\(\int \frac {\cos ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [281]
Optimal result
Integrand size = 25, antiderivative size = 194 \[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {3 \left (a^2-2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{7/2} f}+\frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}
\]
[Out]
3/8*(a^2-2*a*b+5*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(7/2)/f+1/8*(3*a-5*b)*cos(f*x+e)
*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/8*(a
-3*b)*b*(3*a+5*b)*tan(f*x+e)/a^3/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)
Rubi [A] (verified)
Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4231, 425, 541, 12, 385, 209}
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {b (a-3 b) (3 a+5 b) \tan (e+f x)}{8 a^3 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {(3 a-5 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {3 \left (a^2-2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 a^{7/2} f}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f \sqrt {a+b \tan ^2(e+f x)+b}}
\]
[In]
Int[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(3*(a^2 - 2*a*b + 5*b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*a^(7/2)*f) + ((3*a
- 5*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*Sqrt[a + b + b*Tan[e + f*x]^2]) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*
a*f*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((a - 3*b)*b*(3*a + 5*b)*Tan[e + f*x])/(8*a^3*(a + b)*f*Sqrt[a + b + b*T
an[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 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 385
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 425
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]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]
Rule 541
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 4231
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*}
\text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {-3 a+b-4 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 )}{4 a f} \\ & = \frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {3 a^2+5 b^2+2 (3 a-5 b) b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 a^2 f} \\ & = \frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {3 (a+b) \left (a^2-2 a b+5 b^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b) f} \\ & = \frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (3 \left (a^2-2 a b+5 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 a^3 f} \\ & = \frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (3 \left (a^2-2 a b+5 b^2\right )\right ) \text {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 )}{8 a^3 f} \\ & = \frac {3 \left (a^2-2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{7/2} f}+\frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \\
\end{align*}
Mathematica [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 16.18 (sec) , antiderivative size = 2046, normalized size of antiderivative = 10.55
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Result too large to show}
\]
[In]
Integrate[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^10*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)*((a + b)*Appe
llF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f
*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/
(a + b)])*Sin[e + f*x]^2)*((a*(a + b)*AppellF1[1/2, -3, 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)^2*((a + b)*Appel
lF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*
x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(
a + b)])*Sin[e + f*x]^2)) + ((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*C
os[e + f*x]^7)/(2*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*((a + b)*AppellF1[1/2, -3, 3/2
, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e +
f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 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, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5*
Sin[e + f*x]^2)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*((a + b)*AppellF1[1/2, -3, 3/2,
3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e +
f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e +
f*x]^2)) + ((a + b)*Cos[e + f*x]^6*Sin[e + f*x]*((a*f*AppellF1[3/2, -3, 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) - 2*f*AppellF1[3/2, -2, 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]))/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]
^2)*((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/
2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a
*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) - ((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e +
f*x]^2)/(a + b)]*Cos[e + f*x]^6*Sin[e + f*x]*(2*f*(a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*
x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 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] + (a + b)*((a*f*AppellF1[3/2, -3, 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) - 2*f*AppellF1[3/2, -2, 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]) + Sin[e + f*x]^2*(a*((3*a*f*AppellF1[5/2, -3, 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) - (18*f*AppellF1[5/2, -2, 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) - 2*(a + b)*((9*a*f*AppellF1[5/2, -2, 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)) - (12*f*AppellF1[5/2, -1, 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)*((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]
+ (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 3/
2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)^2) + (a*(a + b)*AppellF1[1/2, -3, 3/2, 3/
2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^6*Sin[e + f*x]*Sin[2*(e + f*x)])/(2*(a + 2*b + a*C
os[2*(e + f*x)])^(3/2)*(a + b - a*Sin[e + f*x]^2)*((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[
e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*
AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2))))
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(1142\) vs. \(2(174)=348\).
Time = 6.40 (sec) , antiderivative size = 1143, normalized size of antiderivative =
5.89
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| method | result | size |
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| default |
\(\text {Expression too large to display}\) |
\(1143\) |
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[In]
int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/8/f/(a+b)/a^3/(-a)^(1/2)*(b+a*cos(f*x+e)^2)*(2*(-a)^(1/2)*a^3*cos(f*x+e)^4*sin(f*x+e)+2*(-a)^(1/2)*a^2*b*cos
(f*x+e)^4*sin(f*x+e)+3*(-a)^(1/2)*a^3*cos(f*x+e)^2*sin(f*x+e)-2*(-a)^(1/2)*a^2*b*cos(f*x+e)^2*sin(f*x+e)-5*(-a
)^(1/2)*a*b^2*cos(f*x+e)^2*sin(f*x+e)+3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)
+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*a^3*cos(f*x+e)-3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*(
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b*c
os(f*x+e)+9*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^2*cos(f*x+e)+15*
ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*cos(f*x+e)+3*(-a)^(1/2)*a^2*
b*sin(f*x+e)-4*(-a)^(1/2)*a*b^2*sin(f*x+e)-15*(-a)^(1/2)*b^3*sin(f*x+e)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2
)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^3-3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-
4*sin(f*x+e)*a)*a^2*b+9*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^2+15
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f
*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*b^3)/(a+b*sec(f*x+e)^2)^(3/2)*s
ec(f*x+e)^3
Fricas [A] (verification not implemented)
none
Time = 1.48 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.18
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (a^{3} b - a^{2} b^{2} + 3 \, a b^{3} + 5 \, b^{4} + {\left (a^{4} - a^{3} b + 3 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) - 8 \, {\left (2 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{64 \, {\left ({\left (a^{6} + a^{5} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b + a^{4} b^{2}\right )} f\right )}}, -\frac {3 \, {\left (a^{3} b - a^{2} b^{2} + 3 \, a b^{3} + 5 \, b^{4} + {\left (a^{4} - a^{3} b + 3 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left (2 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{6} + a^{5} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b + a^{4} b^{2}\right )} f\right )}}\right ]
\]
[In]
integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/64*(3*(a^3*b - a^2*b^2 + 3*a*b^3 + 5*b^4 + (a^4 - a^3*b + 3*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(-a)*lo
g(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*(1
6*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*(2*(a^4 + a^3*b)*cos(f*x + e)^5 + (3*a^4 - 2*a^3*b - 5*a^2*b^2)*cos(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 - 15*a
*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^6 + a^5*b)*f*cos(f*x + e)^2
+ (a^5*b + a^4*b^2)*f), -1/32*(3*(a^3*b - a^2*b^2 + 3*a*b^3 + 5*b^4 + (a^4 - a^3*b + 3*a^2*b^2 + 5*a*b^3)*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*(2*(a^4 + a^3*b)*cos(f*x + e)^5 + (3*a^4 - 2*a^3*b - 5*a^2*b^2)*co
s(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 - 15*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(
f*x + e))/((a^6 + a^5*b)*f*cos(f*x + e)^2 + (a^5*b + a^4*b^2)*f)]
Sympy [F]
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx
\]
[In]
integrate(cos(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Integral(cos(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
Maxima [F]
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(cos(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
Giac [F]
\[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
sage0*x
Mupad [F(-1)]
Timed out. \[
\int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x
\]
[In]
int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2),x)
[Out]
int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2), x)