3.239 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=356 \[ \frac{4 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 A c+b B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 A c+b B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{2 \left (b x^2+c x^4\right )^{3/2} (9 A c+b B)}{9 b x^{3/2}}+\frac{4}{15} \sqrt{x} \sqrt{b x^2+c x^4} (9 A c+b B)+\frac{8 b x^{3/2} \left (b+c x^2\right ) (9 A c+b B)}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}} \]

[Out]

(8*b*(b*B + 9*A*c)*x^(3/2)*(b + c*x^2))/(15*Sqrt[c]*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (4*(b*B + 9*A
*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/15 + (2*(b*B + 9*A*c)*(b*x^2 + c*x^4)^(3/2))/(9*b*x^(3/2)) - (2*A*(b*x^2 + c*
x^4)^(5/2))/(b*x^(11/2)) - (8*b^(5/4)*(b*B + 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c
]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4]) + (4*b^(5/4)*(b*
B + 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[
x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.445831, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2038, 2021, 2032, 329, 305, 220, 1196} \[ \frac{4 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 A c+b B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 A c+b B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{2 \left (b x^2+c x^4\right )^{3/2} (9 A c+b B)}{9 b x^{3/2}}+\frac{4}{15} \sqrt{x} \sqrt{b x^2+c x^4} (9 A c+b B)+\frac{8 b x^{3/2} \left (b+c x^2\right ) (9 A c+b B)}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(9/2),x]

[Out]

(8*b*(b*B + 9*A*c)*x^(3/2)*(b + c*x^2))/(15*Sqrt[c]*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (4*(b*B + 9*A
*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/15 + (2*(b*B + 9*A*c)*(b*x^2 + c*x^4)^(3/2))/(9*b*x^(3/2)) - (2*A*(b*x^2 + c*
x^4)^(5/2))/(b*x^(11/2)) - (8*b^(5/4)*(b*B + 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c
]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4]) + (4*b^(5/4)*(b*
B + 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[
x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4])

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx &=-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}-\frac{\left (2 \left (-\frac{b B}{2}-\frac{9 A c}{2}\right )\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx}{b}\\ &=\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}+\frac{1}{3} (2 (b B+9 A c)) \int \frac{\sqrt{b x^2+c x^4}}{\sqrt{x}} \, dx\\ &=\frac{4}{15} (b B+9 A c) \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}+\frac{1}{15} (4 b (b B+9 A c)) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{4}{15} (b B+9 A c) \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}+\frac{\left (4 b (b B+9 A c) x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{15 \sqrt{b x^2+c x^4}}\\ &=\frac{4}{15} (b B+9 A c) \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}+\frac{\left (8 b (b B+9 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{b x^2+c x^4}}\\ &=\frac{4}{15} (b B+9 A c) \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}+\frac{\left (8 b^{3/2} (b B+9 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{c} \sqrt{b x^2+c x^4}}-\frac{\left (8 b^{3/2} (b B+9 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{b}}}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{c} \sqrt{b x^2+c x^4}}\\ &=\frac{8 b (b B+9 A c) x^{3/2} \left (b+c x^2\right )}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{4}{15} (b B+9 A c) \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}-\frac{8 b^{5/4} (b B+9 A c) x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{4 b^{5/4} (b B+9 A c) x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0737532, size = 85, normalized size = 0.24 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \left (\frac{x^2 (9 A c+b B) \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{b}\right )}{\sqrt{\frac{c x^2}{b}+1}}-\frac{3 A \left (b+c x^2\right )^2}{b}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(9/2),x]

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*((-3*A*(b + c*x^2)^2)/b + ((b*B + 9*A*c)*x^2*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x
^2)/b)])/Sqrt[1 + (c*x^2)/b]))/(3*x^(3/2))

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Maple [A]  time = 0.02, size = 429, normalized size = 1.2 \begin{align*}{\frac{2}{45\, \left ( c{x}^{2}+b \right ) ^{2}c} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 5\,B{c}^{3}{x}^{6}+108\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){b}^{2}c-54\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){b}^{2}c+12\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){b}^{3}-6\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){b}^{3}+9\,A{x}^{4}{c}^{3}+16\,B{x}^{4}b{c}^{2}-36\,A{x}^{2}b{c}^{2}+11\,B{x}^{2}{b}^{2}c-45\,A{b}^{2}c \right ){x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(9/2),x)

[Out]

2/45*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(5*B*c^3*x^6+108*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2
)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2
))^(1/2),1/2*2^(1/2))*b^2*c-54*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(
1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c+12*
B*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))
^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^3-6*B*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2)
)^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2
))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^3+9*A*x^4*c^3+16*B*x^4*b*c^2-36*A*x^2*b*c^2+11*B*x^2*b^2*c-45*A*b^2*c)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B c x^{4} +{\left (B b + A c\right )} x^{2} + A b\right )} \sqrt{c x^{4} + b x^{2}}}{x^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

integral((B*c*x^4 + (B*b + A*c)*x^2 + A*b)*sqrt(c*x^4 + b*x^2)/x^(5/2), 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)*(c*x**4+b*x**2)**(3/2)/x**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(9/2), x)