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

Optimal. Leaf size=319 \[ \frac {\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {a \sin (e+f x) \cos ^4(e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 f}-\frac {2 (a-3 (a+b)) \sin (e+f x) \cos ^2(e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 f}-\frac {b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a f \left (-a \sin ^2(e+f x)+a+b\right )} \]

[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)

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Rubi [A]  time = 0.64, antiderivative size = 395, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4148, 6722, 1974, 416, 528, 524, 426, 424, 421, 419} \[ \frac {\left (8 a^2+13 a b+3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a \cos ^2(e+f x)+b}}+\frac {a \sin (e+f x) \cos ^4(e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}{5 f \sqrt {a \cos ^2(e+f x)+b}}-\frac {2 (a-3 (a+b)) \sin (e+f x) \cos ^2(e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}{15 f \sqrt {a \cos ^2(e+f x)+b}}-\frac {b (a+b) (4 a+3 b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a+b \sec ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 a f \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

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

Rule 416

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

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Mathematica [C]  time = 8.98, size = 350, normalized size = 1.10 \[ \frac {\cos ^3(e+f x) \csc (2 (e+f x)) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (a \left (a \sqrt {-\frac {1}{a+b}} \sin ^2(2 (e+f x)) \sqrt {a \cos (2 (e+f x))+a+2 b} (3 a \cos (2 (e+f x))+11 a+12 b)+16 i \sqrt {2} b (2 a+3 b) \sqrt {\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {-\frac {a \cos ^2(e+f x)}{b}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {1}{a+b}} \sqrt {\cos (2 (e+f x)) a+a+2 b}}{\sqrt {2}}\right )|\frac {a+b}{b}\right )\right )-8 i \sqrt {2} b \left (8 a^2+13 a b+3 b^2\right ) \sqrt {\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {-\frac {a \cos ^2(e+f x)}{b}} E\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {1}{a+b}} \sqrt {\cos (2 (e+f x)) a+a+2 b}}{\sqrt {2}}\right )|\frac {a+b}{b}\right )\right )}{30 a^2 f \sqrt {-\frac {1}{a+b}} (a \cos (2 (e+f x))+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[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]*b*(8*a^2 + 13*a*b + 3*b^2)*Sqrt[
-((a*Cos[e + f*x]^2)/b)]*EllipticE[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqrt[2]]
, (a + b)/b]*Sqrt[(a*Sin[e + f*x]^2)/(a + b)] + a*((16*I)*Sqrt[2]*b*(2*a + 3*b)*Sqrt[-((a*Cos[e + f*x]^2)/b)]*
EllipticF[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqrt[2]], (a + b)/b]*Sqrt[(a*Sin[
e + f*x]^2)/(a + b)] + a*Sqrt[-(a + 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[-(a + b)^(-1)]*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (f x + e\right )^{5} \sec \left (f x + e\right )^{2} + a \cos \left (f x + e\right )^{5}\right )} \sqrt {b \sec \left (f x + e\right )^{2} + a}, x\right ) \]

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((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)

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

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

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maple [C]  time = 2.10, size = 6396, normalized size = 20.05 \[ \text {output too large to display} \]

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

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

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

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

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)

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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

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