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

Optimal. Leaf size=239 \[ \frac{4 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (b B-3 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{231 c^{9/4} \sqrt{b x^2+c x^4}}-\frac{8 b^2 \sqrt{b x^2+c x^4} (b B-3 A c)}{231 c^2 \sqrt{x}}-\frac{4 b x^{3/2} \sqrt{b x^2+c x^4} (b B-3 A c)}{77 c}-\frac{2 \left (b x^2+c x^4\right )^{3/2} (b B-3 A c)}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}} \]

[Out]

(-8*b^2*(b*B - 3*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^2*Sqrt[x]) - (4*b*(b*B - 3*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^4])
/(77*c) - (2*(b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(33*c*Sqrt[x]) + (2*B*(b*x^2 + c*x^4)^(5/2))/(15*c*x^(5/2))
+ (4*b^(11/4)*(b*B - 3*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(9/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.369354, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2039, 2021, 2024, 2032, 329, 220} \[ -\frac{8 b^2 \sqrt{b x^2+c x^4} (b B-3 A c)}{231 c^2 \sqrt{x}}+\frac{4 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (b B-3 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{231 c^{9/4} \sqrt{b x^2+c x^4}}-\frac{4 b x^{3/2} \sqrt{b x^2+c x^4} (b B-3 A c)}{77 c}-\frac{2 \left (b x^2+c x^4\right )^{3/2} (b B-3 A c)}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(b*B - 3*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^2*Sqrt[x]) - (4*b*(b*B - 3*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^4])
/(77*c) - (2*(b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(33*c*Sqrt[x]) + (2*B*(b*x^2 + c*x^4)^(5/2))/(15*c*x^(5/2))
+ (4*b^(11/4)*(b*B - 3*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(9/4)*Sqrt[b*x^2 + c*x^4])

Rule 2039

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

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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 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 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 ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx &=\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac{\left (2 \left (\frac{5 b B}{2}-\frac{15 A c}{2}\right )\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx}{15 c}\\ &=-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac{(2 b (b B-3 A c)) \int \sqrt{x} \sqrt{b x^2+c x^4} \, dx}{11 c}\\ &=-\frac{4 b (b B-3 A c) x^{3/2} \sqrt{b x^2+c x^4}}{77 c}-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac{\left (4 b^2 (b B-3 A c)\right ) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx}{77 c}\\ &=-\frac{8 b^2 (b B-3 A c) \sqrt{b x^2+c x^4}}{231 c^2 \sqrt{x}}-\frac{4 b (b B-3 A c) x^{3/2} \sqrt{b x^2+c x^4}}{77 c}-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}+\frac{\left (4 b^3 (b B-3 A c)\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{231 c^2}\\ &=-\frac{8 b^2 (b B-3 A c) \sqrt{b x^2+c x^4}}{231 c^2 \sqrt{x}}-\frac{4 b (b B-3 A c) x^{3/2} \sqrt{b x^2+c x^4}}{77 c}-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}+\frac{\left (4 b^3 (b B-3 A c) x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{231 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{8 b^2 (b B-3 A c) \sqrt{b x^2+c x^4}}{231 c^2 \sqrt{x}}-\frac{4 b (b B-3 A c) x^{3/2} \sqrt{b x^2+c x^4}}{77 c}-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}+\frac{\left (8 b^3 (b B-3 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 )}{231 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{8 b^2 (b B-3 A c) \sqrt{b x^2+c x^4}}{231 c^2 \sqrt{x}}-\frac{4 b (b B-3 A c) x^{3/2} \sqrt{b x^2+c x^4}}{77 c}-\frac{2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}+\frac{4 b^{11/4} (b B-3 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 )}{231 c^{9/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.142065, size = 115, normalized size = 0.48 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \left (5 b^2 (b B-3 A c) \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{b}\right )-\left (b+c x^2\right )^2 \sqrt{\frac{c x^2}{b}+1} \left (-15 A c+5 b B-11 B c x^2\right )\right )}{165 c^2 \sqrt{x} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]*(5*b*B - 15*A*c - 11*B*c*x^2)) + 5*b^2*(b*B - 3*
A*c)*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/b)]))/(165*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.014, size = 307, normalized size = 1.3 \begin{align*} -{\frac{2}{1155\, \left ( c{x}^{2}+b \right ) ^{2}{c}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -77\,B{x}^{9}{c}^{5}-105\,A{x}^{7}{c}^{5}-196\,B{x}^{7}b{c}^{4}+30\,A\sqrt{-bc}\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}c-300\,A{x}^{5}b{c}^{4}-10\,B\sqrt{-bc}\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}^{4}-131\,B{x}^{5}{b}^{2}{c}^{3}-255\,A{x}^{3}{b}^{2}{c}^{3}+8\,B{x}^{3}{b}^{3}{c}^{2}-60\,Ax{b}^{3}{c}^{2}+20\,Bx{b}^{4}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^(3/2),x)

[Out]

-2/1155*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(-77*B*x^9*c^5-105*A*x^7*c^5-196*B*x^7*b*c^4+30*A*(-b*c)^(1/2)
*((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*c-300*A*x^5*b*c^4-10*B*(-b*c)^(1/2)*(
(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^4-131*B*x^5*b^2*c^3-255*A*x^3*b^2*c^3+8*B
*x^3*b^3*c^2-60*A*x*b^3*c^2+20*B*x*b^4*c)/c^3

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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{3}{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^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B c x^{4} +{\left (B b + A c\right )} x^{2} + A b\right )} \sqrt{c x^{4} + b x^{2}} \sqrt{x}, 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^(3/2),x, algorithm="fricas")

[Out]

integral((B*c*x^4 + (B*b + A*c)*x^2 + A*b)*sqrt(c*x^4 + b*x^2)*sqrt(x), 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**(3/2),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{3}{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^(3/2),x, algorithm="giac")

[Out]

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