3.847 \(\int \frac{(a+b x^2)^2}{(e x)^{11/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=438 \[ \frac{\sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}-\frac{2 \sqrt{c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^5 \sqrt{e x}}+\frac{2 \sqrt{d} \sqrt{e x} \sqrt{c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^6 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a \sqrt{c+d x^2} (18 b c-7 a d)}{45 c^2 e^3 (e x)^{5/2}} \]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(9*c*e*(e*x)^(9/2)) - (2*a*(18*b*c - 7*a*d)*Sqrt[c + d*x^2])/(45*c^2*e^3*(e*x)^(5/2))
- (2*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*Sqrt[c + d*x^2])/(15*c^3*e^5*Sqrt[e*x]) + (2*Sqrt[d]*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*c^3*e^6*(Sqrt[c] + Sqrt[d]*x)) - (2*d^(1/4)*(15*b^2*c^
2 - 18*a*b*c*d + 7*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan
[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*c^(11/4)*e^(11/2)*Sqrt[c + d*x^2]) + (d^(1/4)*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*(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])/(15*c^(11/4)*e^(11/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.423363, antiderivative size = 438, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {462, 453, 325, 329, 305, 220, 1196} \[ -\frac{2 \sqrt{c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^5 \sqrt{e x}}+\frac{2 \sqrt{d} \sqrt{e x} \sqrt{c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^6 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}-\frac{2 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a \sqrt{c+d x^2} (18 b c-7 a d)}{45 c^2 e^3 (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(11/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(9*c*e*(e*x)^(9/2)) - (2*a*(18*b*c - 7*a*d)*Sqrt[c + d*x^2])/(45*c^2*e^3*(e*x)^(5/2))
- (2*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*Sqrt[c + d*x^2])/(15*c^3*e^5*Sqrt[e*x]) + (2*Sqrt[d]*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*c^3*e^6*(Sqrt[c] + Sqrt[d]*x)) - (2*d^(1/4)*(15*b^2*c^
2 - 18*a*b*c*d + 7*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan
[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*c^(11/4)*e^(11/2)*Sqrt[c + d*x^2]) + (d^(1/4)*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*(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])/(15*c^(11/4)*e^(11/2)*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 325

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

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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt{c+d x^2}} \, dx &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}+\frac{2 \int \frac{\frac{1}{2} a (18 b c-7 a d)+\frac{9}{2} b^2 c x^2}{(e x)^{7/2} \sqrt{c+d x^2}} \, dx}{9 c e^2}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a (18 b c-7 a d) \sqrt{c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac{\left (4 \left (-\frac{45}{4} b^2 c^2+\frac{3}{4} a d (18 b c-7 a d)\right )\right ) \int \frac{1}{(e x)^{3/2} \sqrt{c+d x^2}} \, dx}{45 c^2 e^4}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a (18 b c-7 a d) \sqrt{c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac{2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt{c+d x^2}}{15 c^3 e^5 \sqrt{e x}}-\frac{\left (4 d \left (-\frac{45}{4} b^2 c^2+\frac{3}{4} a d (18 b c-7 a d)\right )\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{45 c^3 e^6}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a (18 b c-7 a d) \sqrt{c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac{2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt{c+d x^2}}{15 c^3 e^5 \sqrt{e x}}-\frac{\left (8 d \left (-\frac{45}{4} b^2 c^2+\frac{3}{4} a d (18 b c-7 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{45 c^3 e^7}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a (18 b c-7 a d) \sqrt{c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac{2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt{c+d x^2}}{15 c^3 e^5 \sqrt{e x}}-\frac{\left (8 \sqrt{d} \left (-\frac{45}{4} b^2 c^2+\frac{3}{4} a d (18 b c-7 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{45 c^{5/2} e^6}+\frac{\left (8 \sqrt{d} \left (-\frac{45}{4} b^2 c^2+\frac{3}{4} a d (18 b c-7 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{d} x^2}{\sqrt{c} e}}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{45 c^{5/2} e^6}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{9 c e (e x)^{9/2}}-\frac{2 a (18 b c-7 a d) \sqrt{c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac{2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt{c+d x^2}}{15 c^3 e^5 \sqrt{e x}}+\frac{2 \sqrt{d} \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{15 c^3 e^6 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 \sqrt [4]{d} \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}+\frac{\sqrt [4]{d} \left (15 b^2 c^2-a d (18 b c-7 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 )}{15 c^{11/4} e^{11/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.182344, size = 155, normalized size = 0.35 \[ \frac{2 \sqrt{e x} \left (d x^6 \sqrt{\frac{d x^2}{c}+1} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{d x^2}{c}\right )-\left (c+d x^2\right ) \left (a^2 \left (5 c^2-7 c d x^2+21 d^2 x^4\right )+18 a b c x^2 \left (c-3 d x^2\right )+45 b^2 c^2 x^4\right )\right )}{45 c^3 e^6 x^5 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(11/2)*Sqrt[c + d*x^2]),x]

[Out]

(2*Sqrt[e*x]*(-((c + d*x^2)*(45*b^2*c^2*x^4 + 18*a*b*c*x^2*(c - 3*d*x^2) + a^2*(5*c^2 - 7*c*d*x^2 + 21*d^2*x^4
))) + d*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*x^6*Sqrt[1 + (d*x^2)/c]*Hypergeometric2F1[1/2, 3/4, 7/4, -((d*x^
2)/c)]))/(45*c^3*e^6*x^5*Sqrt[c + d*x^2])

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Maple [A]  time = 0.034, size = 667, normalized size = 1.5 \begin{align*}{\frac{1}{45\,{x}^{4}{e}^{5}{c}^{3}} \left ( 42\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{a}^{2}c{d}^{2}-108\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}ab{c}^{2}d+90\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{b}^{2}{c}^{3}-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 ){x}^{4}{a}^{2}c{d}^{2}+54\,\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}^{4}ab{c}^{2}d-45\,\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}^{4}{b}^{2}{c}^{3}-42\,{x}^{6}{a}^{2}{d}^{3}+108\,{x}^{6}abc{d}^{2}-90\,{x}^{6}{b}^{2}{c}^{2}d-28\,{x}^{4}{a}^{2}c{d}^{2}+72\,{x}^{4}ab{c}^{2}d-90\,{x}^{4}{b}^{2}{c}^{3}+4\,{x}^{2}{a}^{2}{c}^{2}d-36\,{x}^{2}ab{c}^{3}-10\,{a}^{2}{c}^{3} \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)^(11/2)/(d*x^2+c)^(1/2),x)

[Out]

1/45/x^4*(42*((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)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a^2*c*d^2-108*((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
icE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b*c^2*d+90*((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)*EllipticE(((d*x+(-c*d)^(1/2))/(
-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c^3-21*((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))*x^4*a^2*c*d^2+54*((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^4*a*b*c^2*d-45*((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^4*b^2*c^3-42*x^6*a^2*d^3+108*x^6*a*b*c*d^2-
90*x^6*b^2*c^2*d-28*x^4*a^2*c*d^2+72*x^4*a*b*c^2*d-90*x^4*b^2*c^3+4*x^2*a^2*c^2*d-36*x^2*a*b*c^3-10*a^2*c^3)/(
d*x^2+c)^(1/2)/e^5/(e*x)^(1/2)/c^3

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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{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(11/2)), 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^{6} x^{8} + c e^{6} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(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^6*x^8 + c*e^6*x^6), 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)**(11/2)/(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{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(11/2)), x)