3.282 \(\int \frac {\cos ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=271 \[ \frac {(5 a-7 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 a^{9/2} f}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a f \sqrt {a+b \tan ^2(e+f x)+b}} \]

[Out]

1/16*(5*a^3-9*a^2*b+15*a*b^2-35*b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(9/2)/f+1/48*(15*
a^2-22*a*b+35*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/24*(5*a-7*b)*cos(f*x+e)^3*sin(f*x+
e)/a^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/6*cos(f*x+e)^5*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/48*b*(15*a^3-
17*a^2*b+25*a*b^2+105*b^3)*tan(f*x+e)/a^4/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4146, 414, 527, 12, 377, 203} \[ \frac {\left (-9 a^2 b+5 a^3+15 a b^2-35 b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 a^{9/2} f}+\frac {b \left (-17 a^2 b+15 a^3+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {(5 a-7 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a f \sqrt {a+b \tan ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

((5*a^3 - 9*a^2*b + 15*a*b^2 - 35*b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(9
/2)*f) + ((15*a^2 - 22*a*b + 35*b^2)*Cos[e + f*x]*Sin[e + f*x])/(48*a^3*f*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((
5*a - 7*b)*Cos[e + f*x]^3*Sin[e + f*x])/(24*a^2*f*Sqrt[a + b + b*Tan[e + f*x]^2]) + (Cos[e + f*x]^5*Sin[e + f*
x])/(6*a*f*Sqrt[a + b + b*Tan[e + f*x]^2]) + (b*(15*a^3 - 17*a^2*b + 25*a*b^2 + 105*b^3)*Tan[e + f*x])/(48*a^4
*(a + b)*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 414

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

Rule 527

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

Rule 4146

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

Rubi steps

\begin {align*} \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {-5 a+b-6 b x^2}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {15 a^2-2 a b+7 b^2+4 (5 a-7 b) b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {-15 a^3-3 a^2 b-a b^2+35 b^3-2 b \left (15 a^2-22 a b+35 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int -\frac {3 (a+b) \left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 a^4 (a+b) f}\\ &=\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 a^4 f}\\ &=\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^4 f}\\ &=\frac {\left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{9/2} f}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 19.73, size = 2068, normalized size = 7.63 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(3*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^14*Sin[e + f*x
])/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b*Sec[e + f*x]^2)^(3/2)*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*
AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin
[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*
x]^2)/(a + b)])*Sin[e + f*x]^2)*((3*a*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(
a + b)]*Cos[e + f*x]^9*Sin[e + f*x]^2)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)^2*(3*(a
+ b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2
, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e
+ f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]
^2)/(a + b)]*Cos[e + f*x]^9)/(2*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*Appel
lF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin[e +
f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)
/(a + b)])*Sin[e + f*x]^2)) - (12*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a +
b)]*Cos[e + f*x]^7*Sin[e + f*x]^2)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*A
ppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin[
e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x
]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b)*Cos[e + f*x]^8*Sin[e + f*x]*((a*f*AppellF1[3/2, -4, 5/2, 5/2, Sin
[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(a + b) - (8*f*AppellF1[3/2, -3, 3/2, 5/2,
Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/3))/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f
*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a
+ b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2,
-3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) - (3*(a + b)*AppellF1[1/2, -4, 3/
2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^8*Sin[e + f*x]*(2*f*(3*a*AppellF1[3/2, -4, 5/
2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a
*Sin[e + f*x]^2)/(a + b)])*Cos[e + f*x]*Sin[e + f*x] + 3*(a + b)*((a*f*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x
]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(a + b) - (8*f*AppellF1[3/2, -3, 3/2, 5/2, Sin[e +
f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/3) + Sin[e + f*x]^2*(3*a*((3*a*f*AppellF1[5/2,
-4, 7/2, 7/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(a + b) - (24*f*AppellF1
[5/2, -3, 5/2, 7/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/5) - 8*(a + b)*((9*
a*f*AppellF1[5/2, -3, 5/2, 7/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(5*(a +
b)) - (18*f*AppellF1[5/2, -2, 3/2, 7/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x]
)/5))))/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*AppellF1[1/2, -4, 3/2, 3
/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e +
f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e +
f*x]^2)^2) + (3*a*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]
^8*Sin[e + f*x]*Sin[2*(e + f*x)])/(2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*(a + b - a*Sin[e + f*x]^2)*(3*(a + b
)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3/2, -4, 5/2, 5/2, S
in[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e +
f*x]^2)/(a + b)])*Sin[e + f*x]^2))))

