\(\int \cos ^5(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [246]

Optimal result

Integrand size = 25, antiderivative size = 319 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {2 (a-3 (a+b)) \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 f}+\frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}+\frac {\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 a f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{15 a f \left (a+b-a \sin ^2(e+f x)\right )} \]

[Out]

-2/15*(-2*a-3*b)*cos(f*x+e)^2*sin(f*x+e)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/f+1/5*a*cos(f*x+e)^4*sin(f*
x+e)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/f+1/15*(8*a^2+13*a*b+3*b^2)*EllipticE(sin(f*x+e),(a/(a+b))^(1/2
))*(cos(f*x+e)^2)^(1/2)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/a/f/(1-a*sin(f*x+e)^2/(a+b))^(1/2)-1/15*b*(a
+b)*(4*a+3*b)*EllipticF(sin(f*x+e),(a/(a+b))^(1/2))*(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)/a/f/(a+b-a*sin(f*x+e)^2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4233, 1985, 1986, 427, 542, 538, 437, 435, 432, 430} \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right )}{15 a f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{15 a f \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {a \sin (e+f x) \cos ^4(e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 f}-\frac {2 (a-3 (a+b)) \sin (e+f x) \cos ^2(e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 f} \]

[In]

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

[Out]

(-2*(a - 3*(a + b))*Cos[e + f*x]^2*Sin[e + f*x]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)])/(15*f) + (a*C
os[e + f*x]^4*Sin[e + f*x]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)])/(5*f) + ((8*a^2 + 13*a*b + 3*b^2)*
Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)
])/(15*a*f*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]) - (b*(a + b)*(4*a + 3*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSi
n[Sin[e + f*x]], a/(a + b)]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b
)])/(15*a*f*(a + b - a*Sin[e + f*x]^2))

Rule 427

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

Rule 430

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

Rule 432

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

Rule 435

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

Rule 437

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

Rule 538

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 542

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 1985

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u*((b + a*c + a*d*x^n)/(c + d*x^n))^p
, x] /; FreeQ[{a, b, c, d, n, p}, x]

Rule 1986

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^(r_.))^(p_), x_Symbol] :> Dist[Simp
[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))], Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)
^(p*r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]

Rule 4233

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \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 {\text {Subst}\left (\int \left (1-x^2\right )^2 \left (\frac {a+b-a x^2}{1-x^2}\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {Subst}\left (\int \sqrt {1-x^2} \left (a+b-a x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a+b-a \sin ^2(e+f x)}} \\ & = \frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {Subst}\left (\int \frac {\sqrt {1-x^2} \left ((a+b) (a-5 (a+b))-2 a (a-3 (a+b)) x^2\right )}{\sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{5 f \sqrt {a+b-a \sin ^2(e+f x)}} \\ & = -\frac {2 (a-3 (a+b)) \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 f}+\frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {Subst}\left (\int \frac {a (a+b) (8 a+9 b)-a \left (8 a^2+13 a b+3 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 a f \sqrt {a+b-a \sin ^2(e+f x)}} \\ & = -\frac {2 (a-3 (a+b)) \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 f}+\frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}-\frac {\left (b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 a f \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {\left (\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {Subst}\left (\int \frac {\sqrt {a+b-a x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 a f \sqrt {a+b-a \sin ^2(e+f x)}} \\ & = -\frac {2 (a-3 (a+b)) \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 f}+\frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}+\frac {\left (\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}\right ) \text {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 f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {\left (b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}\right ) \text {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 f \left (a+b-a \sin ^2(e+f x)\right )} \\ & = -\frac {2 (a-3 (a+b)) \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 f}+\frac {a \cos ^4(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{5 f}+\frac {\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{15 a f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{15 a f \left (a+b-a \sin ^2(e+f x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 18.48 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.12 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\cos ^3(e+f x) \csc (2 (e+f x)) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-8 i \sqrt {2} \left (8 a^3+21 a^2 b+16 a b^2+3 b^3\right ) \sqrt {-\frac {a \cos ^2(e+f x)}{b}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 (e+f x))}}{\sqrt {2}}\right )|\frac {b}{a+b}\right ) \sqrt {\frac {a \sin ^2(e+f x)}{a+b}}+a \left (8 i \sqrt {2} \left (8 a^2+17 a b+9 b^2\right ) \sqrt {-\frac {a \cos ^2(e+f x)}{b}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 (e+f x))}}{\sqrt {2}}\right ),\frac {b}{a+b}\right ) \sqrt {\frac {a \sin ^2(e+f x)}{a+b}}+a \sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 (e+f x))} (11 a+12 b+3 a \cos (2 (e+f x))) \sin ^2(2 (e+f x))\right )\right )}{30 a^2 \sqrt {-\frac {1}{b}} f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \]

[In]

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

[Out]

(Cos[e + f*x]^3*Csc[2*(e + f*x)]*(a + b*Sec[e + f*x]^2)^(3/2)*((-8*I)*Sqrt[2]*(8*a^3 + 21*a^2*b + 16*a*b^2 + 3
*b^3)*Sqrt[-((a*Cos[e + f*x]^2)/b)]*EllipticE[I*ArcSinh[(Sqrt[-b^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqr
t[2]], b/(a + b)]*Sqrt[(a*Sin[e + f*x]^2)/(a + b)] + a*((8*I)*Sqrt[2]*(8*a^2 + 17*a*b + 9*b^2)*Sqrt[-((a*Cos[e
+ f*x]^2)/b)]*EllipticF[I*ArcSinh[(Sqrt[-b^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqrt[2]], b/(a + b)]*Sqr
t[(a*Sin[e + f*x]^2)/(a + b)] + a*Sqrt[-b^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(11*a + 12*b + 3*a*Cos[2*(e
+ f*x)])*Sin[2*(e + f*x)]^2)))/(30*a^2*Sqrt[-b^(-1)]*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.03 (sec) , antiderivative size = 8848, normalized size of antiderivative = 27.74

method	result	size
default	\(\text {Expression too large to display}\)	\(8848\)

[In]

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

[Out]

result too large to display

Fricas [F]

\[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{5} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{5} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{5} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]

[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)