*cos(f*x+e)^6+2*sin(f*x+e)^4*b^3-9*sin(f*x+e)^2*cos(f*x+e)
^2*a*b^2+10*a^3*cos(f*x+e)^4-24*a^2*b*cos(f*x+e)^4-20*sin(f*x+e)^2*a*b^2-15*a^3*cos(f*x+e)^2+5*a^2*b*cos(f*x+e
)^2-30*a^2*b)*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(a+b*tan(f*x+e)^2)^(3/2)*sec(f*x+e)^3

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 9 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 5 \, a^{2} b - 11 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b + 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 2*(5*a^3 - 12*a^2*b + 9*a*b^2 - 2*b^3)*cos(f*x + e)^
5 + (15*a^3 - 5*a^2*b - 11*a*b^2 + b^3)*cos(f*x + e)^3 + 2*(15*a^2*b + 10*a*b^2 - b^3)*cos(f*x + e))*sqrt(((a
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((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. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (183) = 366\).

Time = 0.26 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.95 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, b^{3}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {3 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {30 \, b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {15 \, b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}}{15 \, f} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*b^3/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)) + 15*s
qrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^2 - 2*a*b + b^2) + 3*((a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x +
e)^5 - 5*(a - b + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 15*sqrt(a - b + b/cos(f*x + e)^2)*b^2*cos(f*x +
e))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - 10*((a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 6*sqrt
(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 30*b^2/((a^3 - 3*a^2*b + 3*a*b^2
- b^3)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)) + 15*b/((a^2 - 2*a*b + b^2)*sqrt(a - b + b/cos(f*x + e)^2)
*cos(f*x + e)))/f

Giac [F]

\[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^(3/2), x)