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fricas [A]  time = 7.33, size = 941, normalized size = 3.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/384*(3*(5*a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 20*a*b^4 - 35*b^5 + (5*a^5 - 4*a^4*b + 6*a^3*b^2 - 20*a^2*b^3 - 3
5*a*b^4)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 -
14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^
2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2
*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*(a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 2*a^4*b - 7*a^3*b^2)*cos(f*
x + e)^5 + (15*a^5 - 7*a^4*b + 13*a^3*b^2 + 35*a^2*b^3)*cos(f*x + e)^3 + (15*a^4*b - 17*a^3*b^2 + 25*a^2*b^3 +
105*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x +
e)^2 + (a^6*b + a^5*b^2)*f), -1/192*(3*(5*a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 20*a*b^4 - 35*b^5 + (5*a^5 - 4*a^4*
b + 6*a^3*b^2 - 20*a^2*b^3 - 35*a*b^4)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b
)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a
^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(8*(a^5 + a^4*b)*cos(f*
x + e)^7 + 2*(5*a^5 - 2*a^4*b - 7*a^3*b^2)*cos(f*x + e)^5 + (15*a^5 - 7*a^4*b + 13*a^3*b^2 + 35*a^2*b^3)*cos(f
*x + e)^3 + (15*a^4*b - 17*a^3*b^2 + 25*a^2*b^3 + 105*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2)*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^5*b^2)*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)

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maple [C]  time = 2.76, size = 3171, normalized size = 11.70 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

1/48/f*(b+a*cos(f*x+e)^2)*(-210*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+co
s(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b
))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)
+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^4*sin(f*x+e)-1
0*a^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4+30*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)
*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos
(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(
f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/
(a+b))^(1/2))*a^4*sin(f*x+e)+8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^4+10*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^4+15*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a^4-15*((2*I*a^(
1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a^4+105*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*b^4+4*(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a^3*b-13*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e
)^2*a^2*b^2-35*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b^3-8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2)*cos(f*x+e)^6*a^3*b+8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^3*b-4*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^3*b-14*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2*b^2+14*((2*I*
a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a^2*b^2-8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^6*
a^4-15*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-
2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+co
s(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*
a*b-b^2)/(a+b)^2)^(1/2))*a^4*sin(f*x+e)+105*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x
+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(
f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/
2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^4*sin(f*x+e)-15*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+
b))^(1/2)*a^3*b+17*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*c
os(f*x+e)^3*a^3*b+13*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a^2*b^2+35*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)*cos(f*x+e)^3*a*b^3+7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a^3*b+15*((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*a^3*b-17*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*a^2*b^2+25*
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*a*b^3-25*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3+12*2
^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(
1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e)
)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)
/(a+b)^2)^(1/2))*a^3*b*sin(f*x+e)-18*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/
(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))
/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1
/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b^2*sin(f*x+e)-24*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f
*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2
)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2))*a^3*b*sin(f*x+e)-120*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*c
os(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(
1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2
*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)
)*a*b^3*sin(f*x+e)+36*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/
(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*E
llipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b
),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b^2*sin(f*x+e)-105*((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+60*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x
+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(
f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/
2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^3*sin(f*x+e))*sin(f*x+e)/(-1+cos(f*x+e))/cos
(f*x+e)^3/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/a^4/(a+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2),x)

[Out]

int(cos(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**6/(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral(cos(e + f*x)**6/(a + b*sec(e + f*x)**2)**(3/2), x)

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