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

Optimal. Leaf size=255 \[ -\frac{\sqrt{c} \sqrt{d} \sqrt{a+b x^2} (b c-3 a d) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{3 a^2 \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{2 \sqrt{b} \sqrt{c+d x^2} (b c-2 a d) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{3 a^{3/2} \sqrt{a+b x^2} (b c-a d)^2 \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac{b x \sqrt{c+d x^2}}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.146316, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {414, 525, 418, 411} \[ -\frac{\sqrt{c} \sqrt{d} \sqrt{a+b x^2} (b c-3 a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a^2 \sqrt{c+d x^2} (b c-a d)^2 \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{2 \sqrt{b} \sqrt{c+d x^2} (b c-2 a d) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{3 a^{3/2} \sqrt{a+b x^2} (b c-a d)^2 \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac{b x \sqrt{c+d x^2}}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

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{1}{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}} \, dx &=\frac{b x \sqrt{c+d x^2}}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{-2 b c+3 a d-b d x^2}{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}} \, dx}{3 a (b c-a d)}\\ &=\frac{b x \sqrt{c+d x^2}}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}-\frac{(d (b c-3 a d)) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{3 a (b c-a d)^2}+\frac{(2 b (b c-2 a d)) \int \frac{\sqrt{c+d x^2}}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a (b c-a d)^2}\\ &=\frac{b x \sqrt{c+d x^2}}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac{2 \sqrt{b} (b c-2 a d) \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{3 a^{3/2} (b c-a d)^2 \sqrt{a+b x^2} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac{\sqrt{c} \sqrt{d} (b c-3 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^2 (b c-a d)^2 \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 = 0.592447, size = 261, normalized size = 1.02 \[ \frac{-i \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (3 a^2 d^2-5 a b c d+2 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+b x \sqrt{\frac{b}{a}} \left (c+d x^2\right ) \left (-5 a^2 d+a b \left (3 c-4 d x^2\right )+2 b^2 c x^2\right )-2 i b c \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (2 a d-b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{3 a^2 \sqrt{\frac{b}{a}} \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[b/a]*x*(c + d*x^2)*(-5*a^2*d + 2*b^2*c*x^2 + a*b*(3*c - 4*d*x^2)) - (2*I)*b*c*(-(b*c) + 2*a*d)*(a + b*
x^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*(2*b^2*c^2 - 5
*a*b*c*d + 3*a^2*d^2)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a
*d)/(b*c)])/(3*a^2*Sqrt[b/a]*(b*c - a*d)^2*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.028, size = 752, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(b^3*d*x^8 + (b^3*c + 3*a*b^2*d)*x^6 + 3*(a*b^2*c + a^2*b*d)*x^4 + a^
3*c + (3*a^2*b*c + a^3*d)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/((a + b*x**2)**(5/2)*sqrt(c + d*x**2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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