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

Optimal. Leaf size=214 \[ -\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (a \cos ^2(e+f x)+b\right )}+\frac {(a+3 b) (3 a+5 b) \cos ^3(e+f x)}{12 a^4 b f}-\frac {\cos ^5(e+f x)}{5 a^3 f}-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (a \cos ^2(e+f x)+b\right )^2}+\frac {\sqrt {b} \left (15 a^2+70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{11/2} f}-\frac {\left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)}{2 a^5 f} \]

[Out]

-1/2*(3*a^2+14*a*b+13*b^2)*cos(f*x+e)/a^5/f+1/12*(a+3*b)*(3*a+5*b)*cos(f*x+e)^3/a^4/b/f-1/5*cos(f*x+e)^5/a^3/f
-1/4*(a+b)^2*cos(f*x+e)^7/a^2/b/f/(b+a*cos(f*x+e)^2)^2-1/8*b*(a+b)*(3*a+11*b)*cos(f*x+e)/a^5/f/(b+a*cos(f*x+e)
^2)+1/8*(15*a^2+70*a*b+63*b^2)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(11/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4133, 463, 455, 1810, 205} \[ -\frac {\left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)}{2 a^5 f}+\frac {\sqrt {b} \left (15 a^2+70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{11/2} f}-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (a \cos ^2(e+f x)+b\right )^2}+\frac {(a+3 b) (3 a+5 b) \cos ^3(e+f x)}{12 a^4 b f}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (a \cos ^2(e+f x)+b\right )}-\frac {\cos ^5(e+f x)}{5 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(15*a^2 + 70*a*b + 63*b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(11/2)*f) - ((3*a^2 + 14*a*b
+ 13*b^2)*Cos[e + f*x])/(2*a^5*f) + ((a + 3*b)*(3*a + 5*b)*Cos[e + f*x]^3)/(12*a^4*b*f) - Cos[e + f*x]^5/(5*a^
3*f) - ((a + b)^2*Cos[e + f*x]^7)/(4*a^2*b*f*(b + a*Cos[e + f*x]^2)^2) - (b*(a + b)*(3*a + 11*b)*Cos[e + f*x])
/(8*a^5*f*(b + a*Cos[e + f*x]^2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4133

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*} \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (1-x^2\right )^2}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (-4 a^2+7 (a+b)^2-4 a b x^2\right )}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 a^2 b f}\\ &=-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-a b^2 (a+b) (3 a+11 b)+2 a^2 b (a+b) (3 a+11 b) x^2-2 a^3 (a+b) (3 a+11 b) x^4+8 a^4 b x^6}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^6 b f}\\ &=-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (4 a b \left (3 a^2+14 a b+13 b^2\right )-2 a^2 (a+3 b) (3 a+5 b) x^2+8 a^3 b x^4+\frac {-15 a^3 b^2-70 a^2 b^3-63 a b^4}{b+a x^2}\right ) \, dx,x,\cos (e+f x)\right )}{8 a^6 b f}\\ &=-\frac {\left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)}{2 a^5 f}+\frac {(a+3 b) (3 a+5 b) \cos ^3(e+f x)}{12 a^4 b f}-\frac {\cos ^5(e+f x)}{5 a^3 f}-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )}+\frac {\left (b \left (15 a^2+70 a b+63 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^5 f}\\ &=\frac {\sqrt {b} \left (15 a^2+70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{11/2} f}-\frac {\left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)}{2 a^5 f}+\frac {(a+3 b) (3 a+5 b) \cos ^3(e+f x)}{12 a^4 b f}-\frac {\cos ^5(e+f x)}{5 a^3 f}-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 10.56, size = 1641, normalized size = 7.67 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-900*a^(11/2)*b^(3/2)*Cos[e + f*x] - 109000*a^(9/2)*b^(5/2)*Co
s[e + f*x] - 936000*a^(7/2)*b^(7/2)*Cos[e + f*x] - 2803072*a^(5/2)*b^(9/2)*Cos[e + f*x] - 3763200*a^(3/2)*b^(1
1/2)*Cos[e + f*x] - 1935360*Sqrt[a]*b^(13/2)*Cos[e + f*x] - 900*a^(11/2)*b^(3/2)*Cos[e + f*x]*Cos[2*(e + f*x)]
+ 900*a^(9/2)*b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 24000*a^(7/2)*b^(5/2)*Cos[e + f*x]*(a + 2
*b + a*Cos[2*(e + f*x)]) + 43200*a^(5/2)*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 225*a^5*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]*Sq
rt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 115200*a^2*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 537600*a*b^4*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 483840*b^5*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 225*a^5*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 115200*a^2*b^3*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*S
qrt[(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]*Ta
n[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 537600*a*b^4*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 483840*b^5*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]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 225*a^5*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*
(a + 2*b + a*Cos[2*(e + f*x)])^2 - 225*a^5*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b +
a*Cos[2*(e + f*x)])^2 + 19200*a^(5/2)*b^(5/2)*Cos[e]*Cos[f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 20352*a^(9/2
)*b^(5/2)*Cos[e + f*x]*Cos[4*(e + f*x)] - 115712*a^(7/2)*b^(7/2)*Cos[e + f*x]*Cos[4*(e + f*x)] - 129024*a^(5/2
)*b^(9/2)*Cos[e + f*x]*Cos[4*(e + f*x)] + 2048*a^(9/2)*b^(5/2)*Cos[e + f*x]*Cos[6*(e + f*x)] + 4608*a^(7/2)*b^
(7/2)*Cos[e + f*x]*Cos[6*(e + f*x)] - 384*a^(9/2)*b^(5/2)*Cos[e + f*x]*Cos[8*(e + f*x)] - 19200*a^(5/2)*b^(5/2
)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sin[e]*Sin[f*x] - 32496*a^(9/2)*b^(5/2)*Csc[e + f*x]*Sin[4*(e + f*x)] - 252
080*a^(7/2)*b^(7/2)*Csc[e + f*x]*Sin[4*(e + f*x)] - 577024*a^(5/2)*b^(9/2)*Csc[e + f*x]*Sin[4*(e + f*x)] - 403
200*a^(3/2)*b^(11/2)*Csc[e + f*x]*Sin[4*(e + f*x)]))/(491520*a^(11/2)*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3)

