3.93 \(\int \frac{\sqrt{2+x^2}}{(1+x^2)^{5/2} (a+b x^2)} \, dx\)

Optimal. Leaf size=215 \[ -\frac{\sqrt{2} \sqrt{x^2+2} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{3 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}+\frac{2 b^2 \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)^2}+\frac{x \sqrt{x^2+2}}{3 \left (x^2+1\right )^{3/2} (a-b)}+\frac{\sqrt{2} \sqrt{x^2+2} (a-2 b) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.144995, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {546, 539, 526, 525, 418, 411} \[ \frac{2 b^2 \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)^2}+\frac{x \sqrt{x^2+2}}{3 \left (x^2+1\right )^{3/2} (a-b)}-\frac{\sqrt{2} \sqrt{x^2+2} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}+\frac{\sqrt{2} \sqrt{x^2+2} (a-2 b) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]

[Out]

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

Rule 546

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[b^2/(b*c
- a*d)^2, Int[((c + d*x^2)^(q + 2)*(e + f*x^2)^r)/(a + b*x^2), x], x] - Dist[d/(b*c - a*d)^2, Int[(c + d*x^2)^
q*(e + f*x^2)^r*(2*b*c - a*d + b*d*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r}, x] && LtQ[q, -1]

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 526

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

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, 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 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{\sqrt{2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx &=-\frac{\int \frac{\sqrt{2+x^2} \left (-a+2 b+b x^2\right )}{\left (1+x^2\right )^{5/2}} \, dx}{(a-b)^2}+\frac{b^2 \int \frac{\sqrt{2+x^2}}{\sqrt{1+x^2} \left (a+b x^2\right )} \, dx}{(a-b)^2}\\ &=\frac{x \sqrt{2+x^2}}{3 (a-b) \left (1+x^2\right )^{3/2}}+\frac{2 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 (a-b)^2 \sqrt{\frac{1+x^2}{2+x^2}} \sqrt{2+x^2}}+\frac{\int \frac{2 (2 a-5 b)+(a-4 b) x^2}{\left (1+x^2\right )^{3/2} \sqrt{2+x^2}} \, dx}{3 (a-b)^2}\\ &=\frac{x \sqrt{2+x^2}}{3 (a-b) \left (1+x^2\right )^{3/2}}+\frac{2 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 (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}{(a-b)^2}-\frac{2 \int \frac{1}{\sqrt{1+x^2} \sqrt{2+x^2}} \, dx}{3 (a-b)}\\ &=\frac{x \sqrt{2+x^2}}{3 (a-b) \left (1+x^2\right )^{3/2}}+\frac{\sqrt{2} (a-2 b) \sqrt{2+x^2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{(a-b)^2 \sqrt{1+x^2} \sqrt{\frac{2+x^2}{1+x^2}}}-\frac{\sqrt{2} \sqrt{2+x^2} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 (a-b) \sqrt{1+x^2} \sqrt{\frac{2+x^2}{1+x^2}}}+\frac{2 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 (a-b)^2 \sqrt{\frac{1+x^2}{2+x^2}} \sqrt{2+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.377982, size = 357, normalized size = 1.66 \[ \frac{-i \sqrt{2} a \left (x^2+1\right )^2 (4 a-7 b) \text{EllipticF}\left (i \sinh ^{-1}(x),\frac{1}{2}\right )+6 a^2 \sqrt{x^2+1} \sqrt{x^2+2} x^3+8 a^2 \sqrt{x^2+1} \sqrt{x^2+2} x-6 i \sqrt{2} b^2 x^4 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-12 i \sqrt{2} b^2 x^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-6 i \sqrt{2} b^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-12 a b \sqrt{x^2+1} \sqrt{x^2+2} x^3-14 a b \sqrt{x^2+1} \sqrt{x^2+2} x+6 i \sqrt{2} a \left (x^2+1\right )^2 (a-2 b) E\left (i \sinh ^{-1}(x)|\frac{1}{2}\right )+3 i \sqrt{2} a b x^4 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )+6 i \sqrt{2} a b x^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )+3 i \sqrt{2} a b \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )}{6 a \left (x^2+1\right )^2 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]

[Out]

(8*a^2*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] - 14*a*b*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] + 6*a^2*x^3*Sqrt[1 + x^2]*Sqrt[2 +
x^2] - 12*a*b*x^3*Sqrt[1 + x^2]*Sqrt[2 + x^2] + (6*I)*Sqrt[2]*a*(a - 2*b)*(1 + x^2)^2*EllipticE[I*ArcSinh[x],
1/2] - I*Sqrt[2]*a*(4*a - 7*b)*(1 + x^2)^2*EllipticF[I*ArcSinh[x], 1/2] + (3*I)*Sqrt[2]*a*b*EllipticPi[b/a, I
*ArcSinh[x], 1/2] - (6*I)*Sqrt[2]*b^2*EllipticPi[b/a, I*ArcSinh[x], 1/2] + (6*I)*Sqrt[2]*a*b*x^2*EllipticPi[b/
a, I*ArcSinh[x], 1/2] - (12*I)*Sqrt[2]*b^2*x^2*EllipticPi[b/a, I*ArcSinh[x], 1/2] + (3*I)*Sqrt[2]*a*b*x^4*Elli
pticPi[b/a, I*ArcSinh[x], 1/2] - (6*I)*Sqrt[2]*b^2*x^4*EllipticPi[b/a, I*ArcSinh[x], 1/2])/(6*a*(a - b)^2*(1 +
x^2)^2)

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Maple [B]  time = 0.033, size = 477, normalized size = 2.2 \begin{align*} -{\frac{1}{3\, \left ( a-b \right ) ^{2}a} \left ( 6\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}{a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}{a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,{x}^{5}{a}^{2}+6\,{x}^{5}ab+6\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){x}^{2}{b}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){b}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-10\,{x}^{3}{a}^{2}+19\,{x}^{3}ab-8\,{a}^{2}x+14\,xab \right ){\frac{1}{\sqrt{{x}^{2}+2}}} \left ({x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x)

[Out]

-1/3*(6*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*x^2*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-I*EllipticF(1/2*I*x*2^(1/2),2
^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)
+I*EllipticF(1/2*I*x*2^(1/2),2^(1/2))*x^2*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2
))*x^2*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)
^(1/2)-3*x^5*a^2+6*x^5*a*b+6*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-I*EllipticF(
1/2*I*x*2^(1/2),2^(1/2))*x^2*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*x^2
*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)+6*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*x^2*b^2*(x^2+2)^(1/2)*(x^2+1)^(
1/2)+I*EllipticF(1/2*I*x*2^(1/2),2^(1/2))*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)+6*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a
,2^(1/2))*b^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-10*x^3*a^2+19*x^3*a*b-8*a^2*x+14*x*a*b)/(x^2+2)^(1/2)/(a-b)^2/a/(x^2
+1)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}}{{\left (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + 2} \sqrt{x^{2} + 1}}{b x^{8} +{\left (a + 3 \, b\right )} x^{6} + 3 \,{\left (a + b\right )} x^{4} +{\left (3 \, a + b\right )} x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^8 + (a + 3*b)*x^6 + 3*(a + b)*x^4 + (3*a + 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+2)**(1/2)/(x**2+1)**(5/2)/(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 (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="giac")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)