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

Optimal. Leaf size=442 \[ -\frac {(e x)^{3/2} \left (7 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (2 b c-7 a d)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{4 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (2 b c-7 a d)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{2 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}-\frac {\sqrt {e x} \sqrt {c+d x^2} \left (a d (2 b c-7 a d)+b^2 c^2\right )}{2 c^3 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {(e x)^{3/2} \left (a d (2 b c-7 a d)+b^2 c^2\right )}{2 c^3 d e^3 \sqrt {c+d x^2}} \]

[Out]

-1/3*(7*a^2*d^2-2*a*b*c*d+b^2*c^2)*(e*x)^(3/2)/c^2/d/e^3/(d*x^2+c)^(3/2)-2*a^2/c/e/(d*x^2+c)^(3/2)/(e*x)^(1/2)
+1/2*(b^2*c^2+a*d*(-7*a*d+2*b*c))*(e*x)^(3/2)/c^3/d/e^3/(d*x^2+c)^(1/2)-1/2*(b^2*c^2+a*d*(-7*a*d+2*b*c))*(e*x)
^(1/2)*(d*x^2+c)^(1/2)/c^3/d^(3/2)/e^2/(c^(1/2)+x*d^(1/2))+1/2*(b^2*c^2+a*d*(-7*a*d+2*b*c))*(cos(2*arctan(d^(1
/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticE(sin(2*
arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2
)^(1/2)/c^(11/4)/d^(7/4)/e^(3/2)/(d*x^2+c)^(1/2)-1/4*(b^2*c^2+a*d*(-7*a*d+2*b*c))*(cos(2*arctan(d^(1/4)*(e*x)^
(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(d^(
1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^
(11/4)/d^(7/4)/e^(3/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {462, 457, 290, 329, 305, 220, 1196} \[ -\frac {(e x)^{3/2} \left (7 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\sqrt {e x} \sqrt {c+d x^2} \left (a d (2 b c-7 a d)+b^2 c^2\right )}{2 c^3 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (2 b c-7 a d)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{4 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (2 b c-7 a d)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{2 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}+\frac {(e x)^{3/2} \left (a d (2 b c-7 a d)+b^2 c^2\right )}{2 c^3 d e^3 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(c*e*Sqrt[e*x]*(c + d*x^2)^(3/2)) - ((b^2*c^2 - 2*a*b*c*d + 7*a^2*d^2)*(e*x)^(3/2))/(3*c^2*d*e^3*(c +
d*x^2)^(3/2)) + ((b^2*c^2 + a*d*(2*b*c - 7*a*d))*(e*x)^(3/2))/(2*c^3*d*e^3*Sqrt[c + d*x^2]) - ((b^2*c^2 + a*d
*(2*b*c - 7*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(2*c^3*d^(3/2)*e^2*(Sqrt[c] + Sqrt[d]*x)) + ((b^2*c^2 + a*d*(2*b*
c - 7*a*d))*(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])/(2*c^(11/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2]) - ((b^2*c^2 + a*d*(2*b*c - 7*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])/(4*c^(11/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2])

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 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 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 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 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 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 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)^{3/2} \left (c+d x^2\right )^{5/2}} \, dx &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}+\frac {2 \int \frac {\sqrt {e x} \left (\frac {1}{2} a (2 b c-7 a d)+\frac {1}{2} b^2 c x^2\right )}{\left (c+d x^2\right )^{5/2}} \, dx}{c e^2}\\ &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) (e x)^{3/2}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) \int \frac {\sqrt {e x}}{\left (c+d x^2\right )^{3/2}} \, dx}{2 c^2 d e^2}\\ &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) (e x)^{3/2}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) (e x)^{3/2}}{2 c^3 d e^3 \sqrt {c+d x^2}}-\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{4 c^3 d e^2}\\ &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) (e x)^{3/2}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) (e x)^{3/2}}{2 c^3 d e^3 \sqrt {c+d x^2}}-\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{2 c^3 d e^3}\\ &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) (e x)^{3/2}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) (e x)^{3/2}}{2 c^3 d e^3 \sqrt {c+d x^2}}-\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{2 c^{5/2} d^{3/2} e^2}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\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 )}{2 c^{5/2} d^{3/2} e^2}\\ &=-\frac {2 a^2}{c e \sqrt {e x} \left (c+d x^2\right )^{3/2}}-\frac {\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) (e x)^{3/2}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) (e x)^{3/2}}{2 c^3 d e^3 \sqrt {c+d x^2}}-\frac {\left (b^2 c^2+a d (2 b c-7 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{2 c^3 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {\left (b^2 c^2+a d (2 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 )}{2 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}-\frac {\left (b^2 c^2+a d (2 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 )}{4 c^{11/4} d^{7/4} e^{3/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 161, normalized size = 0.36 \[ \frac {x \left (-x^2 \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \left (-7 a^2 d^2+2 a b c d+b^2 c^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {d x^2}{c}\right )+a^2 (-d) \left (12 c^2+35 c d x^2+21 d^2 x^4\right )+2 a b c d x^2 \left (5 c+3 d x^2\right )+b^2 c^2 x^2 \left (c+3 d x^2\right )\right )}{6 c^3 d (e x)^{3/2} \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d^{3} e^{2} x^{8} + 3 \, c d^{2} e^{2} x^{6} + 3 \, c^{2} d e^{2} x^{4} + c^{3} e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(3/2)/(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^2*x^8 + 3*c*d^2*e^2*x^6 + 3*c^2*d*e^2*x^
4 + c^3*e^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 1187, normalized size = 2.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(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)*(-1/(-c*d)^(
1/2)*d*x)^(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*d^3-12*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*Ellipti
cE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b*c^2*d^2-6*((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)*(-1/(-c*d)^(1/2)*d*x)^(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^3*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)*(-1/(-c*d)^(1/2)*d*x)^(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*d^3+6*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b*c^2*d
^2+3*((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)*(-1/(-c*d)^(1/2)
*d*x)^(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^3*d+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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE((
(d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c^2*d^2-12*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)
^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^3*d-6*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b
^2*c^4-21*((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)*(-1/(-c*d)^
(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c^2*d^2+6*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF
(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^3*d+3*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^
(1/2))^(1/2),1/2*2^(1/2))*b^2*c^4-42*a^2*d^4*x^4+12*a*b*c*d^3*x^4+6*b^2*c^2*d^2*x^4-70*a^2*c*d^3*x^2+20*a*b*c^
2*d^2*x^2+2*b^2*c^3*d*x^2-24*a^2*c^2*d^2)/d^2/c^3/e/(e*x)^(1/2)/(d*x^2+c)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{3/2}\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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