3.846 \(\int \frac{(a+b x^2)^2}{(e x)^{9/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=193 \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-14 a b c d+21 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{21 c^{9/4} \sqrt [4]{d} e^{9/2} \sqrt{c+d x^2}}-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}-\frac{2 a \sqrt{c+d x^2} (14 b c-5 a d)}{21 c^2 e^3 (e x)^{3/2}} \]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(7*c*e*(e*x)^(7/2)) - (2*a*(14*b*c - 5*a*d)*Sqrt[c + d*x^2])/(21*c^2*e^3*(e*x)^(3/2))
+ ((21*b^2*c^2 - 14*a*b*c*d + 5*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*Elli
pticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(21*c^(9/4)*d^(1/4)*e^(9/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.168304, antiderivative size = 193, 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, 453, 329, 220} \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-14 a b c d+21 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{21 c^{9/4} \sqrt [4]{d} e^{9/2} \sqrt{c+d x^2}}-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}-\frac{2 a \sqrt{c+d x^2} (14 b c-5 a d)}{21 c^2 e^3 (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(9/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(7*c*e*(e*x)^(7/2)) - (2*a*(14*b*c - 5*a*d)*Sqrt[c + d*x^2])/(21*c^2*e^3*(e*x)^(3/2))
+ ((21*b^2*c^2 - 14*a*b*c*d + 5*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*Elli
pticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(21*c^(9/4)*d^(1/4)*e^(9/2)*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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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)^{9/2} \sqrt{c+d x^2}} \, dx &=-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}+\frac{2 \int \frac{\frac{1}{2} a (14 b c-5 a d)+\frac{7}{2} b^2 c x^2}{(e x)^{5/2} \sqrt{c+d x^2}} \, dx}{7 c e^2}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}-\frac{2 a (14 b c-5 a d) \sqrt{c+d x^2}}{21 c^2 e^3 (e x)^{3/2}}-\frac{\left (4 \left (-\frac{21}{4} b^2 c^2+\frac{1}{4} a d (14 b c-5 a d)\right )\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{21 c^2 e^4}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}-\frac{2 a (14 b c-5 a d) \sqrt{c+d x^2}}{21 c^2 e^3 (e x)^{3/2}}-\frac{\left (8 \left (-\frac{21}{4} b^2 c^2+\frac{1}{4} a d (14 b c-5 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{21 c^2 e^5}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{7 c e (e x)^{7/2}}-\frac{2 a (14 b c-5 a d) \sqrt{c+d x^2}}{21 c^2 e^3 (e x)^{3/2}}+\frac{\left (21 b^2 c^2-a d (14 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 )}{21 c^{9/4} \sqrt [4]{d} e^{9/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.18989, size = 159, normalized size = 0.82 \[ \frac{x^{9/2} \left (\frac{2 a \left (c+d x^2\right ) \left (-3 a c+5 a d x^2-14 b c x^2\right )}{c^2 x^{7/2}}+\frac{2 i x \sqrt{\frac{c}{d x^2}+1} \left (5 a^2 d^2-14 a b c d+21 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{c^2 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{21 (e x)^{9/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(9/2)*Sqrt[c + d*x^2]),x]

[Out]

(x^(9/2)*((2*a*(c + d*x^2)*(-3*a*c - 14*b*c*x^2 + 5*a*d*x^2))/(c^2*x^(7/2)) + ((2*I)*(21*b^2*c^2 - 14*a*b*c*d
+ 5*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(c^2*Sqrt[(I*S
qrt[c])/Sqrt[d]])))/(21*(e*x)^(9/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.03, size = 370, normalized size = 1.9 \begin{align*}{\frac{1}{21\,{x}^{3}d{c}^{2}{e}^{4}} \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}^{3}{a}^{2}{d}^{2}-14\,\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}^{3}abcd+21\,\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}^{3}{b}^{2}{c}^{2}+10\,{x}^{4}{a}^{2}{d}^{3}-28\,{x}^{4}abc{d}^{2}+4\,{x}^{2}{a}^{2}c{d}^{2}-28\,{x}^{2}ab{c}^{2}d-6\,{a}^{2}{c}^{2}d \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)^(9/2)/(d*x^2+c)^(1/2),x)

[Out]

1/21/(d*x^2+c)^(1/2)/x^3*(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
^3*a^2*d^2-14*((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^3*a*b*c*d+21*
((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^3*b^2*c^2+10*x^4*a^2*d^3-28
*x^4*a*b*c*d^2+4*x^2*a^2*c*d^2-28*x^2*a*b*c^2*d-6*a^2*c^2*d)/d/c^2/e^4/(e*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{\sqrt{d x^{2} + c} \left (e x\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(9/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(9/2)), 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 e^{5} x^{7} + c e^{5} x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(9/2)/(d*x^2+c)^(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*e^5*x^7 + c*e^5*x^5), 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/(e*x)**(9/2)/(d*x**2+c)**(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}}{\sqrt{d x^{2} + c} \left (e x\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(9/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(9/2)), x)