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

Optimal. Leaf size=213 \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 c^{9/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}-\frac{\sqrt{e x} (5 a d+7 b c) (b c-a d)}{6 c^2 d^2 e \sqrt{c+d x^2}}+\frac{\sqrt{e x} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}} \]

[Out]

((b*c - a*d)^2*Sqrt[e*x])/(3*c*d^2*e*(c + d*x^2)^(3/2)) - ((b*c - a*d)*(7*b*c + 5*a*d)*Sqrt[e*x])/(6*c^2*d^2*e
*Sqrt[c + d*x^2]) + ((5*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqr
t[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(9/4)*d^(9/4)*Sqrt[e]*Sqrt[c
+ d*x^2])

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Rubi [A]  time = 0.162365, antiderivative size = 213, 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 = {463, 457, 329, 220} \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{9/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}-\frac{\sqrt{e x} (5 a d+7 b c) (b c-a d)}{6 c^2 d^2 e \sqrt{c+d x^2}}+\frac{\sqrt{e x} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[e*x])/(3*c*d^2*e*(c + d*x^2)^(3/2)) - ((b*c - a*d)*(7*b*c + 5*a*d)*Sqrt[e*x])/(6*c^2*d^2*e
*Sqrt[c + d*x^2]) + ((5*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqr
t[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(9/4)*d^(9/4)*Sqrt[e]*Sqrt[c
+ d*x^2])

Rule 463

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

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}{\sqrt{e x} \left (c+d x^2\right )^{5/2}} \, dx &=\frac{(b c-a d)^2 \sqrt{e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{\frac{1}{2} \left (-6 a^2 d^2+(b c-a d)^2\right )-3 b^2 c d x^2}{\sqrt{e x} \left (c+d x^2\right )^{3/2}} \, dx}{3 c d^2}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) (7 b c+5 a d) \sqrt{e x}}{6 c^2 d^2 e \sqrt{c+d x^2}}+\frac{\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{12 c^2 d^2}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) (7 b c+5 a d) \sqrt{e x}}{6 c^2 d^2 e \sqrt{c+d x^2}}+\frac{\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{6 c^2 d^2 e}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) (7 b c+5 a d) \sqrt{e x}}{6 c^2 d^2 e \sqrt{c+d x^2}}+\frac{\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\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 )}{12 c^{9/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.266587, size = 169, normalized size = 0.79 \[ \frac{x \left (\frac{i \sqrt{x} \sqrt{\frac{c}{d x^2}+1} \left (5 a^2 d^2+2 a b c d+5 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}+5 a^2 d^2+\frac{2 c (b c-a d)^2}{c+d x^2}+2 a b c d-7 b^2 c^2\right )}{6 c^2 d^2 \sqrt{e x} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-7*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2 + (2*c*(b*c - a*d)^2)/(c + d*x^2) + (I*(5*b^2*c^2 + 2*a*b*c*d + 5*a^2*d
^2)*Sqrt[1 + c/(d*x^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/Sqrt[(I*Sqrt[c])/
Sqrt[d]]))/(6*c^2*d^2*Sqrt[e*x]*Sqrt[c + d*x^2])

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Maple [B]  time = 0.025, size = 660, normalized size = 3.1 \begin{align*}{\frac{1}{12\,{c}^{2}{d}^{3}} \left ( 5\,\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 ) \sqrt{-cd}{x}^{2}{a}^{2}{d}^{3}+2\,\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 ) \sqrt{-cd}{x}^{2}abc{d}^{2}+5\,\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 ) \sqrt{-cd}{x}^{2}{b}^{2}{c}^{2}d+5\,\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 ) \sqrt{-cd}{a}^{2}c{d}^{2}+2\,\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 ) \sqrt{-cd}ab{c}^{2}d+5\,\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 ) \sqrt{-cd}{b}^{2}{c}^{3}+10\,{x}^{3}{a}^{2}{d}^{4}+4\,{x}^{3}abc{d}^{3}-14\,{x}^{3}{b}^{2}{c}^{2}{d}^{2}+14\,x{a}^{2}c{d}^{3}-4\,xab{c}^{2}{d}^{2}-10\,x{b}^{2}{c}^{3}d \right ){\frac{1}{\sqrt{ex}}} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(5*((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))*(-c*d)^(1/2)*x^2*a^2*d^3+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)*El
lipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*x^2*a*b*c*d^2+5*((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))*(-c*d)^(1/2)*x^2*b^2*c^2*d+5*((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))*(-c*d)^(1/2)*a^2*c*d^2+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))*(-c*d)^(1/2)*a*b*c^2*d+5*((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))
*(-c*d)^(1/2)*b^2*c^3+10*x^3*a^2*d^4+4*x^3*a*b*c*d^3-14*x^3*b^2*c^2*d^2+14*x*a^2*c*d^3-4*x*a*b*c^2*d^2-10*x*b^
2*c^3*d)/(e*x)^(1/2)/c^2/d^3/(d*x^2+c)^(3/2)

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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{5}{2}} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*sqrt(e*x)), 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^{3} e x^{7} + 3 \, c d^{2} e x^{5} + 3 \, c^{2} d e x^{3} + c^{3} e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(5/2)/(e*x)^(1/2),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^3*e*x^7 + 3*c*d^2*e*x^5 + 3*c^2*d*e*x^3 + c^
3*e*x), 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((b*x**2+a)**2/(d*x**2+c)**(5/2)/(e*x)**(1/2),x)

[Out]

Timed out

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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{5}{2}} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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