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

Optimal. Leaf size=436 \[ \frac{\sqrt [4]{c} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{30 d^{15/4} \sqrt{c+d x^2}}+\frac{e^2 \sqrt{e x} \sqrt{c+d x^2} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right )}{15 d^{7/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{\sqrt [4]{c} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{15/4} \sqrt{c+d x^2}}-\frac{e (e x)^{3/2} \sqrt{c+d x^2} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right )}{45 c d^3}+\frac{(e x)^{7/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e} \]

[Out]

((b*c - a*d)^2*(e*x)^(7/2))/(c*d^2*e*Sqrt[c + d*x^2]) - ((77*b^2*c^2 - 126*a*b*c*d + 45*a^2*d^2)*e*(e*x)^(3/2)
*Sqrt[c + d*x^2])/(45*c*d^3) + (2*b^2*(e*x)^(7/2)*Sqrt[c + d*x^2])/(9*d^2*e) + ((77*b^2*c^2 - 126*a*b*c*d + 45
*a^2*d^2)*e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) - (c^(1/4)*(77*b^2*c^2 - 126*a*b*c
*d + 45*a^2*d^2)*e^(5/2)*(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*d^(15/4)*Sqrt[c + d*x^2]) + (c^(1/4)*(77*b^2*c^2 - 126*a*b*c*d
+ 45*a^2*d^2)*e^(5/2)*(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])/(30*d^(15/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.378029, antiderivative size = 436, 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 = {463, 459, 321, 329, 305, 220, 1196} \[ \frac{e^2 \sqrt{e x} \sqrt{c+d x^2} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right )}{15 d^{7/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\sqrt [4]{c} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{30 d^{15/4} \sqrt{c+d x^2}}-\frac{\sqrt [4]{c} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{15/4} \sqrt{c+d x^2}}-\frac{e (e x)^{3/2} \sqrt{c+d x^2} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right )}{45 c d^3}+\frac{(e x)^{7/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(e*x)^(7/2))/(c*d^2*e*Sqrt[c + d*x^2]) - ((77*b^2*c^2 - 126*a*b*c*d + 45*a^2*d^2)*e*(e*x)^(3/2)
*Sqrt[c + d*x^2])/(45*c*d^3) + (2*b^2*(e*x)^(7/2)*Sqrt[c + d*x^2])/(9*d^2*e) + ((77*b^2*c^2 - 126*a*b*c*d + 45
*a^2*d^2)*e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) - (c^(1/4)*(77*b^2*c^2 - 126*a*b*c
*d + 45*a^2*d^2)*e^(5/2)*(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*d^(15/4)*Sqrt[c + d*x^2]) + (c^(1/4)*(77*b^2*c^2 - 126*a*b*c*d
+ 45*a^2*d^2)*e^(5/2)*(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])/(30*d^(15/4)*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 459

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{(e x)^{5/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\int \frac{(e x)^{5/2} \left (\frac{1}{2} \left (-2 a^2 d^2+7 (b c-a d)^2\right )-b^2 c d x^2\right )}{\sqrt{c+d x^2}} \, dx}{c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e}-\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) \int \frac{(e x)^{5/2}}{\sqrt{c+d x^2}} \, dx}{18 c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{45 c d^3}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e}+\frac{\left (\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^2\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{30 d^3}\\ &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{45 c d^3}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e}+\frac{\left (\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 d^3}\\ &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{45 c d^3}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e}+\frac{\left (\sqrt{c} \left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 d^{7/2}}-\frac{\left (\sqrt{c} \left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^2\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 )}{15 d^{7/2}}\\ &=\frac{(b c-a d)^2 (e x)^{7/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{45 c d^3}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d^2 e}+\frac{\left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^2 \sqrt{e x} \sqrt{c+d x^2}}{15 d^{7/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{\sqrt [4]{c} \left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^{5/2} \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 d^{15/4} \sqrt{c+d x^2}}+\frac{\sqrt [4]{c} \left (77 b^2 c^2-126 a b c d+45 a^2 d^2\right ) e^{5/2} \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 )}{30 d^{15/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.138021, size = 133, normalized size = 0.31 \[ \frac{e (e x)^{3/2} \left (3 \sqrt{\frac{c}{d x^2}+1} \left (45 a^2 d^2-126 a b c d+77 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )-45 a^2 d^2+18 a b d \left (7 c+2 d x^2\right )+b^2 \left (-77 c^2-22 c d x^2+10 d^2 x^4\right )\right )}{45 d^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(e*x)^(3/2)*(-45*a^2*d^2 + 18*a*b*d*(7*c + 2*d*x^2) + b^2*(-77*c^2 - 22*c*d*x^2 + 10*d^2*x^4) + 3*(77*b^2*c
^2 - 126*a*b*c*d + 45*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(45*d^3*S
qrt[c + d*x^2])

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Maple [A]  time = 0.036, size = 618, normalized size = 1.4 \begin{align*}{\frac{{e}^{2}}{90\,x{d}^{4}}\sqrt{ex} \left ( 20\,{x}^{6}{b}^{2}{d}^{3}+270\,\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 ){a}^{2}c{d}^{2}-756\,\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 ) ab{c}^{2}d+462\,\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 ){b}^{2}{c}^{3}-135\,\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 ){a}^{2}c{d}^{2}+378\,\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 ) ab{c}^{2}d-231\,\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 ){b}^{2}{c}^{3}+72\,{x}^{4}ab{d}^{3}-44\,{x}^{4}{b}^{2}c{d}^{2}-90\,{x}^{2}{a}^{2}{d}^{3}+252\,{x}^{2}abc{d}^{2}-154\,{x}^{2}{b}^{2}{c}^{2}d \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90/x*e^2*(e*x)^(1/2)*(20*x^6*b^2*d^3+270*((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))
*a^2*c*d^2-756*((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))*a*b*c^2*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)*Elliptic
E(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^3-135*((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))*a^2*c*d^2+378*((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))*a*b
*c^2*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))*b^2*c^3+72*x^4*a*b*d^3-44*x^4*b
^2*c*d^2-90*x^2*a^2*d^3+252*x^2*a*b*c*d^2-154*x^2*b^2*c^2*d)/(d*x^2+c)^(1/2)/d^4

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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 (e x\right )^{\frac{5}{2}}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} e^{2} x^{6} + 2 \, a b e^{2} x^{4} + a^{2} e^{2} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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