3.54 \(\int (a+b x^2) \sqrt {2+d x^2} \sqrt {3+f x^2} \, dx\)

Optimal. Leaf size=356 \[ \frac {x \sqrt {d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt {f x^2+3}}-\frac {\sqrt {2} \sqrt {d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}-\frac {\sqrt {2} \sqrt {d x^2+2} (-10 a d f+3 b d+2 b f) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{5 d f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+\frac {x \sqrt {d x^2+2} \sqrt {f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac {b x \left (d x^2+2\right )^{3/2} \sqrt {f x^2+3}}{5 d} \]

[Out]

1/15*(5*a*d*f*(3*d+2*f)-2*b*(9*d^2-6*d*f+4*f^2))*x*(d*x^2+2)^(1/2)/d^2/f/(f*x^2+3)^(1/2)-1/15*(5*a*d*f*(3*d+2*
f)-2*b*(9*d^2-6*d*f+4*f^2))*(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^2/f^(3/2)/((d*x^2+2)/(f*x^2+3))^(1/2)/(f*x^2+3)^(1/2)-1/5*(
-10*a*d*f+3*b*d+2*b*f)*(1/(3*f*x^2+9))^(1/2)*(3*f*x^2+9)^(1/2)*EllipticF(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)+1/5*b*x*(d*x
^2+2)^(3/2)*(f*x^2+3)^(1/2)/d+1/15*(5*a*d*f+3*b*d-4*b*f)*x*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)/d/f

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Rubi [A]  time = 0.32, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {528, 531, 418, 492, 411} \[ \frac {x \sqrt {d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt {f x^2+3}}-\frac {\sqrt {2} \sqrt {d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}-\frac {\sqrt {2} \sqrt {d x^2+2} (-10 a d f+3 b d+2 b f) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{5 d f^{3/2} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+\frac {x \sqrt {d x^2+2} \sqrt {f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac {b x \left (d x^2+2\right )^{3/2} \sqrt {f x^2+3}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*x*Sqrt[2 + d*x^2])/(15*d^2*f*Sqrt[3 + f*x^2]) + ((3*b*d -
4*b*f + 5*a*d*f)*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2])/(15*d*f) + (b*x*(2 + d*x^2)^(3/2)*Sqrt[3 + f*x^2])/(5*d)
- (Sqrt[2]*(5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sq
rt[3]], 1 - (3*d)/(2*f)])/(15*d^2*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(3*b*d + 2
*b*f - 10*a*d*f)*Sqrt[2 + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(5*d*f^(3/2)*Sqrt[(2
+ d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

Rule 411

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

Rule 418

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[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 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 531

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 \left (a+b x^2\right ) \sqrt {2+d x^2} \sqrt {3+f x^2} \, dx &=\frac {b x \left (2+d x^2\right )^{3/2} \sqrt {3+f x^2}}{5 d}+\frac {\int \frac {\sqrt {2+d x^2} \left (-3 (2 b-5 a d)+(3 b d-4 b f+5 a d f) x^2\right )}{\sqrt {3+f x^2}} \, dx}{5 d}\\ &=\frac {(3 b d-4 b f+5 a d f) x \sqrt {2+d x^2} \sqrt {3+f x^2}}{15 d f}+\frac {b x \left (2+d x^2\right )^{3/2} \sqrt {3+f x^2}}{5 d}+\frac {\int \frac {-6 (3 b d+2 b f-10 a d f)+\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{15 d f}\\ &=\frac {(3 b d-4 b f+5 a d f) x \sqrt {2+d x^2} \sqrt {3+f x^2}}{15 d f}+\frac {b x \left (2+d x^2\right )^{3/2} \sqrt {3+f x^2}}{5 d}-\frac {(2 (3 b d+2 b f-10 a d f)) \int \frac {1}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{5 d f}+\frac {\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac {x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx}{15 d f}\\ &=\frac {\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt {2+d x^2}}{15 d^2 f \sqrt {3+f x^2}}+\frac {(3 b d-4 b f+5 a d f) x \sqrt {2+d x^2} \sqrt {3+f x^2}}{15 d f}+\frac {b x \left (2+d x^2\right )^{3/2} \sqrt {3+f x^2}}{5 d}-\frac {\sqrt {2} (3 b d+2 b f-10 a d 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 )}{5 d f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}-\frac {\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac {\sqrt {2+d x^2}}{\left (3+f x^2\right )^{3/2}} \, dx}{5 d^2 f}\\ &=\frac {\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt {2+d x^2}}{15 d^2 f \sqrt {3+f x^2}}+\frac {(3 b d-4 b f+5 a d f) x \sqrt {2+d x^2} \sqrt {3+f x^2}}{15 d f}+\frac {b x \left (2+d x^2\right )^{3/2} \sqrt {3+f x^2}}{5 d}-\frac {\sqrt {2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \sqrt {2+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt {\frac {2+d x^2}{3+f x^2}} \sqrt {3+f x^2}}-\frac {\sqrt {2} (3 b d+2 b f-10 a d 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 )}{5 d 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]  time = 0.35, size = 186, normalized size = 0.52 \[ \frac {i \sqrt {3} \left (2 b \left (9 d^2-6 d f+4 f^2\right )-5 a d f (3 d+2 f)\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+\sqrt {d} f x \sqrt {d x^2+2} \sqrt {f x^2+3} \left (5 a d f+3 b d \left (f x^2+1\right )+2 b f\right )+i \sqrt {3} (3 d-2 f) (5 a d f-6 b d+2 b f) F\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )}{15 d^{3/2} f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*f*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2]*(2*b*f + 5*a*d*f + 3*b*d*(1 + f*x^2)) + I*Sqrt[3]*(-5*a*d*f*(3*d
+ 2*f) + 2*b*(9*d^2 - 6*d*f + 4*f^2))*EllipticE[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)] + I*Sqrt[3]*(3*d
- 2*f)*(-6*b*d + 2*b*f + 5*a*d*f)*EllipticF[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)])/(15*d^(3/2)*f^2)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )} \sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}, x\right ) \]

