3.6.48 \(\int \frac {A+B x^3}{(e x)^{3/2} \sqrt {a+b x^3}} \, dx\) [548]

Optimal. Leaf size=542 \[ -\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}}+\frac {\left (1+\sqrt {3}\right ) (2 A b+a B) \sqrt {e x} \sqrt {a+b x^3}}{a b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}-\frac {\sqrt [4]{3} (2 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{a^{2/3} b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {\left (1-\sqrt {3}\right ) (2 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} a^{2/3} b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

-2*A*(b*x^3+a)^(1/2)/a/e/(e*x)^(1/2)+(2*A*b+B*a)*(1+3^(1/2))*(e*x)^(1/2)*(b*x^3+a)^(1/2)/a/b^(2/3)/e^2/(a^(1/3
)+b^(1/3)*x*(1+3^(1/2)))-3^(1/4)*(2*A*b+B*a)*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b
^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticE((1-(a
^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(e*x)^(1/2)*
((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/a^(2/3)/b^(2/3)/e^2/(b*x^3+a
)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)-1/6*(2*A*b+B*a)*(a^(1/3)+b^(1/
3)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2
)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/
2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(1-3^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/
3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/a^(2/3)/b^(2/3)/e^2/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/
(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.38, antiderivative size = 542, 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 = {464, 335, 314, 231, 1895} \begin {gather*} -\frac {\left (1-\sqrt {3}\right ) \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+2 A b) F\left (\text {ArcCos}\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 )}{2 \sqrt [4]{3} a^{2/3} b^{2/3} e^2 \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 {\sqrt [4]{3} \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+2 A b) E\left (\text {ArcCos}\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 )}{a^{2/3} b^{2/3} e^2 \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 {\left (1+\sqrt {3}\right ) \sqrt {e x} \sqrt {a+b x^3} (a B+2 A b)}{a b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}-\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[a + b*x^3])/(a*e*Sqrt[e*x]) + ((1 + Sqrt[3])*(2*A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^3])/(a*b^(2/3)*e^
2*(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)) - (3^(1/4)*(2*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]*EllipticE[ArcCos[(a^(1/3) + (1 - Sqr
t[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(a^(2/3)*b^(2/3)*e^2*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]) - ((1 - Sqrt[3])*(2*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])/(2*3^(1/4)*a^(2/3)*b^(2/3)*e^2*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 231

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]/(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)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 314

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

Rule 335

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 464

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

Rule 1895

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/
a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqrt[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d
*s*x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/
(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]))*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r
*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^3}{(e x)^{3/2} \sqrt {a+b x^3}} \, dx &=-\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}}+\frac {(2 A b+a B) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^3}} \, dx}{a e^3}\\ &=-\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}}+\frac {(2 (2 A b+a B)) \text {Subst}\left (\int \frac {x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{a e^4}\\ &=-\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}}-\frac {(2 A b+a B) \text {Subst}\left (\int \frac {\left (-1+\sqrt {3}\right ) a^{2/3} e^2-2 b^{2/3} x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{a b^{2/3} e^4}-\frac {\left (\left (1-\sqrt {3}\right ) (2 A b+a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{\sqrt [3]{a} b^{2/3} e^2}\\ &=-\frac {2 A \sqrt {a+b x^3}}{a e \sqrt {e x}}+\frac {\left (1+\sqrt {3}\right ) (2 A b+a B) \sqrt {e x} \sqrt {a+b x^3}}{a b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}-\frac {\sqrt [4]{3} (2 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}} E\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 )}{a^{2/3} b^{2/3} e^2 \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 {\left (1-\sqrt {3}\right ) (2 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 )}{2 \sqrt [4]{3} a^{2/3} b^{2/3} e^2 \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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 83, normalized size = 0.15 \begin {gather*} \frac {x \left (-10 A \left (a+b x^3\right )+2 (2 A b+a B) x^3 \sqrt {1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};-\frac {b x^3}{a}\right )\right )}{5 a (e x)^{3/2} \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-10*A*(a + b*x^3) + 2*(2*A*b + a*B)*x^3*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/2, 5/6, 11/6, -((b*x^3)/a)
]))/(5*a*(e*x)^(3/2)*Sqrt[a + b*x^3])

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 5385, normalized size = 9.94

method	result	size
risch	\(\text {Expression too large to display}\)	\(1119\)
elliptic	\(\text {Expression too large to display}\)	\(1132\)
default	\(\text {Expression too large to display}\)	\(5385\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 2.11, size = 97, normalized size = 0.18 \begin {gather*} \frac {A \Gamma \left (- \frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{6}, \frac {1}{2} \\ \frac {5}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {5}{6}\right )} + \frac {B x^{\frac {5}{2}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} e^{\frac {3}{2}} \Gamma \left (\frac {11}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*gamma(-1/6)*hyper((-1/6, 1/2), (5/6,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*e**(3/2)*sqrt(x)*gamma(5/6)) + B
*x**(5/2)*gamma(5/6)*hyper((1/2, 5/6), (11/6,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*e**(3/2)*gamma(11/6))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {B\,x^3+A}{{\left (e\,x\right )}^{3/2}\,\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)^(3/2)*(a + b*x^3)^(1/2)),x)

[Out]

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

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