3.228 \(\int \sec ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=372 \[ -\frac {\left (2 a^2-3 a b-8 b^2\right ) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 b^2 f}+\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 b^2 f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {(a+4 b) \tan (e+f x) \sec (e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 b f}+\frac {\tan (e+f x) \sec ^3(e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 f}-\frac {(a-8 b) (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 b f \left (-a \sin ^2(e+f x)+a+b\right )} \]

[Out]

-1/15*(2*a^2-3*a*b-8*b^2)*sin(f*x+e)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/b^2/f+1/15*(2*a^2-3*a*b-8*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)/b^2/f/(1-
a*sin(f*x+e)^2/(a+b))^(1/2)-1/15*(a-8*b)*(a+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)/b/f/(a+b-a*sin(f*x+e)^2)+1/15*(a+4*b)*sec
(f*x+e)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)*tan(f*x+e)/b/f+1/5*sec(f*x+e)^3*(sec(f*x+e)^2*(a+b-a*sin(f*x
+e)^2))^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.69, antiderivative size = 471, normalized size of antiderivative = 1.27, 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, 412, 527, 524, 426, 424, 421, 419} \[ -\frac {\left (2 a^2-3 a b-8 b^2\right ) \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}{15 b^2 f \sqrt {a \cos ^2(e+f x)+b}}+\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 b^2 f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a \cos ^2(e+f x)+b}}+\frac {\tan (e+f x) \sec ^3(e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}{5 f \sqrt {a \cos ^2(e+f x)+b}}+\frac {(a+4 b) \tan (e+f x) \sec (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}}{15 b f \sqrt {a \cos ^2(e+f x)+b}}-\frac {(a-8 b) (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a+b \sec ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac {a}{a+b}\right )}{15 b f \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-((2*a^2 - 3*a*b - 8*b^2)*Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]*Sqrt[a + b - a*Sin[e + f*x]^2])/(15*b^2*f*Sq
rt[b + a*Cos[e + f*x]^2]) + ((2*a^2 - 3*a*b - 8*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[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*b^2*f*Sqrt[b + a*Cos[e + f*x]^2]*Sqrt[1
- (a*Sin[e + f*x]^2)/(a + b)]) - ((a - 8*b)*(a + b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[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*b*f*Sqrt[b + a*Cos[e + f*x]^2]*Sqrt
[a + b - a*Sin[e + f*x]^2]) + ((a + 4*b)*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x]^2
]*Tan[e + f*x])/(15*b*f*Sqrt[b + a*Cos[e + f*x]^2]) + (Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b -
a*Sin[e + f*x]^2]*Tan[e + f*x])/(5*f*Sqrt[b + a*Cos[e + f*x]^2])

Rule 412

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(n*(p
+ 1) + 1) + d*(n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p,
-1] && LtQ[0, 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 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 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 \sec ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+\frac {b}{1-x^2}}}{\left (1-x^2\right )^3} \, 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 \frac {\sqrt {b+a \left (1-x^2\right )}}{\left (1-x^2\right )^{7/2}} \, 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 \frac {\sqrt {a+b-a x^2}}{\left (1-x^2\right )^{7/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {b+a \cos ^2(e+f x)}}\\ &=\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (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 {-4 (a+b)+3 a x^2}{\left (1-x^2\right )^{5/2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{5 f \sqrt {b+a \cos ^2(e+f x)}}\\ &=\frac {(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (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-8 b) (a+b)+a (a+4 b) x^2}{\left (1-x^2\right )^{3/2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt {b+a \cos ^2(e+f x)}}\\ &=-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt {b+a \cos ^2(e+f x)}}+\frac {(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (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 {-2 a (a-2 b) (a+b)+a \left (2 a^2-3 a b-8 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {b+a \cos ^2(e+f x)}}\\ &=-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt {b+a \cos ^2(e+f x)}}+\frac {(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt {b+a \cos ^2(e+f x)}}-\frac {\left ((a-8 b) (a+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 b f \sqrt {b+a \cos ^2(e+f x)}}-\frac {\left (\left (-2 a^2+3 a b+8 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 b^2 f \sqrt {b+a \cos ^2(e+f x)}}\\ &=-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt {b+a \cos ^2(e+f x)}}+\frac {(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt {b+a \cos ^2(e+f x)}}-\frac {\left (\left (-2 a^2+3 a b+8 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 b^2 f \sqrt {b+a \cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {\left ((a-8 b) (a+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 b f \sqrt {b+a \cos ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}\\ &=-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\left (2 a^2-3 a b-8 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 b^2 f \sqrt {b+a \cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {(a-8 b) (a+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 b f \sqrt {b+a \cos ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)}}+\frac {(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt {b+a \cos ^2(e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \sqrt {a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt {b+a \cos ^2(e+f x)}}\\ \end {align*}

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Mathematica [F]  time = 27.03, size = 0, normalized size = 0.00 \[ \int \sec ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

Integrate[Sec[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2], x]

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.29, size = 6562, normalized size = 17.64 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^5,x)

[Out]

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{5}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**5*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sec(e + f*x)**2)*sec(e + f*x)**5, x)

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