\(\int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [116]

Optimal result

Integrand size = 25, antiderivative size = 144 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^3 f}+\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{5 (a-b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3745, 473, 464, 270} \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{15 f (a-b)^3}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{5 f (a-b)}+\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{15 f (a-b)^2} \]

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(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 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{5 (a-b) f}+\frac {\text {Subst}\left (\int \frac {-2 (5 a-3 b)+5 (a-b) x^2}{x^4 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f} \\ & = \frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{5 (a-b) f}+\frac {\left (15 a^2-10 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f} \\ & = -\frac {\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^3 f}+\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{5 (a-b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cos (e+f x) \left (-89 a^2+34 a b-9 b^2+4 \left (7 a^2-10 a b+3 b^2\right ) \cos (2 (e+f x))-3 (a-b)^2 \cos (4 (e+f x))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{120 \sqrt {2} (a-b)^3 f} \]

[In]

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

[Out]

(Cos[e + f*x]*(-89*a^2 + 34*a*b - 9*b^2 + 4*(7*a^2 - 10*a*b + 3*b^2)*Cos[2*(e + f*x)] - 3*(a - b)^2*Cos[4*(e +
f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(120*Sqrt[2]*(a - b)^3*f)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (3 \sin \left (f x +e \right )^{4} b^{2}+6 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a b +3 a^{2} \cos \left (f x +e \right )^{4}-10 a b \sin \left (f x +e \right )^{2}-10 \cos \left (f x +e \right )^{2} a^{2}+15 a^{2}\right ) \sec \left (f x +e \right )}{15 f \left (a -b \right )^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(131\)

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {{\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (5 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 2*(5*a^2 - 8*a*b + 3*b^2)*cos(f*x + e)^3 + (15*a^2 - 10*a*b + 3*
b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)

Sympy [F(-1)]

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

[In]

integrate(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a - b} + \frac {3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 10 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2} - 2 \, a b + b^{2}}}{15 \, f} \]

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^(1/2), x)