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

Optimal. Leaf size=436 \[ \frac {(a+6 b) \sin (e+f x) \cos ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{5 a^2 f (a+b) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac {4 b \left (a^2-2 a b+12 b^2\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a^4 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac {\left (4 a^2-5 a b-24 b^2\right ) \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{15 a^3 f (a+b) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac {\left (8 a^3-9 a^2 b+16 a b^2+48 b^3\right ) \left (-a \sin ^2(e+f x)+a+b\right ) E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a^4 f (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac {b \sin (e+f x) \cos ^4(e+f x)}{a f (a+b) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[Out]

-b*cos(f*x+e)^4*sin(f*x+e)/a/(a+b)/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+1/15*(4*a^2-5*a*b-24*b^2)*sin(f
*x+e)*(a+b-a*sin(f*x+e)^2)/a^3/(a+b)/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+1/5*(a+6*b)*cos(f*x+e)^2*sin(
f*x+e)*(a+b-a*sin(f*x+e)^2)/a^2/(a+b)/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+1/15*(8*a^3-9*a^2*b+16*a*b^2
+48*b^3)*EllipticE(sin(f*x+e),(a/(a+b))^(1/2))*(a+b-a*sin(f*x+e)^2)/a^4/(a+b)/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+
e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/(1-a*sin(f*x+e)^2/(a+b))^(1/2)-4/15*b*(a^2-2*a*b+12*b^2)*EllipticF(sin(f*x+e)
,(a/(a+b))^(1/2))*(1-a*sin(f*x+e)^2/(a+b))^(1/2)/a^4/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2)
)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.75, antiderivative size = 509, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4148, 6722, 1974, 413, 528, 524, 426, 424, 421, 419} \[ \frac {\left (4 a^2-5 a b-24 b^2\right ) \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b}}{15 a^3 f (a+b) \sqrt {a+b \sec ^2(e+f x)}}-\frac {4 b \left (a^2-2 a b+12 b^2\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a^4 f \sqrt {\cos ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (-9 a^2 b+8 a^3+16 a b^2+48 b^3\right ) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a^4 f (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+6 b) \sin (e+f x) \cos ^2(e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b}}{5 a^2 f (a+b) \sqrt {a+b \sec ^2(e+f x)}}-\frac {b \sin (e+f x) \cos ^4(e+f x) \sqrt {a \cos ^2(e+f x)+b}}{a f (a+b) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*Cos[e + f*x]^4*Sqrt[b + a*Cos[e + f*x]^2]*Sin[e + f*x])/(a*(a + b)*f*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a +
b - a*Sin[e + f*x]^2])) + ((4*a^2 - 5*a*b - 24*b^2)*Sqrt[b + a*Cos[e + f*x]^2]*Sin[e + f*x]*Sqrt[a + b - a*Sin
[e + f*x]^2])/(15*a^3*(a + b)*f*Sqrt[a + b*Sec[e + f*x]^2]) + ((a + 6*b)*Cos[e + f*x]^2*Sqrt[b + a*Cos[e + f*x
]^2]*Sin[e + f*x]*Sqrt[a + b - a*Sin[e + f*x]^2])/(5*a^2*(a + b)*f*Sqrt[a + b*Sec[e + f*x]^2]) + ((8*a^3 - 9*a
^2*b + 16*a*b^2 + 48*b^3)*Sqrt[b + a*Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[a + b - a
*Sin[e + f*x]^2])/(15*a^4*(a + b)*f*Sqrt[Cos[e + f*x]^2]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[1 - (a*Sin[e + f*x]^2
)/(a + b)]) - (4*b*(a^2 - 2*a*b + 12*b^2)*Sqrt[b + a*Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], a/(a + b)
]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(15*a^4*f*Sqrt[Cos[e + f*x]^2]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b -
a*Sin[e + f*x]^2])

Rule 413

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

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 528

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

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

Rule 4148

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

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

\begin {align*} \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a+\frac {b}{1-x^2}\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sqrt {b+a \cos ^2(e+f x)} \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{7/2}}{\left (b+a \left (1-x^2\right )\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=\frac {\sqrt {b+a \cos ^2(e+f x)} \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{7/2}}{\left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}-\frac {\sqrt {b+a \cos ^2(e+f x)} \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2} \left (-a-b+(a+6 b) x^2\right )}{\sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {(a+6 b) \cos ^2(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{5 a^2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\sqrt {b+a \cos ^2(e+f x)} \operatorname {Subst}\left (\int \frac {\sqrt {1-x^2} \left (2 (2 a-3 b) (a+b)+\left (-4 a^2+5 a b+24 b^2\right ) x^2\right )}{\sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{5 a^2 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {\left (4 a^2-5 a b-24 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 a^3 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+6 b) \cos ^2(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{5 a^2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\sqrt {b+a \cos ^2(e+f x)} \operatorname {Subst}\left (\int \frac {-(a+b) \left (8 a^2-13 a b+24 b^2\right )+\left (8 a^3-9 a^2 b+16 a b^2+48 b^3\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 a^3 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {\left (4 a^2-5 a b-24 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 a^3 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+6 b) \cos ^2(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{5 a^2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\left (4 b \left (a^2-2 a b+12 b^2\right ) \sqrt {b+a \cos ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 a^4 f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (\left (8 a^3-9 a^2 b+16 a b^2+48 b^3\right ) \sqrt {b+a \cos ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b-a x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 a^4 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {\left (4 a^2-5 a b-24 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 a^3 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+6 b) \cos ^2(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{5 a^2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (\left (8 a^3-9 a^2 b+16 a b^2+48 b^3\right ) \sqrt {b+a \cos ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {a x^2}{a+b}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 a^4 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {\left (4 b \left (a^2-2 a b+12 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1-\frac {a x^2}{a+b}}} \, dx,x,\sin (e+f x)\right )}{15 a^4 f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}\\ &=-\frac {b \cos ^4(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {\left (4 a^2-5 a b-24 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 a^3 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+6 b) \cos ^2(e+f x) \sqrt {b+a \cos ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{5 a^2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (8 a^3-9 a^2 b+16 a b^2+48 b^3\right ) \sqrt {b+a \cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right ) \sqrt {a+b-a \sin ^2(e+f x)}}{15 a^4 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {4 b \left (a^2-2 a b+12 b^2\right ) \sqrt {b+a \cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{15 a^4 f \sqrt {\cos ^2(e+f x)} \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 15.47, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{5}}{b^{2} \sec \left (f x + e\right )^{4} + 2 \, a b \sec \left (f x + e\right )^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e)^2 + a)*cos(f*x + e)^5/(b^2*sec(f*x + e)^4 + 2*a*b*sec(f*x + e)^2 + a^2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 2.76, size = 15199, normalized size = 34.86 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (e+f\,x\right )}^5}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________