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

Optimal. Leaf size=234 \[ -\frac{2 \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (b^2 c^2-7 a d (a d+2 b c)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}-\frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (b^2 c^2-7 a d (a d+2 b c)\right )}{21 c d e^3}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.190562, antiderivative size = 234, 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, 459, 279, 329, 220} \[ -\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}-\frac{2 \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (b^2 c^2-7 a d (a d+2 b c)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (b^2 c^2-7 a d (a d+2 b c)\right )}{21 c d e^3}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^2*c^2 - 7*a*d*(2*b*c + a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(21*c*d*e^3) - (2*a^2*(c + d*x^2)^(3/2))/(3*c*e
*(e*x)^(3/2)) + (2*b^2*Sqrt[e*x]*(c + d*x^2)^(3/2))/(7*d*e^3) - (2*(b^2*c^2 - 7*a*d*(2*b*c + 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])/(21*c^(1/4)*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 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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && 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{\left (a+b x^2\right )^2 \sqrt{c+d x^2}}{(e x)^{5/2}} \, dx &=-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac{2 \int \frac{\left (\frac{3}{2} a (2 b c+a d)+\frac{3}{2} b^2 c x^2\right ) \sqrt{c+d x^2}}{\sqrt{e x}} \, dx}{3 c e^2}\\ &=-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac{\left (b^2 c^2-7 a d (2 b c+a d)\right ) \int \frac{\sqrt{c+d x^2}}{\sqrt{e x}} \, dx}{7 c d e^2}\\ &=-\frac{2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{21 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac{\left (2 \left (b^2 c^2-7 a d (2 b c+a d)\right )\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{21 d e^2}\\ &=-\frac{2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{21 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac{\left (4 \left (b^2 c^2-7 a d (2 b c+a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{21 d e^3}\\ &=-\frac{2 \left (b^2 c^2-7 a d (2 b c+a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{21 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{3/2}}{7 d e^3}-\frac{2 \left (b^2 c^2-7 a d (2 b c+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 )}{21 \sqrt [4]{c} d^{5/4} e^{5/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.184193, size = 171, normalized size = 0.73 \[ \frac{x^{5/2} \left (\frac{2 \left (c+d x^2\right ) \left (-7 a^2 d+14 a b d x^2+b^2 x^2 \left (2 c+3 d x^2\right )\right )}{d x^{3/2}}+\frac{4 i x \sqrt{\frac{c}{d x^2}+1} \left (7 a^2 d^2+14 a b c d-b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{d \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{21 (e x)^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*((2*(c + d*x^2)*(-7*a^2*d + 14*a*b*d*x^2 + b^2*x^2*(2*c + 3*d*x^2)))/(d*x^(3/2)) + ((4*I)*(-(b^2*c^2)
+ 14*a*b*c*d + 7*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/
(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(21*(e*x)^(5/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.028, size = 383, normalized size = 1.6 \begin{align*}{\frac{2}{21\,x{e}^{2}{d}^{2}} \left ( 7\,\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{a}^{2}{d}^{2}+14\,\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 ) xabcd-\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{b}^{2}{c}^{2}+3\,{x}^{6}{b}^{2}{d}^{3}+14\,{x}^{4}ab{d}^{3}+5\,{x}^{4}{b}^{2}c{d}^{2}-7\,{x}^{2}{a}^{2}{d}^{3}+14\,{x}^{2}abc{d}^{2}+2\,{x}^{2}{b}^{2}{c}^{2}d-7\,{a}^{2}c{d}^{2} \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(5/2),x)

[Out]

2/21/(d*x^2+c)^(1/2)/x*(7*(-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*a
^2*d^2+14*(-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*a*b*c*d-(-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*b^2*c^2+3*x^6*b^2*d^3+14*x^4*a*b*d^3
+5*x^4*b^2*c*d^2-7*x^2*a^2*d^3+14*x^2*a*b*c*d^2+2*x^2*b^2*c^2*d-7*a^2*c*d^2)/e^2/(e*x)^(1/2)/d^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} \sqrt{d x^{2} + c}}{\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*(d*x^2+c)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/(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}}{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*(d*x^2+c)^(1/2)/(e*x)^(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)/(e^3*x^3), x)

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Sympy [C]  time = 17.4519, size = 153, normalized size = 0.65 \begin{align*} \frac{a^{2} \sqrt{c} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{a b \sqrt{c} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{b^{2} \sqrt{c} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/(e*x)**(5/2),x)

[Out]

a**2*sqrt(c)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*x**(3/2)*gamma(1/4)
) + a*b*sqrt(c)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), d*x**2*exp_polar(I*pi)/c)/(e**(5/2)*gamma(5/4))
+ b**2*sqrt(c)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*gamma(9/4)
)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c}}{\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*(d*x^2+c)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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