3.950 \(\int \frac{x^4 (a+b x^2)^{3/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=429 \[ -\frac{c^{3/2} \sqrt{a+b x^2} \left (a^2 d^2-11 a b c d+8 b^2 c^2\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{35 b d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (a^2 d^2-11 a b c d+8 b^2 c^2\right )}{35 b d^3}-\frac{2 x \sqrt{a+b x^2} (2 b c-a d) \left (-a^2 d^2-4 a b c d+4 b^2 c^2\right )}{35 b^2 d^3 \sqrt{c+d x^2}}+\frac{2 \sqrt{c} \sqrt{a+b x^2} (2 b c-a d) \left (-a^2 d^2-4 a b c d+4 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b^2 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{2 x^3 \sqrt{a+b x^2} \sqrt{c+d x^2} (3 b c-4 a d)}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d} \]

[Out]

(-2*(2*b*c - a*d)*(4*b^2*c^2 - 4*a*b*c*d - a^2*d^2)*x*Sqrt[a + b*x^2])/(35*b^2*d^3*Sqrt[c + d*x^2]) + ((8*b^2*
c^2 - 11*a*b*c*d + a^2*d^2)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(35*b*d^3) - (2*(3*b*c - 4*a*d)*x^3*Sqrt[a + b*
x^2]*Sqrt[c + d*x^2])/(35*d^2) + (b*x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(7*d) + (2*Sqrt[c]*(2*b*c - a*d)*(4*b
^2*c^2 - 4*a*b*c*d - a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b^2
*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (c^(3/2)*(8*b^2*c^2 - 11*a*b*c*d + a^2*d^2)*
Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a
*(c + d*x^2))]*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.518254, antiderivative size = 429, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {477, 582, 531, 418, 492, 411} \[ \frac{x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (a^2 d^2-11 a b c d+8 b^2 c^2\right )}{35 b d^3}-\frac{2 x \sqrt{a+b x^2} (2 b c-a d) \left (-a^2 d^2-4 a b c d+4 b^2 c^2\right )}{35 b^2 d^3 \sqrt{c+d x^2}}-\frac{c^{3/2} \sqrt{a+b x^2} \left (a^2 d^2-11 a b c d+8 b^2 c^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{2 \sqrt{c} \sqrt{a+b x^2} (2 b c-a d) \left (-a^2 d^2-4 a b c d+4 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b^2 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{2 x^3 \sqrt{a+b x^2} \sqrt{c+d x^2} (3 b c-4 a d)}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^(3/2))/Sqrt[c + d*x^2],x]

[Out]

(-2*(2*b*c - a*d)*(4*b^2*c^2 - 4*a*b*c*d - a^2*d^2)*x*Sqrt[a + b*x^2])/(35*b^2*d^3*Sqrt[c + d*x^2]) + ((8*b^2*
c^2 - 11*a*b*c*d + a^2*d^2)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(35*b*d^3) - (2*(3*b*c - 4*a*d)*x^3*Sqrt[a + b*
x^2]*Sqrt[c + d*x^2])/(35*d^2) + (b*x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(7*d) + (2*Sqrt[c]*(2*b*c - a*d)*(4*b
^2*c^2 - 4*a*b*c*d - a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b^2
*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (c^(3/2)*(8*b^2*c^2 - 11*a*b*c*d + a^2*d^2)*
Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(35*b*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a
*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 477

