3.1.89 \(\int \frac {(1+x^2)^{3/2} \sqrt {2+x^2}}{a+b x^2} \, dx\) [89]

Optimal. Leaf size=242 \[ -\frac {(a-2 b) x \sqrt {2+x^2}}{b^2 \sqrt {1+x^2}}+\frac {x \sqrt {1+x^2} \sqrt {2+x^2}}{3 b}+\frac {\sqrt {2} (a-2 b) \sqrt {2+x^2} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {(3 a-7 b) \sqrt {2+x^2} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {2} b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}+\frac {(a-2 b) (a-b) \sqrt {2+x^2} \Pi \left (1-\frac {b}{a};\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} a b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}} \]

[Out]

-(a-2*b)*x*(x^2+2)^(1/2)/b^2/(x^2+1)^(1/2)+1/3*x*(x^2+1)^(1/2)*(x^2+2)^(1/2)/b-1/6*(3*a-7*b)*(1/(x^2+1))^(1/2)
*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*(x^2+2)^(1/2)/b^2*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)+1/2*(a-2*b)*(a-b)*(1
/(x^2+1))^(1/2)*EllipticPi(x/(x^2+1)^(1/2),1-b/a,1/2*2^(1/2))*(x^2+2)^(1/2)/a/b^2*2^(1/2)/((x^2+2)/(x^2+1))^(1
/2)+(a-2*b)*(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*(x^2+2)^(1/2)/b^2/((x^2+2)/(x^2+1
))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 239, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {557, 553, 542, 545, 429, 506, 422} \begin {gather*} -\frac {\sqrt {2} \sqrt {x^2+2} (3 a-5 b) F\left (\text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{3 b^2 \sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}+\frac {\sqrt {2} \sqrt {x^2+2} (a-2 b) E\left (\text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{b^2 \sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}+\frac {2 \sqrt {x^2+1} (a-b)^2 \Pi \left (1-\frac {2 b}{a};\left .\text {ArcTan}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a b^2 \sqrt {\frac {x^2+1}{x^2+2}} \sqrt {x^2+2}}-\frac {x \sqrt {x^2+2} (a-2 b)}{b^2 \sqrt {x^2+1}}+\frac {x \sqrt {x^2+1} \sqrt {x^2+2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)^(3/2)*Sqrt[2 + x^2])/(a + b*x^2),x]

[Out]

-(((a - 2*b)*x*Sqrt[2 + x^2])/(b^2*Sqrt[1 + x^2])) + (x*Sqrt[1 + x^2]*Sqrt[2 + x^2])/(3*b) + (Sqrt[2]*(a - 2*b
)*Sqrt[2 + x^2]*EllipticE[ArcTan[x], 1/2])/(b^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) - (Sqrt[2]*(3*a - 5*b
)*Sqrt[2 + x^2]*EllipticF[ArcTan[x], 1/2])/(3*b^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) + (2*(a - b)^2*Sqrt
[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/Sqrt[2]], -1])/(a*b^2*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + 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]

Rule 553

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[c*(Sqrt[e +
f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), Ar
cTan[Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 557

Int[(((c_) + (d_.)*(x_)^2)^(3/2)*Sqrt[(e_) + (f_.)*(x_)^2])/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[(b*c - a*
d)^2/b^2, Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] + Dist[d/b^2, Int[(2*b*c - a*d + b*d*x^2)*
(Sqrt[e + f*x^2]/Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c] && PosQ[f/e]

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right )^{3/2} \sqrt {2+x^2}}{a+b x^2} \, dx &=\frac {\int \frac {\sqrt {2+x^2} \left (-a+2 b+b x^2\right )}{\sqrt {1+x^2}} \, dx}{b^2}+\frac {(a-b)^2 \int \frac {\sqrt {2+x^2}}{\sqrt {1+x^2} \left (a+b x^2\right )} \, dx}{b^2}\\ &=\frac {x \sqrt {1+x^2} \sqrt {2+x^2}}{3 b}+\frac {2 (a-b)^2 \sqrt {1+x^2} \Pi \left (1-\frac {2 b}{a};\left .\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a b^2 \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}}+\frac {\int \frac {-2 (3 a-5 b)-3 (a-2 b) x^2}{\sqrt {1+x^2} \sqrt {2+x^2}} \, dx}{3 b^2}\\ &=\frac {x \sqrt {1+x^2} \sqrt {2+x^2}}{3 b}+\frac {2 (a-b)^2 \sqrt {1+x^2} \Pi \left (1-\frac {2 b}{a};\left .\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a b^2 \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}}-\frac {(2 (3 a-5 b)) \int \frac {1}{\sqrt {1+x^2} \sqrt {2+x^2}} \, dx}{3 b^2}-\frac {(a-2 b) \int \frac {x^2}{\sqrt {1+x^2} \sqrt {2+x^2}} \, dx}{b^2}\\ &=-\frac {(a-2 b) x \sqrt {2+x^2}}{b^2 \sqrt {1+x^2}}+\frac {x \sqrt {1+x^2} \sqrt {2+x^2}}{3 b}-\frac {\sqrt {2} (3 a-5 b) \sqrt {2+x^2} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}+\frac {2 (a-b)^2 \sqrt {1+x^2} \Pi \left (1-\frac {2 b}{a};\left .\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a b^2 \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}}+\frac {(a-2 b) \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2}} \, dx}{b^2}\\ &=-\frac {(a-2 b) x \sqrt {2+x^2}}{b^2 \sqrt {1+x^2}}+\frac {x \sqrt {1+x^2} \sqrt {2+x^2}}{3 b}+\frac {\sqrt {2} (a-2 b) \sqrt {2+x^2} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {\sqrt {2} (3 a-5 b) \sqrt {2+x^2} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 b^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}+\frac {2 (a-b)^2 \sqrt {1+x^2} \Pi \left (1-\frac {2 b}{a};\left .\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a b^2 \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.22, size = 204, normalized size = 0.84 \begin {gather*} \frac {a b^2 x \sqrt {1+x^2} \sqrt {2+x^2}+3 i a (a-2 b) b E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-i a \left (3 a^2-9 a b+7 b^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+3 i a^3 \Pi \left (\frac {2 b}{a};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-12 i a^2 b \Pi \left (\frac {2 b}{a};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+15 i a b^2 \Pi \left (\frac {2 b}{a};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-6 i b^3 \Pi \left (\frac {2 b}{a};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )}{3 a b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)^(3/2)*Sqrt[2 + x^2])/(a + b*x^2),x]

