3.546 \(\int \frac{\sqrt{e x} (A+B x^3)}{\sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=83 \[ \frac{\sqrt{e} (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 b^{3/2}}+\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e} \]

[Out]

(B*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*b*e) + ((2*A*b - a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a
+ b*x^3])])/(3*b^(3/2))

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Rubi [A]  time = 0.0673085, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {459, 329, 275, 217, 206} \[ \frac{\sqrt{e} (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 b^{3/2}}+\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*b*e) + ((2*A*b - a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a
+ b*x^3])])/(3*b^(3/2))

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 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{e x} \left (A+B x^3\right )}{\sqrt{a+b x^3}} \, dx &=\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e}-\frac{\left (-3 A b+\frac{3 a B}{2}\right ) \int \frac{\sqrt{e x}}{\sqrt{a+b x^3}} \, dx}{3 b}\\ &=\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e}+\frac{(2 A b-a B) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^6}{e^3}}} \, dx,x,\sqrt{e x}\right )}{b e}\\ &=\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e}+\frac{(2 A b-a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{3 b e}\\ &=\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e}+\frac{(2 A b-a B) \operatorname{Subst}\left (\int \frac{1}{1-\frac{b x^2}{e^3}} \, dx,x,\frac{(e x)^{3/2}}{\sqrt{a+b x^3}}\right )}{3 b e}\\ &=\frac{B (e x)^{3/2} \sqrt{a+b x^3}}{3 b e}+\frac{(2 A b-a B) \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0424473, size = 78, normalized size = 0.94 \[ \frac{\sqrt{e x} \left ((2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a+b x^3}}\right )+\sqrt{b} B x^{3/2} \sqrt{a+b x^3}\right )}{3 b^{3/2} \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(Sqrt[b]*B*x^(3/2)*Sqrt[a + b*x^3] + (2*A*b - a*B)*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a + b*x^3]]))/(3*
b^(3/2)*Sqrt[x])

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Maple [C]  time = 0.044, size = 6424, normalized size = 77.4 \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)*(e*x)^(1/2)/(b*x^3+a)^(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{{\left (B x^{3} + A\right )} \sqrt{e x}}{\sqrt{b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 4.2287, size = 417, normalized size = 5.02 \begin{align*} \left [\frac{4 \, \sqrt{b x^{3} + a} \sqrt{e x} B x -{\left (B a - 2 \, A b\right )} \sqrt{\frac{e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \,{\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt{b x^{3} + a} \sqrt{e x} \sqrt{\frac{e}{b}}\right )}{12 \, b}, \frac{2 \, \sqrt{b x^{3} + a} \sqrt{e x} B x +{\left (B a - 2 \, A b\right )} \sqrt{-\frac{e}{b}} \arctan \left (\frac{2 \, \sqrt{b x^{3} + a} \sqrt{e x} b x \sqrt{-\frac{e}{b}}}{2 \, b e x^{3} + a e}\right )}{6 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*sqrt(b*x^3 + a)*sqrt(e*x)*B*x - (B*a - 2*A*b)*sqrt(e/b)*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2
*b^2*x^4 + a*b*x)*sqrt(b*x^3 + a)*sqrt(e*x)*sqrt(e/b)))/b, 1/6*(2*sqrt(b*x^3 + a)*sqrt(e*x)*B*x + (B*a - 2*A*b
)*sqrt(-e/b)*arctan(2*sqrt(b*x^3 + a)*sqrt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)))/b]

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Sympy [A]  time = 6.05434, size = 107, normalized size = 1.29 \begin{align*} \frac{2 A \sqrt{e} \operatorname{asinh}{\left (\frac{\sqrt{b} \left (e x\right )^{\frac{3}{2}}}{\sqrt{a} e^{\frac{3}{2}}} \right )}}{3 \sqrt{b}} + \frac{B \sqrt{a} \left (e x\right )^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}}{3 b e} - \frac{B a \sqrt{e} \operatorname{asinh}{\left (\frac{\sqrt{b} \left (e x\right )^{\frac{3}{2}}}{\sqrt{a} e^{\frac{3}{2}}} \right )}}{3 b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)*(e*x)**(1/2)/(b*x**3+a)**(1/2),x)

[Out]

2*A*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(3*sqrt(b)) + B*sqrt(a)*(e*x)**(3/2)*sqrt(1 + b*x**
3/a)/(3*b*e) - B*a*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(3*b**(3/2))

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Giac [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)*(e*x)^(1/2)/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

Timed out