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

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

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Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.06, size = 78, normalized size = 0.94 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.53, size = 89, normalized size = 1.07 \begin {gather*} \frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}-\frac {e^2 \sqrt {\frac {b}{e^3}} (2 A b-a B) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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[b/e^3]*e^2*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) + Sqrt
[a + b*x^3]])/(3*b^2)

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fricas [A]  time = 1.28, size = 184, normalized size = 2.22 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.36, size = 72, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b x^{3} e^{4} + a e^{4}} B x^{\frac {3}{2}} e^{\left (-\frac {3}{2}\right )}}{3 \, b} + \frac {{\left (B a e^{5} - 2 \, A b e^{5}\right )} e^{\left (-\frac {9}{2}\right )} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} e^{2} + \sqrt {b x^{3} e^{4} + a e^{4}} \right |}\right )}{3 \, b^{\frac {3}{2}}} \end {gather*}

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]

1/3*sqrt(b*x^3*e^4 + a*e^4)*B*x^(3/2)*e^(-3/2)/b + 1/3*(B*a*e^5 - 2*A*b*e^5)*e^(-9/2)*log(abs(-sqrt(b)*x^(3/2)
*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/b^(3/2)

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maple [C]  time = 1.15, size = 6424, normalized size = 77.40 \begin {gather*} \text {output too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{3} + A\right )} \sqrt {e x}}{\sqrt {b x^{3} + a}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^3+A\right )\,\sqrt {e\,x}}{\sqrt {b\,x^3+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.81, size = 107, normalized size = 1.29 \begin {gather*} \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 {gather*}

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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