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

Optimal. Leaf size=338 \[ -\frac{2 b (2 a-b) (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{15 a^2 f \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\left (8 a^2+3 a b-2 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 a^2 f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{2 (2 a-b) \sin (e+f x) \cos ^2(e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 a f}+\frac{\sin (e+f x) \cos ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right ) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 a f} \]

[Out]

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

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Rubi [A]  time = 0.571286, antiderivative size = 400, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4148, 6722, 1974, 416, 528, 524, 426, 424, 421, 419} \[ \frac{\left (8 a^2+3 a b-2 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 a^2 f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b}}-\frac{2 b (2 a-b) (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 a^2 f \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b}}+\frac{\sin (e+f x) \cos ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )^{3/2} \sqrt{a+b \sec ^2(e+f x)}}{5 a f \sqrt{a \cos ^2(e+f x)+b}}+\frac{2 (2 a-b) \sin (e+f x) \cos ^2(e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{15 a f \sqrt{a \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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 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 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 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 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)])

Rubi steps

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

Mathematica [F]  time = 11.436, size = 0, normalized size = 0. \[ \int \cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [C]  time = 0.747, size = 6392, normalized size = 18.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(f*x + e)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(f*x + e)^5, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(f*x + e)^5, x)