\(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 219 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

-1/5*(5*a^2+20*a*b+16*b^2)*cos(f*x+e)/a^3/f/(a+b*sec(f*x+e)^2)^(3/2)+2/15*(5*a+4*b)*cos(f*x+e)^3/a^2/f/(a+b*se
c(f*x+e)^2)^(3/2)-1/5*cos(f*x+e)^5/a/f/(a+b*sec(f*x+e)^2)^(3/2)-4/15*b*(5*a^2+20*a*b+16*b^2)*sec(f*x+e)/a^4/f/
(a+b*sec(f*x+e)^2)^(3/2)-8/15*b*(5*a^2+20*a*b+16*b^2)*sec(f*x+e)/a^5/f/(a+b*sec(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4219, 473, 464, 277, 198, 197} \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (\frac {4 b (5 a+4 b)}{a^2}+5\right ) \cos (e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-1/5*((5 + (4*b*(5*a + 4*b))/a^2)*Cos[e + f*x])/(a*f*(a + b*Sec[e + f*x]^2)^(3/2)) + (2*(5*a + 4*b)*Cos[e + f*
x]^3)/(15*a^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) - Cos[e + f*x]^5/(5*a*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (4*b*(5*
a^2 + 20*a*b + 16*b^2)*Sec[e + f*x])/(15*a^4*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (8*b*(5*a^2 + 20*a*b + 16*b^2)*
Sec[e + f*x])/(15*a^5*f*Sqrt[a + 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 198

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

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 4219

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^
n)^p/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {-2 (5 a+4 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f} \\ & = \frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\left (5 a^2+20 a b+16 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^2 f} \\ & = -\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (4 b \left (5 a^2+20 a b+16 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^3 f} \\ & = -\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (8 b \left (5 a^2+20 a b+16 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f} \\ & = -\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.11 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (425 a^4+6400 a^3 b+22784 a^2 b^2+32768 a b^3+16384 b^4+48 a \left (11 a^3+150 a^2 b+384 a b^2+256 b^3\right ) \cos (2 (e+f x))+12 a^2 \left (7 a^2+64 a b+64 b^2\right ) \cos (4 (e+f x))-16 a^4 \cos (6 (e+f x))-32 a^3 b \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))\right ) \sec ^5(e+f x)}{3840 a^5 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

[In]

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

[Out]

-1/3840*((a + 2*b + a*Cos[2*(e + f*x)])*(425*a^4 + 6400*a^3*b + 22784*a^2*b^2 + 32768*a*b^3 + 16384*b^4 + 48*a
*(11*a^3 + 150*a^2*b + 384*a*b^2 + 256*b^3)*Cos[2*(e + f*x)] + 12*a^2*(7*a^2 + 64*a*b + 64*b^2)*Cos[4*(e + f*x
)] - 16*a^4*Cos[6*(e + f*x)] - 32*a^3*b*Cos[6*(e + f*x)] + 3*a^4*Cos[8*(e + f*x)])*Sec[e + f*x]^5)/(a^5*f*(a +
b*Sec[e + f*x]^2)^(5/2))

Maple [A] (verified)

Time = 5.67 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (a +b \right )^{7} \left (3 \cos \left (f x +e \right )^{8} a^{4}-10 a^{4} \cos \left (f x +e \right )^{6}-8 a^{3} b \cos \left (f x +e \right )^{6}+15 a^{4} \cos \left (f x +e \right )^{4}+60 \cos \left (f x +e \right )^{4} a^{3} b +48 a^{2} b^{2} \cos \left (f x +e \right )^{4}+60 \cos \left (f x +e \right )^{2} a^{3} b +240 a^{2} b^{2} \cos \left (f x +e \right )^{2}+192 a \,b^{3} \cos \left (f x +e \right )^{2}+40 a^{2} b^{2}+160 a \,b^{3}+128 b^{4}\right ) \sec \left (f x +e \right )^{5}}{15 f \left (\sqrt {-a b}+a \right )^{7} \left (\sqrt {-a b}-a \right )^{7} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(217\)

[In]

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

[Out]

1/15/f*a^2/((-a*b)^(1/2)+a)^7/((-a*b)^(1/2)-a)^7*(b+a*cos(f*x+e)^2)*(a+b)^7*(3*cos(f*x+e)^8*a^4-10*a^4*cos(f*x
+e)^6-8*a^3*b*cos(f*x+e)^6+15*a^4*cos(f*x+e)^4+60*cos(f*x+e)^4*a^3*b+48*a^2*b^2*cos(f*x+e)^4+60*cos(f*x+e)^2*a
^3*b+240*a^2*b^2*cos(f*x+e)^2+192*a*b^3*cos(f*x+e)^2+40*a^2*b^2+160*a*b^3+128*b^4)/(a+b*sec(f*x+e)^2)^(5/2)*se
c(f*x+e)^5

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, a^{4} \cos \left (f x + e\right )^{9} - 2 \, {\left (5 \, a^{4} + 4 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}} \]

[In]

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

[Out]

-1/15*(3*a^4*cos(f*x + e)^9 - 2*(5*a^4 + 4*a^3*b)*cos(f*x + e)^7 + 3*(5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x +
e)^5 + 12*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^3 + 8*(5*a^2*b^2 + 20*a*b^3 + 16*b^4)*cos(f*x + e))*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4}} + \frac {3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 20 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5}} + \frac {5 \, {\left (6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac {10 \, {\left (9 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{4} \cos \left (f x + e\right )^{3}} + \frac {5 \, {\left (12 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{5} \cos \left (f x + e\right )^{3}}}{15 \, f} \]

[In]

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

[Out]

-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 - 10*((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 9*sq
rt(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^4 + (3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 20*(a + b/cos(
f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 90*sqrt(a + b/cos(f*x + e)^2)*b^2*cos(f*x + e))/a^5 + 5*(6*(a + b/cos(f*x
+ e)^2)*b*cos(f*x + e)^2 - b^2)/((a + b/cos(f*x + e)^2)^(3/2)*a^3*cos(f*x + e)^3) + 10*(9*(a + b/cos(f*x + e)
^2)*b^2*cos(f*x + e)^2 - b^3)/((a + b/cos(f*x + e)^2)^(3/2)*a^4*cos(f*x + e)^3) + 5*(12*(a + b/cos(f*x + e)^2)
*b^3*cos(f*x + e)^2 - b^4)/((a + b/cos(f*x + e)^2)^(3/2)*a^5*cos(f*x + e)^3))/f

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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