Integrand size = 26, antiderivative size = 83 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}+\frac {(2 A b-a B) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}} \]
1/3*(2*A*b-B*a)*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))*e^(1/ 2)/b^(3/2)+1/3*B*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b/e
Time = 0.80 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {e x} \left (\sqrt {b} B x^{3/2} \sqrt {a+b x^3}+(2 A b-a B) \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{3 b^{3/2} \sqrt {x}} \]
(Sqrt[e*x]*(Sqrt[b]*B*x^(3/2)*Sqrt[a + b*x^3] + (2*A*b - a*B)*Log[Sqrt[b]* x^(3/2) + Sqrt[a + b*x^3]]))/(3*b^(3/2)*Sqrt[x])
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {959, 851, 807, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(2 A b-a B) \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx}{2 b}+\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {(2 A b-a B) \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{b e}+\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {(2 A b-a B) \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 b e}+\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(2 A b-a B) \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 b e}+\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {e} (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 b^{3/2}}+\frac {B (e x)^{3/2} \sqrt {a+b x^3}}{3 b e}\) |
(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)/e^2])])/(3*b^(3/2))
3.6.46.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
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] - Simp[(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]
Time = 4.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {B \,x^{2} \sqrt {b \,x^{3}+a}\, e}{3 b \sqrt {e x}}+\frac {\left (2 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e \sqrt {\left (b \,x^{3}+a \right ) e x}}{3 b \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\) | \(94\) |
default | \(\frac {\sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (2 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) b e +B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, x -B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a e \right )}{3 \sqrt {\left (b \,x^{3}+a \right ) e x}\, b \sqrt {b e}}\) | \(112\) |
elliptic | \(\text {Expression too large to display}\) | \(1046\) |
1/3*B*x^2/b*(b*x^3+a)^(1/2)*e/(e*x)^(1/2)+1/3*(2*A*b-B*a)/b/(b*e)^(1/2)*ar ctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))*e*((b*x^3+a)*e*x)^(1/2)/(e*x) ^(1/2)/(b*x^3+a)^(1/2)
Time = 0.54 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.22 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\left [\frac {4 \, \sqrt {b x^{3} + a} \sqrt {e x} B x - {\left (B a - 2 \, A b\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right )}{12 \, b}, \frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} B x + {\left (B a - 2 \, A b\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right )}{6 \, b}\right ] \]
[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]
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (73) = 146\).
Time = 1.51 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\begin {cases} \frac {2 \left (\begin {cases} \text {NaN} & \text {for}\: e^{3} = 0 \\\frac {\begin {cases} \frac {B e^{3} \left (e x\right )^{\frac {3}{2}} \sqrt {a + b x^{3}}}{2 b} + \left (A e^{3} - \frac {B a e^{3}}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {A e^{3} \left (e x\right )^{\frac {3}{2}} + \frac {B \left (e x\right )^{\frac {9}{2}}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases}}{3 e^{3}} & \text {otherwise} \end {cases}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((2*Piecewise((nan, Eq(e**3, 0)), (Piecewise((B*e**3*(e*x)**(3/2) *sqrt(a + b*x**3)/(2*b) + (A*e**3 - B*a*e**3/(2*b))*Piecewise((log(2*b*(e* x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True)), Ne(b/e**3, 0)), ((A *e**3*(e*x)**(3/2) + B*(e*x)**(9/2)/3)/sqrt(a), True))/(3*e**3), True))/e, Ne(e, 0)), (0, True))
\[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \sqrt {e x}}{\sqrt {b x^{3} + a}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} B x}{3 \, b {\left | e \right |}^{2}} + \frac {{\left (B a e^{5} - 2 \, A b e^{5}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{3 \, \sqrt {b e} b {\left | e \right |}^{4}} \]
1/3*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*B*x/(b*abs(e)^2) + 1/3*(B*a*e^5 - 2* A*b*e^5)*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a*e^4)))/(sqr t(b*e)*b*abs(e)^4)
Timed out. \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\int \frac {\left (B\,x^3+A\right )\,\sqrt {e\,x}}{\sqrt {b\,x^3+a}} \,d x \]