Integrand size = 22, antiderivative size = 53 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 B x^{3/2}}{3 b}+\frac {2 (A b-a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{3/2}} \] Output:
2/3*B*x^(3/2)/b+2/3*(A*b-B*a)*arctan(b^(1/2)*x^(3/2)/a^(1/2))/a^(1/2)/b^(3 /2)
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 B x^{3/2}}{3 b}-\frac {2 (-A b+a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{3/2}} \] Input:
Integrate[(Sqrt[x]*(A + B*x^3))/(a + b*x^3),x]
Output:
(2*B*x^(3/2))/(3*b) - (2*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]]) /(3*Sqrt[a]*b^(3/2))
Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {959, 851, 807, 218}
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.
\(\displaystyle \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {x}}{b x^3+a}dx}{b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {2 (A b-a B) \int \frac {x}{b x^3+a}d\sqrt {x}}{b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2 (A b-a B) \int \frac {1}{a+b x}dx^{3/2}}{3 b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 (A b-a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{3/2}}+\frac {2 B x^{3/2}}{3 b}\) |
Input:
Int[(Sqrt[x]*(A + B*x^3))/(a + b*x^3),x]
Output:
(2*B*x^(3/2))/(3*b) + (2*(A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(3 *Sqrt[a]*b^(3/2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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 = 0.77 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {2 B \,x^{\frac {3}{2}}}{3 b}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b \sqrt {a b}}\) | \(40\) |
default | \(\frac {2 B \,x^{\frac {3}{2}}}{3 b}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b \sqrt {a b}}\) | \(40\) |
risch | \(\frac {2 B \,x^{\frac {3}{2}}}{3 b}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b \sqrt {a b}}\) | \(40\) |
Input:
int(x^(1/2)*(B*x^3+A)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/3*B*x^(3/2)/b+2/3*(A*b-B*a)/b/(a*b)^(1/2)*arctan(b*x^(3/2)/(a*b)^(1/2))
Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\left [\frac {2 \, B a b x^{\frac {3}{2}} + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right )}{3 \, a b^{2}}, \frac {2 \, {\left (B a b x^{\frac {3}{2}} - {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right )\right )}}{3 \, a b^{2}}\right ] \] Input:
integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="fricas")
Output:
[1/3*(2*B*a*b*x^(3/2) + (B*a - A*b)*sqrt(-a*b)*log((b*x^3 - 2*sqrt(-a*b)*x ^(3/2) - a)/(b*x^3 + a)))/(a*b^2), 2/3*(B*a*b*x^(3/2) - (B*a - A*b)*sqrt(a *b)*arctan(sqrt(a*b)*x^(3/2)/a))/(a*b^2)]
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (51) = 102\).
Time = 4.92 (sec) , antiderivative size = 381, normalized size of antiderivative = 7.19 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + \frac {2 B x^{\frac {3}{2}}}{3}}{b} & \text {for}\: a = 0 \\\frac {A \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} - \frac {A \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} - \frac {A \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} + \frac {A \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} - \frac {B a \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {B a \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {B a \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} - \frac {B a \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 B x^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases} \] Input:
integrate(x**(1/2)*(B*x**3+A)/(b*x**3+a),x)
Output:
Piecewise((zoo*(-2*A/(3*x**(3/2)) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(9/2)/9)/a, Eq(b, 0)), ((-2*A/(3*x**(3/2)) + 2* B*x**(3/2)/3)/b, Eq(a, 0)), (A*log(sqrt(x) - (-a/b)**(1/6))/(3*b*sqrt(-a/b )) - A*log(sqrt(x) + (-a/b)**(1/6))/(3*b*sqrt(-a/b)) - A*log(-4*sqrt(x)*(- a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b*sqrt(-a/b)) + A*log(4*sqrt(x)*(- a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b*sqrt(-a/b)) - B*a*log(sqrt(x) - (-a/b)**(1/6))/(3*b**2*sqrt(-a/b)) + B*a*log(sqrt(x) + (-a/b)**(1/6))/(3*b **2*sqrt(-a/b)) + B*a*log(-4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3) )/(3*b**2*sqrt(-a/b)) - B*a*log(4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)** (1/3))/(3*b**2*sqrt(-a/b)) + 2*B*x**(3/2)/(3*b), True))
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \, B x^{\frac {3}{2}}}{3 \, b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} \] Input:
integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="maxima")
Output:
2/3*B*x^(3/2)/b - 2/3*(B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b )
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \, B x^{\frac {3}{2}}}{3 \, b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} \] Input:
integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="giac")
Output:
2/3*B*x^(3/2)/b - 2/3*(B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b )
Time = 0.80 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2\,B\,x^{3/2}}{3\,b}-\frac {2\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,x^{3/2}\,\left (24\,A^2\,b^3-48\,A\,B\,a\,b^2+24\,B^2\,a^2\,b\right )}{\left (72\,B\,a^2\,b^2-72\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}\right )\,\left (A\,b-B\,a\right )}{3\,\sqrt {a}\,b^{3/2}} \] Input:
int((x^(1/2)*(A + B*x^3))/(a + b*x^3),x)
Output:
(2*B*x^(3/2))/(3*b) - (2*atan((3*a^(1/2)*b^(3/2)*x^(3/2)*(24*A^2*b^3 + 24* B^2*a^2*b - 48*A*B*a*b^2))/((72*B*a^2*b^2 - 72*A*a*b^3)*(A*b - B*a)))*(A*b - B*a))/(3*a^(1/2)*b^(3/2))
Time = 0.21 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \sqrt {x}\, x}{3} \] Input:
int(x^(1/2)*(B*x^3+A)/(b*x^3+a),x)
Output:
(2*sqrt(x)*x)/3