3.4.29 \(\int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [329]

Optimal. Leaf size=220 \[ \frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (2 a^2+7 a b-3 b^2\right ) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a+b) (a+3 b) F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/5*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/b/f+2/15*(a+3*b)*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^
(1/2)/b/f-1/15*(2*a^2+7*a*b-3*b^2)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin
(f*x+e)^2)^(1/2)/b^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+2/15*a*(a+b)*(a+3*b)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*Ellipti
cF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 260, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3271, 427, 542, 538, 437, 435, 432, 430} \begin {gather*} -\frac {\left (2 a^2+7 a b-3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{15 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {2 a (a+b) (a+3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sin (e+f x) \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {2 (a+3 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(2*(a + 3*b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(15*b*f) - (Cos[e + f*x]*Sin[e + f*x]*(a +
b*Sin[e + f*x]^2)^(3/2))/(5*b*f) - ((2*a^2 + 7*a*b - 3*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]
], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(15*b^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (2*a*(a + b)*(
a + 3*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)
/a])/(15*b^2*f*Sqrt[a + b*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 3271

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2
)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !IntegerQ
[p]

Rubi steps

\begin {align*} \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \left (1-x^2\right )^{3/2} \sqrt {a+b x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2} \left (a+5 b-2 (a+3 b) x^2\right )}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{5 b f}\\ &=\frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-a (a+9 b)+\left (2 a^2+7 a b-3 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f}\\ &=\frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {\left (2 a (a+b) (a+3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f}-\frac {\left (\left (2 a^2+7 a b-3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f}\\ &=\frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (\left (2 a^2+7 a b-3 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (2 a (a+b) (a+3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a+b) (a+3 b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.02, size = 199, normalized size = 0.90 \begin {gather*} \frac {-16 a \left (2 a^2+7 a b-3 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+32 a \left (a^2+4 a b+3 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-\sqrt {2} b \left (8 a^2-32 a b-15 b^2-4 (4 a-3 b) b \cos (2 (e+f x))+3 b^2 \cos (4 (e+f x))\right ) \sin (2 (e+f x))}{240 b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(-16*a*(2*a^2 + 7*a*b - 3*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 32*a*(a^2 +
4*a*b + 3*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(8*a^2 - 32*a*b
- 15*b^2 - 4*(4*a - 3*b)*b*Cos[2*(e + f*x)] + 3*b^2*Cos[4*(e + f*x)])*Sin[2*(e + f*x)])/(240*b^2*f*Sqrt[2*a +
b - b*Cos[2*(e + f*x)]])

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Maple [A]
time = 8.16, size = 432, normalized size = 1.96

method	result	size
default	\(\frac {-3 b^{3} \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right )+4 a \,b^{2} \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-a^{2} b +2 a \,b^{2}+3 b^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +6 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-7 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}}{15 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(-3*b^3*sin(f*x+e)*cos(f*x+e)^6+4*a*b^2*sin(f*x+e)*cos(f*x+e)^4+(-a^2*b+2*a*b^2+3*b^3)*cos(f*x+e)^2*sin(f
*x+e)+2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+8*a^2*
(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b+6*a*(cos(f*x+e)^
2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b^2-2*(cos(f*x+e)^2)^(1/2)*(-b
/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3-7*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e
)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+3*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/
a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)/b^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*cos(f*x + e)^4, x)

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Fricas [F]
time = 0.13, size = 27, normalized size = 0.12 \begin {gather*} {\rm integral}\left (\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )^{4}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^4, x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*cos(f*x + e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (e+f\,x\right )}^4\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2)^(1/2), x)

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