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

Optimal. Leaf size=258 \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a d (2 b c-3 a d)+b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{\sqrt{e x} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac{\sqrt{e x} \left (5 a d (2 b c-3 a d)+b^2 c^2\right )}{6 c^3 d e^3 \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.234409, antiderivative size = 258, 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 = {462, 457, 290, 329, 220} \[ -\frac{\sqrt{e x} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a d (2 b c-3 a d)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}+\frac{\sqrt{e x} \left (5 a d (2 b c-3 a d)+b^2 c^2\right )}{6 c^3 d e^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*(c + d*x^2)^(3/2)) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*Sqrt[e*x])/(3*c^2*d*e^3*(c
+ d*x^2)^(3/2)) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*d))*Sqrt[e*x])/(6*c^3*d*e^3*Sqrt[c + d*x^2]) + ((b^2*c^2 + 5
*a*d*(2*b*c - 3*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])/(12*c^(13/4)*d^(5/4)*e^(5/2)*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 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 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 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{\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{5/2}} \, dx &=-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac{2 \int \frac{\frac{3}{2} a (2 b c-3 a d)+\frac{3}{2} b^2 c x^2}{\sqrt{e x} \left (c+d x^2\right )^{5/2}} \, dx}{3 c e^2}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \int \frac{1}{\sqrt{e x} \left (c+d x^2\right )^{3/2}} \, dx}{6 c^2 d e^2}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt{e x}}{6 c^3 d e^3 \sqrt{c+d x^2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{12 c^3 d e^2}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt{e x}}{6 c^3 d e^3 \sqrt{c+d x^2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{6 c^3 d e^3}\\ &=-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{e x}}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 a d)\right ) \sqrt{e x}}{6 c^3 d e^3 \sqrt{c+d x^2}}+\frac{\left (b^2 c^2+5 a d (2 b c-3 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 )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.267107, size = 211, normalized size = 0.82 \[ \frac{x^{5/2} \left (\frac{a^2 (-d) \left (4 c^2+21 c d x^2+15 d^2 x^4\right )+2 a b c d x^2 \left (7 c+5 d x^2\right )+b^2 c^2 x^2 \left (d x^2-c\right )}{c^3 d x^{3/2} \left (c+d x^2\right )}+\frac{i x \sqrt{\frac{c}{d x^2}+1} \left (-15 a^2 d^2+10 a b c d+b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{c^3 d \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{6 (e x)^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 686, normalized size = 2.7 \begin{align*} -{\frac{1}{12\,x{e}^{2}{c}^{3}{d}^{2}} \left ( 15\,\sqrt{-cd}\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 ){x}^{3}{a}^{2}{d}^{3}-10\,\sqrt{-cd}\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 ){x}^{3}abc{d}^{2}-\sqrt{-cd}\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 ){x}^{3}{b}^{2}{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}x{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}xab{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}x{b}^{2}{c}^{3}+30\,{x}^{4}{a}^{2}{d}^{4}-20\,{x}^{4}abc{d}^{3}-2\,{x}^{4}{b}^{2}{c}^{2}{d}^{2}+42\,{x}^{2}{a}^{2}c{d}^{3}-28\,{x}^{2}ab{c}^{2}{d}^{2}+2\,{x}^{2}{b}^{2}{c}^{3}d+8\,{a}^{2}{c}^{2}{d}^{2} \right ){\frac{1}{\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((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(5/2),x)

[Out]

-1/12*(15*(-c*d)^(1/2)*((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^3*a^2*d^3-10*(-c*
d)^(1/2)*((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^3*a*b*c*d^2-(-c*d)^(1/2)*((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^3*b^2*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)*x*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)*x*a*b*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)*x*b^2*c^3+30*x^4*a^2*d^4-20*x^4*a*b*c*d^3-2*x^4*b^2*c^2*d^2+42*x^2*a^2*c*d^3-28*x^2*a*b*c^2
*d^2+2*x^2*b^2*c^3*d+8*a^2*c^2*d^2)/x/e^2/(e*x)^(1/2)/c^3/d^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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(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} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{d^{3} e^{3} x^{9} + 3 \, c d^{2} e^{3} x^{7} + 3 \, c^{2} d e^{3} x^{5} + c^{3} e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(5/2)/(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^3*x^9 + 3*c*d^2*e^3*x^7 + 3*c^2*d*e^3*x^
5 + c^3*e^3*x^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((b*x**2+a)**2/(e*x)**(5/2)/(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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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