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

Optimal. Leaf size=240 \[ -\frac{c^{3/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{231 d^{13/4} \sqrt{c+d x^2}}+\frac{2 e \sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right )}{231 d^3}-\frac{2 b (e x)^{5/2} \sqrt{c+d x^2} (9 b c-22 a d)}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3} \]

[Out]

(2*(77*a^2*d^2 + 5*b*c*(9*b*c - 22*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d^3) - (2*b*(9*b*c - 22*a*d)*(e*x)^
(5/2)*Sqrt[c + d*x^2])/(77*d^2*e) + (2*b^2*(e*x)^(9/2)*Sqrt[c + d*x^2])/(11*d*e^3) - (c^(3/4)*(77*a^2*d^2 + 5*
b*c*(9*b*c - 22*a*d))*e^(3/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcT
an[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(231*d^(13/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.219103, antiderivative size = 240, 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 = {464, 459, 321, 329, 220} \[ -\frac{c^{3/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{231 d^{13/4} \sqrt{c+d x^2}}+\frac{2 e \sqrt{e x} \sqrt{c+d x^2} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right )}{231 d^3}-\frac{2 b (e x)^{5/2} \sqrt{c+d x^2} (9 b c-22 a d)}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(77*a^2*d^2 + 5*b*c*(9*b*c - 22*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d^3) - (2*b*(9*b*c - 22*a*d)*(e*x)^
(5/2)*Sqrt[c + d*x^2])/(77*d^2*e) + (2*b^2*(e*x)^(9/2)*Sqrt[c + d*x^2])/(11*d*e^3) - (c^(3/4)*(77*a^2*d^2 + 5*
b*c*(9*b*c - 22*a*d))*e^(3/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcT
an[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(231*d^(13/4)*Sqrt[c + d*x^2])

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 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 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)^{3/2} \left (a+b x^2\right )^2}{\sqrt{c+d x^2}} \, dx &=\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3}+\frac{2 \int \frac{(e x)^{3/2} \left (\frac{11 a^2 d}{2}-\frac{1}{2} b (9 b c-22 a d) x^2\right )}{\sqrt{c+d x^2}} \, dx}{11 d}\\ &=-\frac{2 b (9 b c-22 a d) (e x)^{5/2} \sqrt{c+d x^2}}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3}-\frac{1}{77} \left (-77 a^2-\frac{5 b c (9 b c-22 a d)}{d^2}\right ) \int \frac{(e x)^{3/2}}{\sqrt{c+d x^2}} \, dx\\ &=\frac{2 \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}-\frac{2 b (9 b c-22 a d) (e x)^{5/2} \sqrt{c+d x^2}}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3}-\frac{\left (c \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{d^2}\right ) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{231 d}\\ &=\frac{2 \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}-\frac{2 b (9 b c-22 a d) (e x)^{5/2} \sqrt{c+d x^2}}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3}-\frac{\left (2 c \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{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 )}{231 d}\\ &=\frac{2 \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}-\frac{2 b (9 b c-22 a d) (e x)^{5/2} \sqrt{c+d x^2}}{77 d^2 e}+\frac{2 b^2 (e x)^{9/2} \sqrt{c+d x^2}}{11 d e^3}-\frac{c^{3/4} \left (77 a^2+\frac{5 b c (9 b c-22 a d)}{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 )}{231 d^{5/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.211704, size = 190, normalized size = 0.79 \[ \frac{(e x)^{3/2} \left (\frac{2 \sqrt{x} \left (c+d x^2\right ) \left (77 a^2 d^2+22 a b d \left (3 d x^2-5 c\right )+3 b^2 \left (15 c^2-9 c d x^2+7 d^2 x^4\right )\right )}{d^3}-\frac{2 i c x \sqrt{\frac{c}{d x^2}+1} \left (77 a^2 d^2-110 a b c d+45 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{d^3 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{231 x^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(3/2)*((2*Sqrt[x]*(c + d*x^2)*(77*a^2*d^2 + 22*a*b*d*(-5*c + 3*d*x^2) + 3*b^2*(15*c^2 - 9*c*d*x^2 + 7*d
^2*x^4)))/d^3 - ((2*I)*c*(45*b^2*c^2 - 110*a*b*c*d + 77*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqr
t[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)))/(231*x^(3/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.026, size = 405, normalized size = 1.7 \begin{align*} -{\frac{e}{231\,x{d}^{4}}\sqrt{ex} \left ( -42\,{x}^{7}{b}^{2}{d}^{4}+77\,\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}-110\,\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+45\,\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}-132\,{x}^{5}ab{d}^{4}+12\,{x}^{5}{b}^{2}c{d}^{3}-154\,{x}^{3}{a}^{2}{d}^{4}+88\,{x}^{3}abc{d}^{3}-36\,{x}^{3}{b}^{2}{c}^{2}{d}^{2}-154\,x{a}^{2}c{d}^{3}+220\,xab{c}^{2}{d}^{2}-90\,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)^(3/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

-1/231*e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-42*x^7*b^2*d^4+77*((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^2-110*((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+45*((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^3-132*x^5*a*b*d^4+12*x^5*b^2*c*d^3-154*x^3*a^2*d^4+88*x^3*a*b*c*d^3-36*x^3*b^2*c^2*d^2-154*x*a^2*c*d^3+22
0*x*a*b*c^2*d^2-90*x*b^2*c^3*d)/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{3}{2}}}{\sqrt{d x^{2} + c}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C]  time = 46.3341, size = 144, normalized size = 0.6 \begin{align*} \frac{a^{2} e^{\frac{3}{2}} 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 \sqrt{c} \Gamma \left (\frac{9}{4}\right )} + \frac{a b e^{\frac{3}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt{c} \Gamma \left (\frac{13}{4}\right )} + \frac{b^{2} e^{\frac{3}{2}} x^{\frac{13}{2}} \Gamma \left (\frac{13}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{13}{4} \\ \frac{17}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt{c} \Gamma \left (\frac{17}{4}\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)**(1/2),x)

[Out]

a**2*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gamma(9/4)) +
a*b*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*gamma(13/4)) +
b**2*e**(3/2)*x**(13/2)*gamma(13/4)*hyper((1/2, 13/4), (17/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gamma(17
/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} \left (e x\right )^{\frac{3}{2}}}{\sqrt{d x^{2} + c}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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