3.5.98 \(\int \cot ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [498]

Optimal. Leaf size=232 \[ \frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f}-\frac {\cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {4 (a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

1/3*(3*a-b)*cot(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/a/f-1/3*cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2)/f+1/3*(7*a-b)*El
lipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/a/f/(1+b*sin(f*x+e)^
2/a)^(1/2)-4/3*(a+b)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(
1/2)/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 232, normalized size of antiderivative = 1.00, 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 = {3275, 484, 594, 538, 437, 435, 432, 430} \begin {gather*} -\frac {4 (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 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(7 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 a f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a - b)*Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a*f) - (Cot[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2])/(3*f
) + ((7*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*a*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (4*(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*f*Sqrt[a + b*Sin[e + f*x]^2])

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 484

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^p*((c + d*x^n)^q/(e*(m + 1))), x] - Dist[n/(e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^(p -
1)*(c + d*x^n)^(q - 1)*Simp[b*c*p + a*d*q + b*d*(p + q)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*
c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && LtQ[m, -1] && GtQ[p, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x
]

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 594

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*g*(m + 1))), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 3275

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

Rubi steps

\begin {align*} \int \cot ^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 \frac {\left (1-x^2\right )^{3/2} \sqrt {a+b x^2}}{x^4} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(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 {\sqrt {1-x^2} \left (\frac {1}{2} (-3 a+b)-2 b x^2\right )}{x^2 \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=\frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f}-\frac {\cot ^3(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 {\frac {1}{2} a (3 a-5 b)+\frac {1}{2} (7 a-b) b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a f}\\ &=\frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f}-\frac {\cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left ((7 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 a f}-\frac {\left (4 (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 f}\\ &=\frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f}-\frac {\cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left ((7 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 a f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left (4 (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 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {(3 a-b) \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a f}-\frac {\cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {(7 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 a f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {4 (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 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.20, size = 197, normalized size = 0.85 \begin {gather*} \frac {-\frac {\left (-8 a^2-4 a b+3 b^2+4 \left (4 a^2+2 a b-b^2\right ) \cos (2 (e+f x))+b (-4 a+b) \cos (4 (e+f x))\right ) \cot (e+f x) \csc ^2(e+f x)}{2 \sqrt {2}}+2 a (7 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 )-8 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 )}{6 a f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/2*((-8*a^2 - 4*a*b + 3*b^2 + 4*(4*a^2 + 2*a*b - b^2)*Cos[2*(e + f*x)] + b*(-4*a + b)*Cos[4*(e + f*x)])*Cot
[e + f*x]*Csc[e + f*x]^2)/Sqrt[2] + 2*a*(7*a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(
b/a)] - 8*a*(a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)])/(6*a*f*Sqrt[2*a + b - b
*Cos[2*(e + f*x)]])

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Maple [A]
time = 10.21, size = 351, normalized size = 1.51

method	result	size
default	\(-\frac {4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} \left (\sin ^{3}\left (f x +e \right )\right )+4 b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \left (\sin ^{3}\left (f x +e \right )\right )-7 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} \left (\sin ^{3}\left (f x +e \right )\right )+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b \left (\sin ^{3}\left (f x +e \right )\right )+4 a b \left (\sin ^{6}\left (f x +e \right )\right )-b^{2} \left (\sin ^{6}\left (f x +e \right )\right )+4 a^{2} \left (\sin ^{4}\left (f x +e \right )\right )-6 a b \left (\sin ^{4}\left (f x +e \right )\right )+b^{2} \left (\sin ^{4}\left (f x +e \right )\right )-5 a^{2} \left (\sin ^{2}\left (f x +e \right )\right )+2 a b \left (\sin ^{2}\left (f x +e \right )\right )+a^{2}}{3 a \sin \left (f x +e \right )^{3} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(4*(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*sin(f*x+e)^
3+4*b*(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*sin(f*x+e)^3-7*
(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*sin(f*x+e)^3+(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*sin(f*x+e)^3+4*a*b*sin(f*
x+e)^6-b^2*sin(f*x+e)^6+4*a^2*sin(f*x+e)^4-6*a*b*sin(f*x+e)^4+b^2*sin(f*x+e)^4-5*a^2*sin(f*x+e)^2+2*a*b*sin(f*
x+e)^2+a^2)/a/sin(f*x+e)^3/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(cot(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)*cot(f*x + e)^4, x)

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Fricas [F]
time = 0.15, size = 27, normalized size = 0.12 \begin {gather*} {\rm integral}\left (\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cot \left (f x + e\right )^{4}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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)*cot(f*x + e)^4, 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 )}} \cot ^{4}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*cot(e + f*x)**4, 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(cot(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)*cot(f*x + e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\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(cot(e + f*x)^4*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

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

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