3.4.30 \(\int \cos ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [330]

Optimal. Leaf size=159 \[ \frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(a-b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (a+b) F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 b f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

1/3*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f-1/3*(a-b)*(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/f/(1+b*sin(f*x+e)^2/a)^(1/2)+1/3*a*(a+b)*(cos(f*x+e)^2)^(1/2)/c
os(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/b/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 199, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3271, 428, 538, 437, 435, 432, 430} \begin {gather*} \frac {a (a+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 )}{3 b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a-b) \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 )}{3 b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 428

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

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Mathematica [A]
time = 0.68, size = 158, normalized size = 0.99 \begin {gather*} \frac {-2 \sqrt {2} a (a-b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 \sqrt {2} a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+b (2 a+b-b \cos (2 (e+f x))) \sin (2 (e+f x))}{6 \sqrt {2} b f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 2*Sqrt[2]*a*(a + b)*
Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + b*(2*a + b - b*Cos[2*(e + f*x)])*Sin[2*(e
+ f*x)])/(6*Sqrt[2]*b*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [A]
time = 6.22, size = 265, normalized size = 1.67

method	result	size
default	\(\frac {-b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -a b \left (\sin ^{3}\left (f x +e \right )\right )+b^{2} \left (\sin ^{3}\left (f x +e \right )\right )+\sin \left (f x +e \right ) a b}{3 b \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-b^2*sin(f*x+e)^5+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*
a^2+a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-(cos(f*x+e)^2)^
(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x
+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b-a*b*sin(f*x+e)^3+b^2*sin(f*x+e)^3+sin(f*x+e)*a*b)/b/c
os(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)^2*(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)^2, x)

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Fricas [F]
time = 0.10, size = 27, normalized size = 0.17 \begin {gather*} {\rm integral}\left (\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )^{2}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(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)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cos ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*cos(e + f*x)**2, x)

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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)^2*(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)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (e+f\,x\right )}^2\,\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)^2*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

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

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