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

Optimal. Leaf size=288 \[ -\frac{2 c^{7/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (11 a^2 d^2+b c (3 b c-10 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{231 d^{13/4} \sqrt{c+d x^2}}+\frac{2 (e x)^{5/2} \sqrt{c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{77 d^2 e}+\frac{4 c e \sqrt{e x} \sqrt{c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{231 d^3}-\frac{2 b (e x)^{5/2} \left (c+d x^2\right )^{3/2} (3 b c-10 a d)}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3} \]

[Out]

(4*c*(11*a^2*d^2 + b*c*(3*b*c - 10*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d^3) + (2*(11*a^2*d^2 + b*c*(3*b*c
- 10*a*d))*(e*x)^(5/2)*Sqrt[c + d*x^2])/(77*d^2*e) - (2*b*(3*b*c - 10*a*d)*(e*x)^(5/2)*(c + d*x^2)^(3/2))/(55*
d^2*e) + (2*b^2*(e*x)^(9/2)*(c + d*x^2)^(3/2))/(15*d*e^3) - (2*c^(7/4)*(11*a^2*d^2 + b*c*(3*b*c - 10*a*d))*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])/(231*d^(13/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.291296, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {464, 459, 279, 321, 329, 220} \[ -\frac{2 c^{7/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (11 a^2 d^2+b c (3 b c-10 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{231 d^{13/4} \sqrt{c+d x^2}}+\frac{2 (e x)^{5/2} \sqrt{c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{77 d^2 e}+\frac{4 c e \sqrt{e x} \sqrt{c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{231 d^3}-\frac{2 b (e x)^{5/2} \left (c+d x^2\right )^{3/2} (3 b c-10 a d)}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*(11*a^2*d^2 + b*c*(3*b*c - 10*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d^3) + (2*(11*a^2*d^2 + b*c*(3*b*c
- 10*a*d))*(e*x)^(5/2)*Sqrt[c + d*x^2])/(77*d^2*e) - (2*b*(3*b*c - 10*a*d)*(e*x)^(5/2)*(c + d*x^2)^(3/2))/(55*
d^2*e) + (2*b^2*(e*x)^(9/2)*(c + d*x^2)^(3/2))/(15*d*e^3) - (2*c^(7/4)*(11*a^2*d^2 + b*c*(3*b*c - 10*a*d))*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])/(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 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 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 (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} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac{2 \int (e x)^{3/2} \sqrt{c+d x^2} \left (\frac{15 a^2 d}{2}-\frac{3}{2} b (3 b c-10 a d) x^2\right ) \, dx}{15 d}\\ &=-\frac{2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac{1}{11} \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) \int (e x)^{3/2} \sqrt{c+d x^2} \, dx\\ &=\frac{2 \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{77 e}-\frac{2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac{1}{77} \left (2 c \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right )\right ) \int \frac{(e x)^{3/2}}{\sqrt{c+d x^2}} \, dx\\ &=\frac{4 c \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}+\frac{2 \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{77 e}-\frac{2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac{\left (2 c^2 \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{231 d}\\ &=\frac{4 c \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}+\frac{2 \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{77 e}-\frac{2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac{\left (4 c^2 \left (11 a^2+\frac{b c (3 b c-10 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{4 c \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{231 d}+\frac{2 \left (11 a^2+\frac{b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{77 e}-\frac{2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac{2 c^{7/4} \left (11 a^2+\frac{b c (3 b c-10 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 = 0.27916, size = 225, normalized size = 0.78 \[ \frac{(e x)^{3/2} \left (\frac{2 \sqrt{x} \left (c+d x^2\right ) \left (55 a^2 d^2 \left (2 c+3 d x^2\right )+10 a b d \left (-10 c^2+6 c d x^2+21 d^2 x^4\right )+b^2 \left (-18 c^2 d x^2+30 c^3+14 c d^2 x^4+77 d^3 x^6\right )\right )}{5 d^3}-\frac{4 i c^2 x \sqrt{\frac{c}{d x^2}+1} \left (11 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{d^3 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{231 x^{3/2} \sqrt{c+d x^2}} \]

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)*(55*a^2*d^2*(2*c + 3*d*x^2) + 10*a*b*d*(-10*c^2 + 6*c*d*x^2 + 21*d^2*x^4)
+ b^2*(30*c^3 - 18*c^2*d*x^2 + 14*c*d^2*x^4 + 77*d^3*x^6)))/(5*d^3) - ((4*I)*c^2*(3*b^2*c^2 - 10*a*b*c*d + 11
*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^3)))/(231*x^(3/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.063, size = 448, normalized size = 1.6 \begin{align*} -{\frac{2\,e}{1155\,x{d}^{4}}\sqrt{ex} \left ( -77\,{x}^{9}{b}^{2}{d}^{5}-210\,{x}^{7}ab{d}^{5}-91\,{x}^{7}{b}^{2}c{d}^{4}+55\,\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}^{2}{d}^{2}-50\,\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}^{3}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}^{4}-165\,{x}^{5}{a}^{2}{d}^{5}-270\,{x}^{5}abc{d}^{4}+4\,{x}^{5}{b}^{2}{c}^{2}{d}^{3}-275\,{x}^{3}{a}^{2}c{d}^{4}+40\,{x}^{3}ab{c}^{2}{d}^{3}-12\,{x}^{3}{b}^{2}{c}^{3}{d}^{2}-110\,x{a}^{2}{c}^{2}{d}^{3}+100\,xab{c}^{3}{d}^{2}-30\,x{b}^{2}{c}^{4}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]

-2/1155*e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-77*x^9*b^2*d^5-210*x^7*a*b*d^5-91*x^7*b^2*c*d^4+55*((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^2*d^2-50*((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^3*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^4-165*x^5*a^2*d^5-270*x^5*a*b*c*d^4+4*x^5*b^2*c^2*d^3-275*x^3*a^2*c*d^
4+40*x^3*a*b*c^2*d^3-12*x^3*b^2*c^3*d^2-110*x*a^2*c^2*d^3+100*x*a*b*c^3*d^2-30*x*b^2*c^4*d)/d^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c} \left (e x\right )^{\frac{3}{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\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}, 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(d*x^2 + c)*sqrt(e*x), x)

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Sympy [C]  time = 73.795, size = 150, normalized size = 0.52 \begin{align*} \frac{a^{2} \sqrt{c} 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 \Gamma \left (\frac{9}{4}\right )} + \frac{a b \sqrt{c} 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 )}}{\Gamma \left (\frac{13}{4}\right )} + \frac{b^{2} \sqrt{c} 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 \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*sqrt(c)*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*gamma(9/4))
+ a*b*sqrt(c)*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), d*x**2*exp_polar(I*pi)/c)/gamma(13/4) +
b**2*sqrt(c)*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*gamma(1
7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c} \left (e x\right )^{\frac{3}{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)^(1/2),x, algorithm="giac")

[Out]

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