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

Optimal. Leaf size=273 \[ \frac{c^{3/2} \sqrt{a+b x^2} (3 b c-a d) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{3 a b \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{2 d x \sqrt{a+b x^2} (2 b c-a d)}{3 b^2 \sqrt{c+d x^2}}-\frac{2 \sqrt{c} \sqrt{d} \sqrt{a+b x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 b^2 \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b} \]

[Out]

(2*d*(2*b*c - a*d)*x*Sqrt[a + b*x^2])/(3*b^2*Sqrt[c + d*x^2]) + (d*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(3*b) -
(2*Sqrt[c]*Sqrt[d]*(2*b*c - a*d)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*b
^2*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (c^(3/2)*(3*b*c - a*d)*Sqrt[a + b*x^2]*EllipticF[A
rcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*a*b*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 = 0.164806, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {416, 531, 418, 492, 411} \[ \frac{2 d x \sqrt{a+b x^2} (2 b c-a d)}{3 b^2 \sqrt{c+d x^2}}-\frac{2 \sqrt{c} \sqrt{d} \sqrt{a+b x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 b^2 \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{c^{3/2} \sqrt{a+b x^2} (3 b c-a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a b \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, 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 (c+d x^2\right )^{3/2}}{\sqrt{a+b x^2}} \, dx &=\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b}+\frac{\int \frac{c (3 b c-a d)+2 d (2 b c-a d) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{3 b}\\ &=\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b}+\frac{(2 d (2 b c-a d)) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{3 b}+\frac{(c (3 b c-a d)) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{3 b}\\ &=\frac{2 d (2 b c-a d) x \sqrt{a+b x^2}}{3 b^2 \sqrt{c+d x^2}}+\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b}+\frac{c^{3/2} (3 b c-a d) \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a b \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 c d (2 b c-a d)) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=\frac{2 d (2 b c-a d) x \sqrt{a+b x^2}}{3 b^2 \sqrt{c+d x^2}}+\frac{d x \sqrt{a+b x^2} \sqrt{c+d x^2}}{3 b}-\frac{2 \sqrt{c} \sqrt{d} (2 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 )}{3 b^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}+\frac{c^{3/2} (3 b c-a d) \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a b \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 = 0.225573, size = 199, normalized size = 0.73 \[ \frac{-i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-b c) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right )+2 i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{3 b \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2) + (2*I)*c*(-2*b*c + a*d)*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*c*(-(b*c) + a*d)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*Elliptic
F[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(3*b*Sqrt[b/a]*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.014, size = 330, normalized size = 1.2 \begin{align*}{\frac{1}{ \left ( 3\,bd{x}^{4}+3\,ad{x}^{2}+3\,bc{x}^{2}+3\,ac \right ) b}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( \sqrt{-{\frac{b}{a}}}{x}^{5}b{d}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{3}a{d}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{3}bcd+ac\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 ) d-\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 ) b{c}^{2}-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 ) acd+4\,\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 ) b{c}^{2}+\sqrt{-{\frac{b}{a}}}xacd \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)*((-b/a)^(1/2)*x^5*b*d^2+(-b/a)^(1/2)*x^3*a*d^2+(-b/a)^(1/2)*x^3*b*c*d+a*c*
((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))*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))*b*c^2-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Ellipt
icE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*c*d+4*((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))*b*c^2+(-b/a)^(1/2)*x*a*c*d)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/b/(-b/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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