\(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [106]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (15 a^2+40 a b+24 b^2\right ) \cos (e+f x)}{15 a^3 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{15 a^4 f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

-1/15*(15*a^2+40*a*b+24*b^2)*cos(f*x+e)/a^3/f/(a+b*sec(f*x+e)^2)^(1/2)+2/15*(5*a+3*b)*cos(f*x+e)^3/a^2/f/(a+b*
sec(f*x+e)^2)^(1/2)-1/5*cos(f*x+e)^5/a/f/(a+b*sec(f*x+e)^2)^(1/2)-2/15*b*(15*a^2+40*a*b+24*b^2)*sec(f*x+e)/a^4
/f/(a+b*sec(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 171, 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 = {4219, 473, 464, 277, 197} \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\left (\frac {8 b (5 a+3 b)}{a^2}+15\right ) \cos (e+f x)}{15 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{15 a^4 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-1/15*((15 + (8*b*(5*a + 3*b))/a^2)*Cos[e + f*x])/(a*f*Sqrt[a + b*Sec[e + f*x]^2]) + (2*(5*a + 3*b)*Cos[e + f*
x]^3)/(15*a^2*f*Sqrt[a + b*Sec[e + f*x]^2]) - Cos[e + f*x]^5/(5*a*f*Sqrt[a + b*Sec[e + f*x]^2]) - (2*b*(15*a^2
+ 40*a*b + 24*b^2)*Sec[e + f*x])/(15*a^4*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 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 )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {-2 (5 a+3 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{5 a f} \\ & = \frac {2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\left (-15 a^2-8 b (5 a+3 b)\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^2 f} \\ & = -\frac {\left (15 a^2+8 b (5 a+3 b)\right ) \cos (e+f x)}{15 a^3 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (2 b \left (-15 a^2-8 b (5 a+3 b)\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^3 f} \\ & = -\frac {\left (15 a^2+8 b (5 a+3 b)\right ) \cos (e+f x)}{15 a^3 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+8 b (5 a+3 b)\right ) \sec (e+f x)}{15 a^4 f \sqrt {a+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.91 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (150 a^3+1528 a^2 b+2944 a b^2+1536 b^3+a \left (125 a^2+544 a b+384 b^2\right ) \cos (2 (e+f x))-2 a^2 (11 a+12 b) \cos (4 (e+f x))+3 a^3 \cos (6 (e+f x))\right ) \sec ^3(e+f x)}{960 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-1/960*((a + 2*b + a*Cos[2*(e + f*x)])*(150*a^3 + 1528*a^2*b + 2944*a*b^2 + 1536*b^3 + a*(125*a^2 + 544*a*b +
384*b^2)*Cos[2*(e + f*x)] - 2*a^2*(11*a + 12*b)*Cos[4*(e + f*x)] + 3*a^3*Cos[6*(e + f*x)])*Sec[e + f*x]^3)/(a^
4*f*(a + b*Sec[e + f*x]^2)^(3/2))

Maple [A] (verified)

Time = 5.04 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {\left (a +b \right )^{6} a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (3 a^{3} \cos \left (f x +e \right )^{6}-10 \cos \left (f x +e \right )^{4} a^{3}-6 \cos \left (f x +e \right )^{4} a^{2} b +15 \cos \left (f x +e \right )^{2} a^{3}+40 \cos \left (f x +e \right )^{2} a^{2} b +24 \cos \left (f x +e \right )^{2} a \,b^{2}+30 a^{2} b +80 a \,b^{2}+48 b^{3}\right ) \sec \left (f x +e \right )^{3}}{15 f \left (\sqrt {-a b}-a \right )^{6} \left (\sqrt {-a b}+a \right )^{6} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(169\)

[In]

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

[Out]

-1/15/f*(a+b)^6*a^2/((-a*b)^(1/2)-a)^6/((-a*b)^(1/2)+a)^6*(b+a*cos(f*x+e)^2)*(3*a^3*cos(f*x+e)^6-10*cos(f*x+e)
^4*a^3-6*cos(f*x+e)^4*a^2*b+15*cos(f*x+e)^2*a^3+40*cos(f*x+e)^2*a^2*b+24*cos(f*x+e)^2*a*b^2+30*a^2*b+80*a*b^2+
48*b^3)/(a+b*sec(f*x+e)^2)^(3/2)*sec(f*x+e)^3

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (3 \, a^{3} \cos \left (f x + e\right )^{7} - 2 \, {\left (5 \, a^{3} + 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} + 40 \, a^{2} b + 24 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b + 40 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}} \]

[In]

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

[Out]

-1/15*(3*a^3*cos(f*x + e)^7 - 2*(5*a^3 + 3*a^2*b)*cos(f*x + e)^5 + (15*a^3 + 40*a^2*b + 24*a*b^2)*cos(f*x + e)
^3 + 2*(15*a^2*b + 40*a*b^2 + 24*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^5*f*cos(f*x
+ e)^2 + a^4*b*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.46 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{3}} + \frac {15 \, b}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )} + \frac {30 \, b^{2}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{3} \cos \left (f x + e\right )} + \frac {15 \, b^{3}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{4} \cos \left (f x + e\right )} + \frac {3 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )\right )}}{a^{4}}}{15 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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