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

Optimal. Leaf size=248 \[ -\frac{e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 c^{5/4} d^{13/4} \sqrt{c+d x^2}}+\frac{e \sqrt{e x} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right )}{6 c d^3 \sqrt{c+d x^2}}+\frac{(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}} \]

[Out]

((b*c - a*d)^2*(e*x)^(5/2))/(3*c*d^2*e*(c + d*x^2)^(3/2)) + ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*e*Sqrt[e*x])/
(6*c*d^3*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(5/2))/(3*d^2*e*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^
2)*e^(3/2)*(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])/(12*c^(5/4)*d^(13/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.194239, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {463, 459, 288, 329, 220} \[ -\frac{e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{5/4} d^{13/4} \sqrt{c+d x^2}}+\frac{e \sqrt{e x} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right )}{6 c d^3 \sqrt{c+d x^2}}+\frac{(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(e*x)^(5/2))/(3*c*d^2*e*(c + d*x^2)^(3/2)) + ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*e*Sqrt[e*x])/
(6*c*d^3*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(5/2))/(3*d^2*e*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^
2)*e^(3/2)*(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])/(12*c^(5/4)*d^(13/4)*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx &=\frac{(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{(e x)^{3/2} \left (\frac{1}{2} \left (-6 a^2 d^2+5 (b c-a d)^2\right )-3 b^2 c d x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}}-\frac{\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) \int \frac{(e x)^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{6 c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e \sqrt{e x}}{6 c d^3 \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}}-\frac{\left (\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{12 c d^3}\\ &=\frac{(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e \sqrt{e x}}{6 c d^3 \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}}-\frac{\left (\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{6 c d^3}\\ &=\frac{(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e \sqrt{e x}}{6 c d^3 \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}}-\frac{\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e^{3/2} \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 )}{12 c^{5/4} d^{13/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.253136, size = 204, normalized size = 0.82 \[ \frac{(e x)^{3/2} \left (\frac{\sqrt{x} \left (a^2 d^2 \left (d x^2-c\right )-2 a b c d \left (5 c+7 d x^2\right )+b^2 c \left (15 c^2+21 c d x^2+4 d^2 x^4\right )\right )}{c d^3 \left (c+d x^2\right )}+\frac{i x \sqrt{\frac{c}{d x^2}+1} \left (a^2 d^2+10 a b c d-15 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{c d^3 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{6 x^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(3/2)*((Sqrt[x]*(a^2*d^2*(-c + d*x^2) - 2*a*b*c*d*(5*c + 7*d*x^2) + b^2*c*(15*c^2 + 21*c*d*x^2 + 4*d^2*
x^4)))/(c*d^3*(c + d*x^2)) + (I*(-15*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh
[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(c*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)))/(6*x^(3/2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.022, size = 674, normalized size = 2.7 \begin{align*}{\frac{e}{12\,cx{d}^{4}} \left ( \sqrt{{ \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{2}\sqrt{{ \left ( -dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{-{dx{\frac{1}{\sqrt{-cd}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-cd}{x}^{2}{a}^{2}{d}^{3}+10\,\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}{x}^{2}abc{d}^{2}-15\,\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}{x}^{2}{b}^{2}{c}^{2}d+\sqrt{{ \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{2}\sqrt{{ \left ( -dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{-{dx{\frac{1}{\sqrt{-cd}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-cd}{a}^{2}c{d}^{2}+10\,\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-15\,\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}+8\,{x}^{5}{b}^{2}c{d}^{3}+2\,{x}^{3}{a}^{2}{d}^{4}-28\,{x}^{3}abc{d}^{3}+42\,{x}^{3}{b}^{2}{c}^{2}{d}^{2}-2\,x{a}^{2}c{d}^{3}-20\,xab{c}^{2}{d}^{2}+30\,x{b}^{2}{c}^{3}d \right ) \sqrt{ex} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(((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)*(-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)*x^2*a^2*d^3+10*((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)*Ell
ipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*x^2*a*b*c*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))*(-c*d)^(1/2)*x^2*b^2*c^2*d+((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+10*((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-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))
*(-c*d)^(1/2)*b^2*c^3+8*x^5*b^2*c*d^3+2*x^3*a^2*d^4-28*x^3*a*b*c*d^3+42*x^3*b^2*c^2*d^2-2*x*a^2*c*d^3-20*x*a*b
*c^2*d^2+30*x*b^2*c^3*d)*e/x*(e*x)^(1/2)/c/d^4/(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 (e x\right )^{\frac{3}{2}}}{{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} e x^{5} + 2 \, a b e x^{3} + a^{2} e x\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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