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 579, normalized size = 2.71 \[ \left [-\frac {48 \, a^{4} \cos \left (f x + e\right )^{9} - 16 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 16 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{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 (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{240 \, {\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 )}}, -\frac {24 \, a^{4} \cos \left (f x + e\right )^{9} - 8 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 8 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{120 \, {\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 )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/240*(48*a^4*cos(f*x + e)^9 - 16*(10*a^4 + 9*a^3*b)*cos(f*x + e)^7 + 16*(15*a^4 + 70*a^3*b + 63*a^2*b^2)*co
s(f*x + e)^5 + 50*(15*a^3*b + 70*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos
(f*x + e)^4 + 15*a^2*b^2 + 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^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*(15*a^2*b^2 + 70*a*b^3
+ 63*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f), -1/120*(24*a^4*cos(f*x
+ e)^9 - 8*(10*a^4 + 9*a^3*b)*cos(f*x + e)^7 + 8*(15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^5 + 25*(15*a^3
*b + 70*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2
+ 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*
x + e)/b) + 15*(15*a^2*b^2 + 70*a*b^3 + 63*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2
+ a^5*b^2*f)]

________________________________________________________________________________________

giac [B]  time = 0.58, size = 837, normalized size = 3.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/(
sqrt(a*b)*a^5) + 30*(9*a^3*b + 33*a^2*b^2 + 39*a*b^3 + 15*b^4 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1)
+ 49*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 23*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 45*b^4*
(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 27*a^2*b^2*(cos(f
*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 45*b^4*(cos(f*x + e)
- 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 11*a^2*b^2*(cos(f*x + e) -
1)^3/(cos(f*x + e) + 1)^3 - 13*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 15*b^4*(cos(f*x + e) - 1)^3/(
cos(f*x + e) + 1)^3)/((a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2*a^5) - 1
6*(8*a^2 + 75*a*b + 90*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 330*a*b*(cos(f*x + e) - 1)/(cos(f*
x + e) + 1) - 360*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
+ 480*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 540*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 270
*a*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 360*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 45*a*b*(c
os(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 90*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/(a^5*((cos(f*x + e
) - 1)/(cos(f*x + e) + 1) - 1)^5))/f