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*(e*x)
^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(b*e*(m + n*(p + q) + 1)), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b x^2\right )^{3/2}}{\sqrt{c+d x^2}} \, dx &=\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}+\frac{\int \frac{x^4 \left (-a (5 b c-7 a d)-2 b (3 b c-4 a d) x^2\right )}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{7 d}\\ &=-\frac{2 (3 b c-4 a d) x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}-\frac{\int \frac{x^2 \left (-6 a b c (3 b c-4 a d)-3 b \left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) x^2\right )}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{35 b d^2}\\ &=\frac{\left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 b d^3}-\frac{2 (3 b c-4 a d) x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}+\frac{\int \frac{-3 a b c \left (8 b^2 c^2-11 a b c d+a^2 d^2\right )-6 b (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right ) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{105 b^2 d^3}\\ &=\frac{\left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 b d^3}-\frac{2 (3 b c-4 a d) x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}-\frac{\left (2 (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right )\right ) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{35 b d^3}-\frac{\left (a c \left (8 b^2 c^2-11 a b c d+a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{35 b d^3}\\ &=-\frac{2 (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right ) x \sqrt{a+b x^2}}{35 b^2 d^3 \sqrt{c+d x^2}}+\frac{\left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 b d^3}-\frac{2 (3 b c-4 a d) x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}-\frac{c^{3/2} \left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b d^{7/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}+\frac{\left (2 c (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right )\right ) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{35 b^2 d^3}\\ &=-\frac{2 (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right ) x \sqrt{a+b x^2}}{35 b^2 d^3 \sqrt{c+d x^2}}+\frac{\left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 b d^3}-\frac{2 (3 b c-4 a d) x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}}{7 d}+\frac{2 \sqrt{c} (2 b c-a d) \left (4 b^2 c^2-4 a b c d-a^2 d^2\right ) \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b^2 d^{7/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}-\frac{c^{3/2} \left (8 b^2 c^2-11 a b c d+a^2 d^2\right ) \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{35 b d^{7/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.607434, size = 305, normalized size = 0.71 \[ \frac{-i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (15 a^2 b c d^2+a^3 d^3-32 a b^2 c^2 d+16 b^3 c^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (a^2 d^2+a b d \left (8 d x^2-11 c\right )+b^2 \left (8 c^2-6 c d x^2+5 d^2 x^4\right )\right )+2 i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (2 a^2 b c d^2+a^3 d^3-12 a b^2 c^2 d+8 b^3 c^3\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{35 b d^4 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^2)^(3/2))/Sqrt[c + d*x^2],x]

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(a^2*d^2 + a*b*d*(-11*c + 8*d*x^2) + b^2*(8*c^2 - 6*c*d*x^2 + 5*d^2*x^4
)) + (2*I)*c*(8*b^3*c^3 - 12*a*b^2*c^2*d + 2*a^2*b*c*d^2 + a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*El
lipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*c*(16*b^3*c^3 - 32*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*Sq
rt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(35*b*Sqrt[b/a]*d^4*Sqrt
[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.023, size = 782, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x)

[Out]

1/35*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(5*(-b/a)^(1/2)*x^9*b^3*d^4+13*(-b/a)^(1/2)*x^7*a*b^2*d^4-(-b/a)^(1/2)*x^
7*b^3*c*d^3+9*(-b/a)^(1/2)*x^5*a^2*b*d^4-4*(-b/a)^(1/2)*x^5*a*b^2*c*d^3+2*(-b/a)^(1/2)*x^5*b^3*c^2*d^2+(-b/a)^
(1/2)*x^3*a^3*d^4-2*(-b/a)^(1/2)*x^3*a^2*b*c*d^3-9*(-b/a)^(1/2)*x^3*a*b^2*c^2*d^2+8*(-b/a)^(1/2)*x^3*b^3*c^3*d
+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*c*d^3+15*((b*x^2+a)/a)^
(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^2*b*c^2*d^2-32*((b*x^2+a)/a)^(1/2)*((d*x
^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*b^2*c^3*d+16*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2
)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^4-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/
a)^(1/2),(a*d/b/c)^(1/2))*a^3*c*d^3-4*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/
c)^(1/2))*a^2*b*c^2*d^2+24*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a
*b^2*c^3*d-16*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^4+(-b/a)
^(1/2)*x*a^3*c*d^3-11*(-b/a)^(1/2)*x*a^2*b*c^2*d^2+8*(-b/a)^(1/2)*x*a*b^2*c^3*d)/b/d^4/(b*d*x^4+a*d*x^2+b*c*x^
2+a*c)/(-b/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)*x^4/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 x^{6} + a x^{4}\right )} \sqrt{b x^{2} + a}}{\sqrt{d x^{2} + c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b*x^6 + a*x^4)*sqrt(b*x^2 + a)/sqrt(d*x^2 + c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (a + b x^{2}\right )^{\frac{3}{2}}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**(3/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**4*(a + b*x**2)**(3/2)/sqrt(c + d*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)*x^4/sqrt(d*x^2 + c), x)