3.1.28 \(\int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [28]

3.1.28.1 Optimal result

Integrand size = 23, antiderivative size = 98 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} (a+b)^2 \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {\cos ^5(e+f x)}{5 a f} \]

output

-(a+b)^2*cos(f*x+e)/a^3/f+1/3*(2*a+b)*cos(f*x+e)^3/a^2/f-1/5*cos(f*x+e)^5/ 
a/f+(a+b)^2*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(7/2)/f
 

3.1.28.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.85 (sec) , antiderivative size = 425, normalized size of antiderivative = 4.34 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \left (15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )-75 a^3 \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )-75 a^3 \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )-8 \sqrt {a} \sqrt {b} \cos (e+f x) \left (89 a^2+220 a b+120 b^2-4 a (7 a+5 b) \cos (2 (e+f x))+3 a^2 \cos (4 (e+f x))\right )\right ) \sec ^2(e+f x)}{1920 a^{7/2} \sqrt {b} f \left (a+b \sec ^2(e+f x)\right )} \]

input

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

output

((a + 2*b + a*Cos[2*(e + f*x)])*(15*(5*a^3 + 64*a^2*b + 128*a*b^2 + 64*b^3 
)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Ta 
n[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan 
[(f*x)/2]))/Sqrt[b]] + 15*(5*a^3 + 64*a^2*b + 128*a*b^2 + 64*b^3)*ArcTan[( 
(-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] 
 + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]) 
)/Sqrt[b]] - 75*a^3*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b 
]] - 75*a^3*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]] - 8*S 
qrt[a]*Sqrt[b]*Cos[e + f*x]*(89*a^2 + 220*a*b + 120*b^2 - 4*a*(7*a + 5*b)* 
Cos[2*(e + f*x)] + 3*a^2*Cos[4*(e + f*x)]))*Sec[e + f*x]^2)/(1920*a^(7/2)* 
Sqrt[b]*f*(a + b*Sec[e + f*x]^2))
 

3.1.28.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4621, 364, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^5}{a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4621

\(\displaystyle -\frac {\int \frac {\cos ^2(e+f x) \left (1-\cos ^2(e+f x)\right )^2}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 364

\(\displaystyle -\frac {\int \left (\frac {\cos ^4(e+f x)}{a}-\frac {(2 a+b) \cos ^2(e+f x)}{a^2}+\frac {(a+b)^2}{a^3}+\frac {-b^3-2 a b^2-a^2 b}{a^3 \left (a \cos ^2(e+f x)+b\right )}\right )d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {\sqrt {b} (a+b)^2 \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2}}+\frac {(a+b)^2 \cos (e+f x)}{a^3}-\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2}+\frac {\cos ^5(e+f x)}{5 a}}{f}\)

input

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

output

-((-((Sqrt[b]*(a + b)^2*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/a^(7/2)) + 
 ((a + b)^2*Cos[e + f*x])/a^3 - ((2*a + b)*Cos[e + f*x]^3)/(3*a^2) + Cos[e 
 + f*x]^5/(5*a))/f)
 

3.1.28.3.1 Defintions of rubi rules used

rule 364

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), 
x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x 
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In 
tegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4621

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-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]
 

3.1.28.4 Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-\frac {a \cos \left (f x +e \right )^{3} b}{3}+a^{2} \cos \left (f x +e \right )+2 a b \cos \left (f x +e \right )+b^{2} \cos \left (f x +e \right )}{a^{3}}+\frac {b \left (a^{2}+2 a b +b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}}{f}\)	\(115\)
default	\(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-\frac {a \cos \left (f x +e \right )^{3} b}{3}+a^{2} \cos \left (f x +e \right )+2 a b \cos \left (f x +e \right )+b^{2} \cos \left (f x +e \right )}{a^{3}}+\frac {b \left (a^{2}+2 a b +b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}}{f}\)	\(115\)
risch	\(-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )}}{16 a f}-\frac {7 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 a^{2} f}-\frac {{\mathrm e}^{i \left (f x +e \right )} b^{2}}{2 a^{3} f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )}}{16 a f}-\frac {7 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{2} f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{2 a^{3} f}-\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f}-\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{a^{3} f}-\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{2 a^{4} f}+\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f}+\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{a^{3} f}+\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{2 a^{4} f}-\frac {\cos \left (5 f x +5 e \right )}{80 f a}+\frac {5 \cos \left (3 f x +3 e \right )}{48 f a}+\frac {\cos \left (3 f x +3 e \right ) b}{12 f \,a^{2}}\)	\(448\)

input

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

output

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

3.1.28.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.34 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {6 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \, a^{3} f}, -\frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{15 \, a^{3} f}\right ] \]

input

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

output

[-1/30*(6*a^2*cos(f*x + e)^5 - 10*(2*a^2 + a*b)*cos(f*x + e)^3 - 15*(a^2 + 
 2*a*b + b^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + 
 e) - b)/(a*cos(f*x + e)^2 + b)) + 30*(a^2 + 2*a*b + b^2)*cos(f*x + e))/(a 
^3*f), -1/15*(3*a^2*cos(f*x + e)^5 - 5*(2*a^2 + a*b)*cos(f*x + e)^3 - 15*( 
a^2 + 2*a*b + b^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) + 15*(a^2 
+ 2*a*b + b^2)*cos(f*x + e))/(a^3*f)]
 

3.1.28.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.28.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \]

input

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

output

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

3.1.28.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (86) = 172\).

Time = 0.31 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.56 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b}{\sqrt {a b} \cos \left (f x + e\right ) + \sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {110 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {160 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {90 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {15 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a^{3} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \]

input

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

output

-1/15*(15*(a^2*b + 2*a*b^2 + b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)* 
cos(f*x + e) + sqrt(a*b)))/(sqrt(a*b)*a^3) - 2*(8*a^2 + 25*a*b + 15*b^2 - 
40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 110*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 + 160*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 - 
90*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 + 15*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)/(a^3*((cos(f*x + e) 
 - 1)/(cos(f*x + e) + 1) - 1)^5))/f
 

3.1.28.9 Mupad [B] (verification not implemented)

Time = 18.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{3\,a^2}+\frac {2}{3\,a}\right )}{f}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a}+\frac {b\,\left (\frac {b}{a^2}+\frac {2}{a}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,{\left (a+b\right )}^2}{a^2\,b+2\,a\,b^2+b^3}\right )\,{\left (a+b\right )}^2}{a^{7/2}\,f} \]

input

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

output

(cos(e + f*x)^3*(b/(3*a^2) + 2/(3*a)))/f - cos(e + f*x)^5/(5*a*f) - (cos(e 
 + f*x)*(1/a + (b*(b/a^2 + 2/a))/a))/f + (b^(1/2)*atan((a^(1/2)*b^(1/2)*co 
s(e + f*x)*(a + b)^2)/(2*a*b^2 + a^2*b + b^3))*(a + b)^2)/(a^(7/2)*f)