3.1.53 \(\int \frac {(a+b x^2) \sqrt {2+d x^2}}{\sqrt {3+f x^2}} \, dx\) [53]

Optimal. Leaf size=262 \[ -\frac {(6 b d-2 b f-3 a d f) x \sqrt {2+d x^2}}{3 d f \sqrt {3+f x^2}}+\frac {b x \sqrt {2+d x^2} \sqrt {3+f x^2}}{3 f}+\frac {\sqrt {2} (6 b d-2 b f-3 a d f) \sqrt {2+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{3 d f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}-\frac {\sqrt {2} (b-a f) \sqrt {2+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}} \]

[Out]

-1/3*(-3*a*d*f+6*b*d-2*b*f)*x*(d*x^2+2)^(1/2)/d/f/(f*x^2+3)^(1/2)+1/3*(-3*a*d*f+6*b*d-2*b*f)*(1/(3*f*x^2+9))^(
1/2)*(3*f*x^2+9)^(1/2)*EllipticE(x*f^(1/2)*3^(1/2)/(3*f*x^2+9)^(1/2),1/2*(4-6*d/f)^(1/2))*2^(1/2)*(d*x^2+2)^(1
/2)/d/f^(3/2)/((d*x^2+2)/(f*x^2+3))^(1/2)/(f*x^2+3)^(1/2)-(-a*f+b)*(1/(3*f*x^2+9))^(1/2)*(3*f*x^2+9)^(1/2)*Ell
ipticF(x*f^(1/2)*3^(1/2)/(3*f*x^2+9)^(1/2),1/2*(4-6*d/f)^(1/2))*2^(1/2)*(d*x^2+2)^(1/2)/f^(3/2)/((d*x^2+2)/(f*
x^2+3))^(1/2)/(f*x^2+3)^(1/2)+1/3*b*x*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)/f

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Rubi [A]
time = 0.12, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {542, 545, 429, 506, 422} \begin {gather*} -\frac {\sqrt {2} \sqrt {d x^2+2} (b-a f) F\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+\frac {\sqrt {2} \sqrt {d x^2+2} (-3 a d f+6 b d-2 b f) E\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{3 d f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}-\frac {x \sqrt {d x^2+2} (-3 a d f+6 b d-2 b f)}{3 d f \sqrt {f x^2+3}}+\frac {b x \sqrt {d x^2+2} \sqrt {f x^2+3}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Sqrt[2 + d*x^2])/Sqrt[3 + f*x^2],x]

[Out]

-1/3*((6*b*d - 2*b*f - 3*a*d*f)*x*Sqrt[2 + d*x^2])/(d*f*Sqrt[3 + f*x^2]) + (b*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2
])/(3*f) + (Sqrt[2]*(6*b*d - 2*b*f - 3*a*d*f)*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)
/(2*f)])/(3*d*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(b - a*f)*Sqrt[2 + d*x^2]*Elli
pticF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[c + d*x^2])), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !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 545

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \sqrt {2+d x^2}}{\sqrt {3+f x^2}} \, dx &=\frac {b x \sqrt {2+d x^2} \sqrt {3+f x^2}}{3 f}+\frac {\int \frac {-6 (b-a f)+(-6 b d+2 b f+3 a d f) x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{3 f}\\ &=\frac {b x \sqrt {2+d x^2} \sqrt {3+f x^2}}{3 f}-\frac {(2 (b-a f)) \int \frac {1}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{f}-\frac {(6 b d-2 b f-3 a d f) \int \frac {x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{3 f}\\ &=-\frac {(6 b d-2 b f-3 a d f) x \sqrt {2+d x^2}}{3 d f \sqrt {3+f x^2}}+\frac {b x \sqrt {2+d x^2} \sqrt {3+f x^2}}{3 f}-\frac {\sqrt {2} (b-a f) \sqrt {2+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}+\frac {(6 b d-2 b f-3 a d f) \int \frac {\sqrt {2+d x^2}}{\left (3+f x^2\right )^{3/2}} \, dx}{d f}\\ &=-\frac {(6 b d-2 b f-3 a d f) x \sqrt {2+d x^2}}{3 d f \sqrt {3+f x^2}}+\frac {b x \sqrt {2+d x^2} \sqrt {3+f x^2}}{3 f}+\frac {\sqrt {2} (6 b d-2 b f-3 a d f) \sqrt {2+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{3 d f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}-\frac {\sqrt {2} (b-a f) \sqrt {2+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.25, size = 142, normalized size = 0.54 \begin {gather*} \frac {b \sqrt {d} f x \sqrt {2+d x^2} \sqrt {3+f x^2}+i \sqrt {3} (6 b d-2 b f-3 a d f) E\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+i \sqrt {3} (3 d-2 f) (-2 b+a f) F\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )}{3 \sqrt {d} f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*Sqrt[2 + d*x^2])/Sqrt[3 + f*x^2],x]

