3.89 \(\int \frac{(1+x^2)^{3/2} \sqrt{2+x^2}}{a+b x^2} \, dx\)

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

[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)]) - ((3*a - 7*b)*Sqrt[2
+ x^2]*EllipticF[ArcTan[x], 1/2])/(3*Sqrt[2]*b^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) + ((a - 2*b)*(a - b
)*Sqrt[2 + x^2]*EllipticPi[1 - b/a, ArcTan[x], 1/2])/(Sqrt[2]*a*b^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)])

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Rubi [A]  time = 0.147457, antiderivative size = 239, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {543, 539, 528, 531, 418, 492, 411} \[ -\frac{x \sqrt{x^2+2} (a-2 b)}{b^2 \sqrt{x^2+1}}-\frac{\sqrt{2} \sqrt{x^2+2} (3 a-5 b) F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 b^2 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}+\frac{\sqrt{2} \sqrt{x^2+2} (a-2 b) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{b^2 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}+\frac{2 \sqrt{x^2+1} (a-b)^2 \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a b^2 \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2}}+\frac{x \sqrt{x^2+1} \sqrt{x^2+2}}{3 b} \]

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 543

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]

Rule 539

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

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 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]

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*}

Mathematica [C]  time = 0.355601, size = 204, normalized size = 0.84 \[ \frac{-i a \left (3 a^2-9 a b+7 b^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+3 i a^3 \Pi \left (\frac{2 b}{a};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )-12 i a^2 b \Pi \left (\frac{2 b}{a};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+a b^2 x \sqrt{x^2+1} \sqrt{x^2+2}+15 i a b^2 \Pi \left (\frac{2 b}{a};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )-6 i b^3 \Pi \left (\frac{2 b}{a};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+3 i a b (a-2 b) E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )}{3 a b^3} \]

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]  time = 0.035, size = 370, normalized size = 1.5 \begin{align*} -{\frac{1}{ \left ( 3\,{x}^{4}+9\,{x}^{2}+6 \right ){b}^{3}a}\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2} \left ( -{x}^{5}a{b}^{2}+3\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{3}-9\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}b+7\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) a{b}^{2}-3\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}b+6\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) a{b}^{2}-3\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){a}^{3}+12\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){a}^{2}b-15\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) a{b}^{2}+6\,i\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){b}^{3}-3\,{x}^{3}a{b}^{2}-2\,xa{b}^{2} \right ) } \end{align*}

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)

[Out]

-1/3*(x^2+1)^(1/2)*(x^2+2)^(1/2)*(-x^5*a*b^2+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*x^3*a*b^2-2*x*a*b^2)/(x^4+3*x^2+2)/b^3/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}{\left (x^{2} + 1\right )}^{\frac{3}{2}}}{b x^{2} + a}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + 2}{\left (x^{2} + 1\right )}^{\frac{3}{2}}}{b x^{2} + a}, x\right ) \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}{\left (x^{2} + 1\right )}^{\frac{3}{2}}}{b x^{2} + a}\,{d x} \end{align*}

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)