3.864 \(\int \frac{(a+b x^2)^2}{(e x)^{7/2} (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=489 \[ -\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{20 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{\sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{10 c^4 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{(e x)^{3/2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{10 c^4 e^5 \sqrt{c+d x^2}}+\frac{(e x)^{3/2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{10 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}} \]

[Out]

(-2*a^2)/(5*c*e*(e*x)^(5/2)*(c + d*x^2)^(3/2)) - (2*a*(10*b*c - 11*a*d))/(5*c^2*e^3*Sqrt[e*x]*(c + d*x^2)^(3/2
)) + ((5*b^2*c^2 - 70*a*b*c*d + 77*a^2*d^2)*(e*x)^(3/2))/(15*c^3*e^5*(c + d*x^2)^(3/2)) + ((5*b^2*c^2 - 70*a*b
*c*d + 77*a^2*d^2)*(e*x)^(3/2))/(10*c^4*e^5*Sqrt[c + d*x^2]) - ((5*b^2*c^2 - 70*a*b*c*d + 77*a^2*d^2)*Sqrt[e*x
]*Sqrt[c + d*x^2])/(10*c^4*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) + ((5*b^2*c^2 - 70*a*b*c*d + 77*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])/(10*c^(15/4)*d^(3/4)*e^(7/2)*Sqrt[c + d*x^2]) - ((5*b^2*c^2 - 70*a*b*c*d + 77*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])/(20*c^(15/4)*d^(3/4)*e^(7/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.469583, antiderivative size = 489, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {462, 453, 290, 329, 305, 220, 1196} \[ -\frac{\sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{10 c^4 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{(e x)^{3/2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{10 c^4 e^5 \sqrt{c+d x^2}}+\frac{(e x)^{3/2} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right )}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}-\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{20 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{10 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(7/2)*(c + d*x^2)^(5/2)),x]

[Out]

(-2*a^2)/(5*c*e*(e*x)^(5/2)*(c + d*x^2)^(3/2)) - (2*a*(10*b*c - 11*a*d))/(5*c^2*e^3*Sqrt[e*x]*(c + d*x^2)^(3/2
)) + ((5*b^2*c^2 - 70*a*b*c*d + 77*a^2*d^2)*(e*x)^(3/2))/(15*c^3*e^5*(c + d*x^2)^(3/2)) + ((5*b^2*c^2 - 70*a*b
*c*d + 77*a^2*d^2)*(e*x)^(3/2))/(10*c^4*e^5*Sqrt[c + d*x^2]) - ((5*b^2*c^2 - 70*a*b*c*d + 77*a^2*d^2)*Sqrt[e*x
]*Sqrt[c + d*x^2])/(10*c^4*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) + ((5*b^2*c^2 - 70*a*b*c*d + 77*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])/(10*c^(15/4)*d^(3/4)*e^(7/2)*Sqrt[c + d*x^2]) - ((5*b^2*c^2 - 70*a*b*c*d + 77*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])/(20*c^(15/4)*d^(3/4)*e^(7/2)*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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -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)^{7/2} \left (c+d x^2\right )^{5/2}} \, dx &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}+\frac{2 \int \frac{\frac{1}{2} a (10 b c-11 a d)+\frac{5}{2} b^2 c x^2}{(e x)^{3/2} \left (c+d x^2\right )^{5/2}} \, dx}{5 c e^2}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}-\frac{\left (4 \left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 a d)\right )\right ) \int \frac{\sqrt{e x}}{\left (c+d x^2\right )^{5/2}} \, dx}{5 c^2 e^4}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}-\frac{\left (2 \left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 a d)\right )\right ) \int \frac{\sqrt{e x}}{\left (c+d x^2\right )^{3/2}} \, dx}{5 c^3 e^4}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{10 c^4 e^5 \sqrt{c+d x^2}}+\frac{\left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 a d)\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{5 c^4 e^4}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{10 c^4 e^5 \sqrt{c+d x^2}}+\frac{\left (2 \left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 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 )}{5 c^4 e^5}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{10 c^4 e^5 \sqrt{c+d x^2}}+\frac{\left (2 \left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 c^{7/2} \sqrt{d} e^4}-\frac{\left (2 \left (-\frac{5}{4} b^2 c^2+\frac{7}{4} a d (10 b c-11 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 )}{5 c^{7/2} \sqrt{d} e^4}\\ &=-\frac{2 a^2}{5 c e (e x)^{5/2} \left (c+d x^2\right )^{3/2}}-\frac{2 a (10 b c-11 a d)}{5 c^2 e^3 \sqrt{e x} \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{15 c^3 e^5 \left (c+d x^2\right )^{3/2}}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) (e x)^{3/2}}{10 c^4 e^5 \sqrt{c+d x^2}}-\frac{\left (5 b^2 c^2-7 a d (10 b c-11 a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{10 c^4 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\left (5 b^2 c^2-7 a d (10 b c-11 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 )}{10 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{\left (5 b^2 c^2-7 a d (10 b c-11 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 )}{20 c^{15/4} d^{3/4} e^{7/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.175168, size = 181, normalized size = 0.37 \[ \frac{x \left (-x^4 \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \left (77 a^2 d^2-70 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{d x^2}{c}\right )+a^2 \left (132 c^2 d x^2-12 c^3+385 c d^2 x^4+231 d^3 x^6\right )-10 a b c x^2 \left (12 c^2+35 c d x^2+21 d^2 x^4\right )+5 b^2 c^2 x^4 \left (5 c+3 d x^2\right )\right )}{30 c^4 (e x)^{7/2} \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(7/2)*(c + d*x^2)^(5/2)),x]

[Out]

(x*(5*b^2*c^2*x^4*(5*c + 3*d*x^2) - 10*a*b*c*x^2*(12*c^2 + 35*c*d*x^2 + 21*d^2*x^4) + a^2*(-12*c^3 + 132*c^2*d
*x^2 + 385*c*d^2*x^4 + 231*d^3*x^6) - (5*b^2*c^2 - 70*a*b*c*d + 77*a^2*d^2)*x^4*(c + d*x^2)*Sqrt[1 + (d*x^2)/c
]*Hypergeometric2F1[1/2, 3/4, 7/4, -((d*x^2)/c)]))/(30*c^4*(e*x)^(7/2)*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.03, size = 1231, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(462*((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^3-420*((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)*Elliptic
E(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b*c^2*d^2+30*((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*d-231*((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^3+210*((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^2-
15*((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*d+462*((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^2*a^2*c^2*d^2-420*((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^2*a*b*c^3*d+30*((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^2*b^2*c^4-231*((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^2*a^2*c^2*d^2+210*((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)*E
llipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b*c^3*d-15*((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^2*b^2*c^4-462*x^6*a^2*d^4+420*x^6*a*b*c*d^3-30*x^6*b^2*c^2*d^2-770*x^4*
a^2*c*d^3+700*x^4*a*b*c^2*d^2-50*x^4*b^2*c^3*d-264*x^2*a^2*c^2*d^2+240*x^2*a*b*c^3*d+24*a^2*c^3*d)/x^2/d/c^4/e
^3/(e*x)^(1/2)/(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}} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(7/2)), 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^{3} e^{4} x^{10} + 3 \, c d^{2} e^{4} x^{8} + 3 \, c^{2} d e^{4} x^{6} + c^{3} e^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(5/2),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^4*x^10 + 3*c*d^2*e^4*x^8 + 3*c^2*d*e^4*x
^6 + c^3*e^4*x^4), 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)**(7/2)/(d*x**2+c)**(5/2),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}} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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