3.855 \(\int \frac{(a+b x^2)^2}{(e x)^{5/2} (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=207 \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-5 a d)+3 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{\sqrt{e x} \left (5 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{3 c^2 d e^3 \sqrt{c+d x^2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}} \]

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*Sqrt[c + d*x^2]) - ((3*b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*Sqrt[e*x])/(3*c^2*d*e^3*Sq
rt[c + d*x^2]) + ((3*b^2*c^2 + a*d*(6*b*c - 5*a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*
x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(6*c^(9/4)*d^(5/4)*e^(5/2)*Sqrt[c + d*x
^2])

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Rubi [A]  time = 0.181223, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {462, 457, 329, 220} \[ -\frac{\sqrt{e x} \left (5 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{3 c^2 d e^3 \sqrt{c+d x^2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-5 a d)+3 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*Sqrt[c + d*x^2]) - ((3*b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*Sqrt[e*x])/(3*c^2*d*e^3*Sq
rt[c + d*x^2]) + ((3*b^2*c^2 + a*d*(6*b*c - 5*a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*
x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(6*c^(9/4)*d^(5/4)*e^(5/2)*Sqrt[c + d*x
^2])

Rule 462

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

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{3/2}} \, dx &=-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}}+\frac{2 \int \frac{\frac{1}{2} a (6 b c-5 a d)+\frac{3}{2} b^2 c x^2}{\sqrt{e x} \left (c+d x^2\right )^{3/2}} \, dx}{3 c e^2}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}}-\frac{\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{6 c^2 d e^2}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}}-\frac{\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 c^2 d e^3}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \sqrt{c+d x^2}}-\frac{\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2+a d (6 b c-5 a d)\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{6 c^{9/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.180499, size = 181, normalized size = 0.87 \[ \frac{x \left (-i x^{5/2} \sqrt{\frac{c}{d x^2}+1} \left (5 a^2 d^2-6 a b c d-3 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )-\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} \left (a^2 d \left (2 c+5 d x^2\right )-6 a b c d x^2+3 b^2 c^2 x^2\right )\right )}{3 c^2 d \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} (e x)^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(3/2)),x]

[Out]

(x*(-(Sqrt[(I*Sqrt[c])/Sqrt[d]]*(3*b^2*c^2*x^2 - 6*a*b*c*d*x^2 + a^2*d*(2*c + 5*d*x^2))) - I*(-3*b^2*c^2 - 6*a
*b*c*d + 5*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1]))/
(3*c^2*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d*(e*x)^(5/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.024, size = 353, normalized size = 1.7 \begin{align*} -{\frac{1}{6\,x{c}^{2}{e}^{2}{d}^{2}} \left ( 5\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) x{a}^{2}{d}^{2}-6\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) xabcd-3\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) x{b}^{2}{c}^{2}+10\,{x}^{2}{a}^{2}{d}^{3}-12\,{x}^{2}abc{d}^{2}+6\,{x}^{2}{b}^{2}{c}^{2}d+4\,{a}^{2}c{d}^{2} \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x)

[Out]

-1/6/x*(5*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*a^2*d^2-6*(-c*d)^
(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2
)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*a*b*c*d-3*(-c*d)^(1/2)*((d*x+(-c*d
)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*Ellipt
icF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*b^2*c^2+10*x^2*a^2*d^3-12*x^2*a*b*c*d^2+6*x^2*b^2*c
^2*d+4*a^2*c*d^2)/(d*x^2+c)^(1/2)/c^2/e^2/(e*x)^(1/2)/d^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(3/2)*(e*x)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{d^{2} e^{3} x^{7} + 2 \, c d e^{3} x^{5} + c^{2} e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(d^2*e^3*x^7 + 2*c*d*e^3*x^5 + c^2*e^3*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{\left (e x\right )^{\frac{5}{2}} \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/((e*x)**(5/2)*(c + d*x**2)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(3/2)*(e*x)^(5/2)), x)