\(\int (a+b \sec ^2(e+f x))^2 \sin ^5(e+f x) \, dx\) [15]

Optimal result

Integrand size = 23, antiderivative size = 97 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=-\frac {\left (a^2-4 a b+b^2\right ) \cos (e+f x)}{f}+\frac {2 a (a-b) \cos ^3(e+f x)}{3 f}-\frac {a^2 \cos ^5(e+f x)}{5 f}+\frac {2 (a-b) b \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4218, 459} \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=-\frac {\left (a^2-4 a b+b^2\right ) \cos (e+f x)}{f}-\frac {a^2 \cos ^5(e+f x)}{5 f}+\frac {2 a (a-b) \cos ^3(e+f x)}{3 f}+\frac {2 b (a-b) \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 4218

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p
)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]

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

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=-\frac {\left (425 a^2-4400 a b+2000 b^2+24 \left (22 a^2-215 a b+120 b^2\right ) \cos (2 (e+f x))+12 \left (7 a^2-60 a b+20 b^2\right ) \cos (4 (e+f x))-16 a^2 \cos (6 (e+f x))+40 a b \cos (6 (e+f x))+3 a^2 \cos (8 (e+f x))\right ) \sec ^3(e+f x)}{1920 f} \]

[In]

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

[Out]

-1/1920*((425*a^2 - 4400*a*b + 2000*b^2 + 24*(22*a^2 - 215*a*b + 120*b^2)*Cos[2*(e + f*x)] + 12*(7*a^2 - 60*a*
b + 20*b^2)*Cos[4*(e + f*x)] - 16*a^2*Cos[6*(e + f*x)] + 40*a*b*Cos[6*(e + f*x)] + 3*a^2*Cos[8*(e + f*x)])*Sec
[e + f*x]^3)/f

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{6}}{3 \cos \left (f x +e \right )^{3}}-\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}-\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\)	\(155\)
default	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{6}}{3 \cos \left (f x +e \right )^{3}}-\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}-\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\)	\(155\)
parts	\(-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {b^{2} \left (\frac {\sin \left (f x +e \right )^{6}}{3 \cos \left (f x +e \right )^{3}}-\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}-\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}+\frac {2 a b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\)	\(160\)
parallelrisch	\(\frac {\left (-256 a^{2}+2560 a b -1280 b^{2}\right ) \cos \left (3 f x +3 e \right )+\left (-528 a^{2}+5160 a b -2880 b^{2}\right ) \cos \left (2 f x +2 e \right )+\left (-84 a^{2}+720 a b -240 b^{2}\right ) \cos \left (4 f x +4 e \right )+\left (16 a^{2}-40 a b \right ) \cos \left (6 f x +6 e \right )-3 a^{2} \cos \left (8 f x +8 e \right )+\left (-768 a^{2}+7680 a b -3840 b^{2}\right ) \cos \left (f x +e \right )-425 a^{2}+4400 a b -2000 b^{2}}{480 f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\)	\(172\)
norman	\(\frac {\frac {16 a^{2}-160 a b +80 b^{2}}{15 f}+\frac {16 \left (5 a^{2}-2 a b -7 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 f}+\frac {2 \left (16 a^{2}-160 a b +80 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 f}-\frac {2 \left (16 a^{2}-160 a b +80 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 f}-\frac {2 \left (16 a^{2}+32 a b +16 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{3 f}-\frac {2 \left (128 a^{2}-320 a b +320 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5}}\)	\(208\)
risch	\(-\frac {a^{2} {\mathrm e}^{5 i \left (f x +e \right )}}{160 f}+\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )} a^{2}}{96 f}-\frac {{\mathrm e}^{3 i \left (f x +e \right )} a b}{12 f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a^{2}}{16 f}+\frac {7 \,{\mathrm e}^{i \left (f x +e \right )} a b}{4 f}-\frac {{\mathrm e}^{i \left (f x +e \right )} b^{2}}{2 f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{16 f}+\frac {7 \,{\mathrm e}^{-i \left (f x +e \right )} a b}{4 f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{2 f}+\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )} a^{2}}{96 f}-\frac {{\mathrm e}^{-3 i \left (f x +e \right )} a b}{12 f}-\frac {a^{2} {\mathrm e}^{-5 i \left (f x +e \right )}}{160 f}-\frac {4 \,{\mathrm e}^{i \left (f x +e \right )} b \left (-3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+3 b \,{\mathrm e}^{4 i \left (f x +e \right )}-6 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}-3 a +3 b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\)	\(313\)

[In]

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

[Out]

1/f*(-1/5*a^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*a*b*(sin(f*x+e)^6/cos(f*x+e)+(8/3+sin(f*x+e)^4+
4/3*sin(f*x+e)^2)*cos(f*x+e))+b^2*(1/3*sin(f*x+e)^6/cos(f*x+e)^3-sin(f*x+e)^6/cos(f*x+e)-(8/3+sin(f*x+e)^4+4/3
*sin(f*x+e)^2)*cos(f*x+e)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=-\frac {3 \, a^{2} \cos \left (f x + e\right )^{8} - 10 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{6} + 15 \, {\left (a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 30 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 5 \, b^{2}}{15 \, f \cos \left (f x + e\right )^{3}} \]

[In]

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

[Out]

-1/15*(3*a^2*cos(f*x + e)^8 - 10*(a^2 - a*b)*cos(f*x + e)^6 + 15*(a^2 - 4*a*b + b^2)*cos(f*x + e)^4 - 30*(a*b
- b^2)*cos(f*x + e)^2 - 5*b^2)/(f*cos(f*x + e)^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=-\frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right ) - \frac {5 \, {\left (6 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{15 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (91) = 182\).

Time = 0.41 (sec) , antiderivative size = 414, normalized size of antiderivative = 4.27 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=\frac {2 \, {\left (\frac {5 \, {\left (6 \, a b - 5 \, b^{2} + \frac {12 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {12 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {6 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {3 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}} + \frac {8 \, a^{2} - 50 \, a b + 15 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {220 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {60 \, b^{2} {\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 {320 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {90 \, 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 {60 \, b^{2} {\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}}}{{\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}\right )}}{15 \, f} \]

[In]

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

[Out]

2/15*(5*(6*a*b - 5*b^2 + 12*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 12*b^2*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + 6*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/((
cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)^3 + (8*a^2 - 50*a*b + 15*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + 220*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 60*b^2*(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 - 320*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 90*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 - 60*b^2*(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)/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5)/f

Mupad [B] (verification not implemented)

Time = 18.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^5(e+f x) \, dx=\frac {\frac {\frac {b^2}{3}+{\cos \left (e+f\,x\right )}^2\,\left (2\,a\,b-2\,b^2\right )}{{\cos \left (e+f\,x\right )}^3}-\cos \left (e+f\,x\right )\,\left (a^2-4\,a\,b+b^2\right )-\frac {a^2\,{\cos \left (e+f\,x\right )}^5}{5}+\frac {2\,a\,{\cos \left (e+f\,x\right )}^3\,\left (a-b\right )}{3}}{f} \]

[In]

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

[Out]

((b^2/3 + cos(e + f*x)^2*(2*a*b - 2*b^2))/cos(e + f*x)^3 - cos(e + f*x)*(a^2 - 4*a*b + b^2) - (a^2*cos(e + f*x
)^5)/5 + (2*a*cos(e + f*x)^3*(a - b))/3)/f