3.952 \(\int \frac{(a+b x^2)^{3/2}}{x^2 \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=244 \[ \frac{2 b \sqrt{c} \sqrt{a+b x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{\sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{a \sqrt{a+b x^2} \sqrt{c+d x^2}}{c x}+\frac{x \sqrt{a+b x^2} (a d+b c)}{c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.160478, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {474, 531, 418, 492, 411} \[ -\frac{a \sqrt{a+b x^2} \sqrt{c+d x^2}}{c x}+\frac{x \sqrt{a+b x^2} (a d+b c)}{c \sqrt{c+d x^2}}+\frac{2 b \sqrt{c} \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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

Mathematica [C]  time = 0.271095, size = 206, normalized size = 0.84 \[ \frac{-i b c x \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-b c) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )-a d \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right )-i b c x \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d+b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{c d x \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*Sqrt[b/a]*d*(a + b*x^2)*(c + d*x^2)) - I*b*c*(b*c + a*d)*x*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*Ellipt
icE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*b*c*(-(b*c) + a*d)*x*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*Elli
pticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(Sqrt[b/a]*c*d*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.018, size = 352, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) cdx}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( -\sqrt{-{\frac{b}{a}}}{x}^{4}ab{d}^{2}+\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) xabcd-\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) x{b}^{2}{c}^{2}+\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) xabcd+\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) x{b}^{2}{c}^{2}-\sqrt{-{\frac{b}{a}}}{x}^{2}{a}^{2}{d}^{2}-\sqrt{-{\frac{b}{a}}}{x}^{2}abcd-\sqrt{-{\frac{b}{a}}}{a}^{2}cd \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-(-b/a)^(1/2)*x^4*a*b*d^2+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x
*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x*a*b*c*d-((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d
/b/c)^(1/2))*x*b^2*c^2+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x*a*b
*c*d+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x*b^2*c^2-(-b/a)^(1/2)*
x^2*a^2*d^2-(-b/a)^(1/2)*x^2*a*b*c*d-(-b/a)^(1/2)*a^2*c*d)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/c/(-b/a)^(1/2)/d/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(3/2)*sqrt(d*x^2 + c)/(d*x^4 + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**2)**(3/2)/(x**2*sqrt(c + d*x**2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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