[Out]

(a*b^2*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] + (3*I)*a*(a - 2*b)*b*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - I*a*(3*a^2 - 9
*a*b + 7*b^2)*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + (3*I)*a^3*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2] - (1
2*I)*a^2*b*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2] + (15*I)*a*b^2*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]
], 2] - (6*I)*b^3*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2])/(3*a*b^3)

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Maple [C] Result contains complex when optimal does not.
time = 0.22, size = 370, normalized size = 1.53

method	result	size
risch	\(\frac {x \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}{3 b}-\frac {\left (\frac {3 i \left (a -2 b \right ) \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (\EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )-\EllipticE \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )\right )}{2 b \sqrt {x^{4}+3 x^{2}+2}}+\frac {i \left (3 a^{2}-12 a b +13 b^{2}\right ) \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{2 b^{2} \sqrt {x^{4}+3 x^{2}+2}}-\frac {3 i \left (a^{3}-4 a^{2} b +5 a \,b^{2}-2 b^{3}\right ) \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b^{2} a \sqrt {x^{4}+3 x^{2}+2}}\right ) \sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}}{3 b \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}\)	\(263\)
default	\(-\frac {\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \left (-a \,b^{2} x^{5}+3 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{3}-9 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} b +7 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a \,b^{2}-3 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticE \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} b +6 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticE \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a \,b^{2}-3 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a^{3}+12 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a^{2} b -15 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a \,b^{2}+6 i \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b^{3}-3 a \,b^{2} x^{3}-2 a \,b^{2} x \right )}{3 \left (x^{4}+3 x^{2}+2\right ) b^{3} a}\)	\(370\)
elliptic	\(\frac {\sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}\, \left (\frac {x \sqrt {x^{4}+3 x^{2}+2}}{3 b}+\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, a \EllipticE \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}\, b^{2}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2}}{2 \sqrt {x^{4}+3 x^{2}+2}\, b^{3}}-\frac {7 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{6 b \sqrt {x^{4}+3 x^{2}+2}}-\frac {4 i a \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b^{2} \sqrt {x^{4}+3 x^{2}+2}}+\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a}{2 \sqrt {x^{4}+3 x^{2}+2}\, b^{2}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{b \sqrt {x^{4}+3 x^{2}+2}}+\frac {5 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b \sqrt {x^{4}+3 x^{2}+2}}-\frac {2 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{a \sqrt {x^{4}+3 x^{2}+2}}+\frac {i a^{2} \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b^{3} \sqrt {x^{4}+3 x^{2}+2}}\right )}{\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}\)	\(513\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(x^2+1)^(1/2)*(x^2+2)^(1/2)*(-a*b^2*x^5+3*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2^(1/2)
)*a^3-9*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2^(1/2))*a^2*b+7*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)
*EllipticF(1/2*I*x*2^(1/2),2^(1/2))*a*b^2-3*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a
^2*b+6*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a*b^2-3*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*
EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*a^3+12*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticPi(1/2*I*x*2^(1/2),2*b/
a,2^(1/2))*a^2*b-15*I*(x^2+1)^(1/2)*(x^2+2)^(1/2)*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*a*b^2+6*I*(x^2+1)^
(1/2)*(x^2+2)^(1/2)*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b^3-3*a*b^2*x^3-2*a*b^2*x)/(x^4+3*x^2+2)/b^3/a

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

[Out]

integrate(sqrt(x^2 + 2)*(x^2 + 1)^(3/2)/(b*x^2 + a), x)

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Fricas [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((x^2+1)^(3/2)*(x^2+2)^(1/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 + 2)*(x^2 + 1)^(3/2)/(b*x^2 + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} + 1\right )^{\frac {3}{2}} \sqrt {x^{2} + 2}}{a + b x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(sqrt(x^2 + 2)*(x^2 + 1)^(3/2)/(b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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