3.563 \(\int \frac{A+B x^3}{\sqrt{e x} (a+b x^3)^{5/2}} \, dx\)

Optimal. Leaf size=297 \[ \frac{2 \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (a B+8 A b) \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{27 \sqrt [4]{3} a^{7/3} b e \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 \sqrt{e x} (a B+8 A b)}{27 a^2 b e \sqrt{a+b x^3}}+\frac{2 \sqrt{e x} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

[Out]

(2*(A*b - a*B)*Sqrt[e*x])/(9*a*b*e*(a + b*x^3)^(3/2)) + (2*(8*A*b + a*B)*Sqrt[e*x])/(27*a^2*b*e*Sqrt[a + b*x^3
]) + (2*(8*A*b + a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3
) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*
b^(1/3)*x)], (2 + Sqrt[3])/4])/(27*3^(1/4)*a^(7/3)*b*e*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 +
Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.222083, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {457, 290, 329, 225} \[ \frac{2 \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (a B+8 A b) F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{27 \sqrt [4]{3} a^{7/3} b e \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 \sqrt{e x} (a B+8 A b)}{27 a^2 b e \sqrt{a+b x^3}}+\frac{2 \sqrt{e x} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(Sqrt[e*x]*(a + b*x^3)^(5/2)),x]

[Out]

(2*(A*b - a*B)*Sqrt[e*x])/(9*a*b*e*(a + b*x^3)^(3/2)) + (2*(8*A*b + a*B)*Sqrt[e*x])/(27*a^2*b*e*Sqrt[a + b*x^3
]) + (2*(8*A*b + a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3
) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*
b^(1/3)*x)], (2 + Sqrt[3])/4])/(27*3^(1/4)*a^(7/3)*b*e*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 +
Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{A+B x^3}{\sqrt{e x} \left (a+b x^3\right )^{5/2}} \, dx &=\frac{2 (A b-a B) \sqrt{e x}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{\left (2 \left (4 A b+\frac{a B}{2}\right )\right ) \int \frac{1}{\sqrt{e x} \left (a+b x^3\right )^{3/2}} \, dx}{9 a b}\\ &=\frac{2 (A b-a B) \sqrt{e x}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{2 (8 A b+a B) \sqrt{e x}}{27 a^2 b e \sqrt{a+b x^3}}+\frac{(2 (8 A b+a B)) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^3}} \, dx}{27 a^2 b}\\ &=\frac{2 (A b-a B) \sqrt{e x}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{2 (8 A b+a B) \sqrt{e x}}{27 a^2 b e \sqrt{a+b x^3}}+\frac{(4 (8 A b+a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^6}{e^3}}} \, dx,x,\sqrt{e x}\right )}{27 a^2 b e}\\ &=\frac{2 (A b-a B) \sqrt{e x}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{2 (8 A b+a B) \sqrt{e x}}{27 a^2 b e \sqrt{a+b x^3}}+\frac{2 (8 A b+a B) \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{27 \sqrt [4]{3} a^{7/3} b e \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.093222, size = 107, normalized size = 0.36 \[ \frac{2 x \left (-2 a^2 B+2 \left (a+b x^3\right ) \sqrt{\frac{b x^3}{a}+1} (a B+8 A b) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{b x^3}{a}\right )+a b \left (11 A+B x^3\right )+8 A b^2 x^3\right )}{27 a^2 b \sqrt{e x} \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(Sqrt[e*x]*(a + b*x^3)^(5/2)),x]

[Out]

(2*x*(-2*a^2*B + 8*A*b^2*x^3 + a*b*(11*A + B*x^3) + 2*(8*A*b + a*B)*(a + b*x^3)*Sqrt[1 + (b*x^3)/a]*Hypergeome
tric2F1[1/6, 1/2, 7/6, -((b*x^3)/a)]))/(27*a^2*b*Sqrt[e*x]*(a + b*x^3)^(3/2))

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Maple [C]  time = 0.059, size = 7077, normalized size = 23.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/(b*x^3+a)^(5/2)/(e*x)^(1/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*sqrt(e*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{b^{3} e x^{10} + 3 \, a b^{2} e x^{7} + 3 \, a^{2} b e x^{4} + a^{3} e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2)/(e*x)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)*sqrt(e*x)/(b^3*e*x^10 + 3*a*b^2*e*x^7 + 3*a^2*b*e*x^4 + a^3*e*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**3+A)/(b*x**3+a)**(5/2)/(e*x)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2)/(e*x)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*sqrt(e*x)), x)