________________________________________________________________________________________

maple [A]  time = 1.02, size = 374, normalized size = 1.75 \[ -\frac {\cos ^{5}\left (f x +e \right )}{5 a^{3} f}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{3 a^{3} f}+\frac {\left (\cos ^{3}\left (f x +e \right )\right ) b}{f \,a^{4}}-\frac {\cos \left (f x +e \right )}{a^{3} f}-\frac {6 b \cos \left (f x +e \right )}{f \,a^{4}}-\frac {6 \cos \left (f x +e \right ) b^{2}}{f \,a^{5}}-\frac {9 b \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \,a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {13 b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{4 f \,a^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {17 b^{3} \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \,a^{4} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {7 b^{2} \cos \left (f x +e \right )}{8 f \,a^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {11 b^{3} \cos \left (f x +e \right )}{4 f \,a^{4} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b^{4} \cos \left (f x +e \right )}{8 f \,a^{5} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {15 b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 f \,a^{3} \sqrt {a b}}+\frac {35 b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{4 f \,a^{4} \sqrt {a b}}+\frac {63 b^{3} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 f \,a^{5} \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*cos(f*x+e)^5/a^3/f+2/3*cos(f*x+e)^3/a^3/f+1/f/a^4*cos(f*x+e)^3*b-cos(f*x+e)/a^3/f-6/f/a^4*b*cos(f*x+e)-6/
f/a^5*cos(f*x+e)*b^2-9/8/f*b/a^2/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3-13/4/f*b^2/a^3/(b+a*cos(f*x+e)^2)^2*cos(f*x
+e)^3-17/8/f*b^3/a^4/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3-7/8/f*b^2/a^3/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)-11/4/f*b^
3/a^4/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)-15/8/f*b^4/a^5/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)+15/8/f*b/a^3/(a*b)^(1/2)*
arctan(a*cos(f*x+e)/(a*b)^(1/2))+35/4/f*b^2/a^4/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+63/8/f*b^3/a^5/(a
*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 204, normalized size = 0.95 \[ -\frac {\frac {15 \, {\left ({\left (9 \, a^{3} b + 26 \, a^{2} b^{2} + 17 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 22 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (f x + e\right )\right )}}{a^{7} \cos \left (f x + e\right )^{4} + 2 \, a^{6} b \cos \left (f x + e\right )^{2} + a^{5} b^{2}} - \frac {15 \, {\left (15 \, a^{2} b + 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} + \frac {8 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )\right )}}{a^{5}}}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*((9*a^3*b + 26*a^2*b^2 + 17*a*b^3)*cos(f*x + e)^3 + (7*a^2*b^2 + 22*a*b^3 + 15*b^4)*cos(f*x + e))/(
a^7*cos(f*x + e)^4 + 2*a^6*b*cos(f*x + e)^2 + a^5*b^2) - 15*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(a*cos(f*x +
e)/sqrt(a*b))/(sqrt(a*b)*a^5) + 8*(3*a^2*cos(f*x + e)^5 - 5*(2*a^2 + 3*a*b)*cos(f*x + e)^3 + 15*(a^2 + 6*a*b +
6*b^2)*cos(f*x + e))/a^5)/f

________________________________________________________________________________________

mupad [B]  time = 4.51, size = 255, normalized size = 1.19 \[ \frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{a^4}+\frac {2}{3\,a^3}\right )}{f}-\frac {\left (\frac {9\,a^3\,b}{8}+\frac {13\,a^2\,b^2}{4}+\frac {17\,a\,b^3}{8}\right )\,{\cos \left (e+f\,x\right )}^3+\left (\frac {7\,a^2\,b^2}{8}+\frac {11\,a\,b^3}{4}+\frac {15\,b^4}{8}\right )\,\cos \left (e+f\,x\right )}{f\,\left (a^7\,{\cos \left (e+f\,x\right )}^4+2\,a^6\,b\,{\cos \left (e+f\,x\right )}^2+a^5\,b^2\right )}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a^3\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a^3}-\frac {3\,b^2}{a^5}+\frac {3\,b\,\left (\frac {3\,b}{a^4}+\frac {2}{a^3}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{15\,a^2\,b+70\,a\,b^2+63\,b^3}\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{8\,a^{11/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(e + f*x)^3*(b/a^4 + 2/(3*a^3)))/f - (cos(e + f*x)^3*((17*a*b^3)/8 + (9*a^3*b)/8 + (13*a^2*b^2)/4) + cos(e
+ f*x)*((11*a*b^3)/4 + (15*b^4)/8 + (7*a^2*b^2)/8))/(f*(a^5*b^2 + a^7*cos(e + f*x)^4 + 2*a^6*b*cos(e + f*x)^2
)) - cos(e + f*x)^5/(5*a^3*f) - (cos(e + f*x)*(1/a^3 - (3*b^2)/a^5 + (3*b*((3*b)/a^4 + 2/a^3))/a))/f + (b^(1/2
)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(70*a*b + 15*a^2 + 63*b^2))/(70*a*b^2 + 15*a^2*b + 63*b^3))*(70*a*b + 15*
a^2 + 63*b^2))/(8*a^(11/2)*f)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________