3.849 \(\int \frac{(e x)^{7/2} (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=296 \[ -\frac{5 c^{3/4} e^{7/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{462 d^{17/4} \sqrt{c+d x^2}}+\frac{5 e^3 \sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right )}{231 d^4}-\frac{e (e x)^{5/2} \sqrt{c+d x^2} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right )}{77 c d^3}+\frac{(e x)^{9/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e} \]

[Out]

((b*c - a*d)^2*(e*x)^(9/2))/(c*d^2*e*Sqrt[c + d*x^2]) + (5*(117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*e^3*Sqrt[e
*x]*Sqrt[c + d*x^2])/(231*d^4) - ((117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*e*(e*x)^(5/2)*Sqrt[c + d*x^2])/(77*
c*d^3) + (2*b^2*(e*x)^(9/2)*Sqrt[c + d*x^2])/(11*d^2*e) - (5*c^(3/4)*(117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*
e^(7/2)*(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])/(462*d^(17/4)*Sqrt[c + d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.248713, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {463, 459, 321, 329, 220} \[ \frac{5 e^3 \sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right )}{231 d^4}-\frac{5 c^{3/4} e^{7/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{462 d^{17/4} \sqrt{c+d x^2}}-\frac{e (e x)^{5/2} \sqrt{c+d x^2} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right )}{77 c d^3}+\frac{(e x)^{9/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

((b*c - a*d)^2*(e*x)^(9/2))/(c*d^2*e*Sqrt[c + d*x^2]) + (5*(117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*e^3*Sqrt[e
*x]*Sqrt[c + d*x^2])/(231*d^4) - ((117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*e*(e*x)^(5/2)*Sqrt[c + d*x^2])/(77*
c*d^3) + (2*b^2*(e*x)^(9/2)*Sqrt[c + d*x^2])/(11*d^2*e) - (5*c^(3/4)*(117*b^2*c^2 - 198*a*b*c*d + 77*a^2*d^2)*
e^(7/2)*(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])/(462*d^(17/4)*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 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{(e x)^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 (e x)^{9/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\int \frac{(e x)^{7/2} \left (\frac{1}{2} \left (-2 a^2 d^2+9 (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)^{9/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e}-\frac{\left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) \int \frac{(e x)^{7/2}}{\sqrt{c+d x^2}} \, dx}{22 c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{9/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e (e x)^{5/2} \sqrt{c+d x^2}}{77 c d^3}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e}+\frac{\left (5 \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^2\right ) \int \frac{(e x)^{3/2}}{\sqrt{c+d x^2}} \, dx}{154 d^3}\\ &=\frac{(b c-a d)^2 (e x)^{9/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{5 \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^3 \sqrt{e x} \sqrt{c+d x^2}}{231 d^4}-\frac{\left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e (e x)^{5/2} \sqrt{c+d x^2}}{77 c d^3}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e}-\frac{\left (5 c \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^4\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{462 d^4}\\ &=\frac{(b c-a d)^2 (e x)^{9/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{5 \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^3 \sqrt{e x} \sqrt{c+d x^2}}{231 d^4}-\frac{\left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e (e x)^{5/2} \sqrt{c+d x^2}}{77 c d^3}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e}-\frac{\left (5 c \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{231 d^4}\\ &=\frac{(b c-a d)^2 (e x)^{9/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{5 \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^3 \sqrt{e x} \sqrt{c+d x^2}}{231 d^4}-\frac{\left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e (e x)^{5/2} \sqrt{c+d x^2}}{77 c d^3}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d^2 e}-\frac{5 c^{3/4} \left (117 b^2 c^2-198 a b c d+77 a^2 d^2\right ) e^{7/2} \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 )}{462 d^{17/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.243443, size = 226, normalized size = 0.76 \[ \frac{e^3 \sqrt{e x} \left (\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} \left (77 a^2 d^2 \left (5 c+2 d x^2\right )+66 a b d \left (-15 c^2-6 c d x^2+2 d^2 x^4\right )+3 b^2 \left (78 c^2 d x^2+195 c^3-26 c d^2 x^4+14 d^3 x^6\right )\right )-5 i c \sqrt{x} \sqrt{\frac{c}{d x^2}+1} \left (77 a^2 d^2-198 a b c d+117 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )\right )}{231 d^4 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

(e^3*Sqrt[e*x]*(Sqrt[(I*Sqrt[c])/Sqrt[d]]*(77*a^2*d^2*(5*c + 2*d*x^2) + 66*a*b*d*(-15*c^2 - 6*c*d*x^2 + 2*d^2*
x^4) + 3*b^2*(195*c^3 + 78*c^2*d*x^2 - 26*c*d^2*x^4 + 14*d^3*x^6)) - (5*I)*c*(117*b^2*c^2 - 198*a*b*c*d + 77*a
^2*d^2)*Sqrt[1 + c/(d*x^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1]))/(231*Sqrt[(I
*Sqrt[c])/Sqrt[d]]*d^4*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.037, size = 407, normalized size = 1.4 \begin{align*} -{\frac{{e}^{3}}{462\,x{d}^{5}}\sqrt{ex} \left ( -84\,{x}^{7}{b}^{2}{d}^{4}+385\,\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}{a}^{2}c{d}^{2}-990\,\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}ab{c}^{2}d+585\,\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}{b}^{2}{c}^{3}-264\,{x}^{5}ab{d}^{4}+156\,{x}^{5}{b}^{2}c{d}^{3}-308\,{x}^{3}{a}^{2}{d}^{4}+792\,{x}^{3}abc{d}^{3}-468\,{x}^{3}{b}^{2}{c}^{2}{d}^{2}-770\,x{a}^{2}c{d}^{3}+1980\,xab{c}^{2}{d}^{2}-1170\,x{b}^{2}{c}^{3}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)^(7/2)*(b*x^2+a)^2/(d*x^2+c)^(3/2),x)

[Out]

-1/462*e^3/x*(e*x)^(1/2)*(-84*x^7*b^2*d^4+385*((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)*a^2*c*d^2-990*((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
)*a*b*c^2*d+585*((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)*b^2*c^3-264*x
^5*a*b*d^4+156*x^5*b^2*c*d^3-308*x^3*a^2*d^4+792*x^3*a*b*c*d^3-468*x^3*b^2*c^2*d^2-770*x*a^2*c*d^3+1980*x*a*b*
c^2*d^2-1170*x*b^2*c^3*d)/(d*x^2+c)^(1/2)/d^5

________________________________________________________________________________________

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{7}{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)^(7/2)*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(7/2)/(d*x^2 + c)^(3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} e^{3} x^{7} + 2 \, a b e^{3} x^{5} + a^{2} e^{3} x^{3}\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)^(7/2)*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*e^3*x^7 + 2*a*b*e^3*x^5 + a^2*e^3*x^3)*sqrt(d*x^2 + c)*sqrt(e*x)/(d^2*x^4 + 2*c*d*x^2 + c^2), x)

________________________________________________________________________________________

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)**(7/2)*(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{7}{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)^(7/2)*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(7/2)/(d*x^2 + c)^(3/2), x)