[Out]

(b*Sqrt[d]*f*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2] + I*Sqrt[3]*(6*b*d - 2*b*f - 3*a*d*f)*EllipticE[I*ArcSinh[(Sqrt
[d]*x)/Sqrt[2]], (2*f)/(3*d)] + I*Sqrt[3]*(3*d - 2*f)*(-2*b + a*f)*EllipticF[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (
2*f)/(3*d)])/(3*Sqrt[d]*f^2)

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Maple [A]
time = 0.14, size = 367, normalized size = 1.40

method	result	size
elliptic	\(\frac {\sqrt {\left (f \,x^{2}+3\right ) \left (d \,x^{2}+2\right )}\, \left (\frac {b x \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}}{3 f}+\frac {\left (2 a -\frac {2 b}{f}\right ) \sqrt {3 f \,x^{2}+9}\, \sqrt {2 d \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )}{2 \sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}}-\frac {\left (a d +2 b -\frac {b \left (6 d +4 f \right )}{3 f}\right ) \sqrt {3 f \,x^{2}+9}\, \sqrt {2 d \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )\right )}{\sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}\, d}\right )}{\sqrt {f \,x^{2}+3}\, \sqrt {d \,x^{2}+2}}\)	\(282\)
risch	\(\frac {b x \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}}{3 f}+\frac {\left (-\frac {\left (3 a d f -6 b d +2 b f \right ) \sqrt {3 f \,x^{2}+9}\, \sqrt {2 d \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )\right )}{\sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}\, d}+\frac {3 a f \sqrt {3 f \,x^{2}+9}\, \sqrt {2 d \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )}{\sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}}-\frac {3 b \sqrt {3 f \,x^{2}+9}\, \sqrt {2 d \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )}{\sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}}\right ) \sqrt {\left (f \,x^{2}+3\right ) \left (d \,x^{2}+2\right )}}{3 f \sqrt {f \,x^{2}+3}\, \sqrt {d \,x^{2}+2}}\)	\(346\)
default	\(\frac {\sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, \left (b \,d^{2} f \,x^{5} \sqrt {-f}+3 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {d}{f}}}{2}\right ) a d f \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}+3 b \,d^{2} x^{3} \sqrt {-f}+2 b d f \,x^{3} \sqrt {-f}-6 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {d}{f}}}{2}\right ) b d \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}+2 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {d}{f}}}{2}\right ) b f \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}+3 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {d}{f}}}{2}\right ) b d \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}-2 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {d}{f}}}{2}\right ) b f \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}+6 b d x \sqrt {-f}\right )}{3 \left (d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6\right ) f \sqrt {-f}\, d}\)	\(367\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+2)^(1/2)/(f*x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)*(b*d^2*f*x^5*(-f)^(1/2)+3*2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2
^(1/2)*3^(1/2)*(d/f)^(1/2))*a*d*f*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)+3*b*d^2*x^3*(-f)^(1/2)+2*b*d*f*x^3*(-f)^(1/2
)-6*2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2^(1/2)*3^(1/2)*(d/f)^(1/2))*b*d*(d*x^2+2)^(1/2)*(f*x^2+3)^
(1/2)+2*2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2^(1/2)*3^(1/2)*(d/f)^(1/2))*b*f*(d*x^2+2)^(1/2)*(f*x^2
+3)^(1/2)+3*2^(1/2)*EllipticF(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2^(1/2)*3^(1/2)*(d/f)^(1/2))*b*d*(d*x^2+2)^(1/2)*(f
*x^2+3)^(1/2)-2*2^(1/2)*EllipticF(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2^(1/2)*3^(1/2)*(d/f)^(1/2))*b*f*(d*x^2+2)^(1/2
)*(f*x^2+3)^(1/2)+6*b*d*x*(-f)^(1/2))/(d*f*x^4+3*d*x^2+2*f*x^2+6)/f/(-f)^(1/2)/d

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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((b*x^2+a)*(d*x^2+2)^(1/2)/(f*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*sqrt(d*x^2 + 2)/sqrt(f*x^2 + 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+2)^(1/2)/(f*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right ) \sqrt {d x^{2} + 2}}{\sqrt {f x^{2} + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+2)**(1/2)/(f*x**2+3)**(1/2),x)

[Out]

Integral((a + b*x**2)*sqrt(d*x**2 + 2)/sqrt(f*x**2 + 3), 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((b*x^2+a)*(d*x^2+2)^(1/2)/(f*x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)*sqrt(d*x^2 + 2)/sqrt(f*x^2 + 3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (b\,x^2+a\right )\,\sqrt {d\,x^2+2}}{\sqrt {f\,x^2+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(d*x^2 + 2)^(1/2))/(f*x^2 + 3)^(1/2),x)

[Out]

int(((a + b*x^2)*(d*x^2 + 2)^(1/2))/(f*x^2 + 3)^(1/2), x)

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