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]

integral((b*x^2 + a)*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 \[ \int {\left (b x^{2} + a\right )} \sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}\,{d x} \]

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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maple [B]  time = 0.03, size = 775, normalized size = 2.18 \[ \frac {\sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, \left (3 \sqrt {-f}\, b \,d^{3} f^{2} x^{7}+5 \sqrt {-f}\, a \,d^{3} f^{2} x^{5}+12 \sqrt {-f}\, b \,d^{3} f \,x^{5}+8 \sqrt {-f}\, b \,d^{2} f^{2} x^{5}+15 \sqrt {-f}\, a \,d^{3} f \,x^{3}+10 \sqrt {-f}\, a \,d^{2} f^{2} x^{3}+9 \sqrt {-f}\, b \,d^{3} x^{3}+30 \sqrt {-f}\, b \,d^{2} f \,x^{3}+4 \sqrt {-f}\, b d \,f^{2} x^{3}+30 \sqrt {-f}\, a \,d^{2} f x +15 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, a \,d^{2} f \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+15 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, a \,d^{2} f \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+10 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, a d \,f^{2} \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )-10 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, a d \,f^{2} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+18 \sqrt {-f}\, b \,d^{2} x -18 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b \,d^{2} \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+9 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b \,d^{2} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+12 \sqrt {-f}\, b d f x +12 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b d f \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )-18 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b d f \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )-8 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b \,f^{2} \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+8 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b \,f^{2} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )\right )}{15 \left (d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6\right ) \sqrt {-f}\, d^{2} f} \]

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)

[Out]

1/15*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)*(3*x^7*b*d^3*f^2*(-f)^(1/2)+5*x^5*a*d^3*f^2*(-f)^(1/2)+12*x^5*b*d^3*f*(-f
)^(1/2)+8*x^5*b*d^2*f^2*(-f)^(1/2)+15*x^3*a*d^3*f*(-f)^(1/2)+10*x^3*a*d^2*f^2*(-f)^(1/2)+15*2^(1/2)*EllipticF(
1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*a*d^2*f*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)-10*2^(1/2)*E
llipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*a*d*f^2*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)+15*
2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*a*d^2*f*(d*x^2+2)^(1/2)*(f*x^2+3)^
(1/2)+10*2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*a*d*f^2*(d*x^2+2)^(1/2)*(
f*x^2+3)^(1/2)+9*x^3*b*d^3*(-f)^(1/2)+30*x^3*b*d^2*f*(-f)^(1/2)+4*x^3*b*d*f^2*(-f)^(1/2)+9*2^(1/2)*EllipticF(1
/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d^2*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)-18*2^(1/2)*Elli
pticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d*f*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)+8*2^(1/2
)*EllipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*f^2*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)-18
*2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d^2*(d*x^2+2)^(1/2)*(f*x^2+3)^(
1/2)+12*2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d*f*(d*x^2+2)^(1/2)*(f*x
^2+3)^(1/2)-8*2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*f^2*(d*x^2+2)^(1/2
)*(f*x^2+3)^(1/2)+30*x*a*d^2*f*(-f)^(1/2)+18*x*b*d^2*(-f)^(1/2)+12*x*b*d*f*(-f)^(1/2))/(d*f*x^4+3*d*x^2+2*f*x^
2+6)/d^2/f/(-f)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )} \sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}\,{d x} \]

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

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

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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