Integrand size = 22, antiderivative size = 73 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{9/2}}{9 b}-\frac {2 \sqrt {a} (A b-a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}} \] Output:
2/3*(A*b-B*a)*x^(3/2)/b^2+2/9*B*x^(9/2)/b-2/3*a^(1/2)*(A*b-B*a)*arctan(b^( 1/2)*x^(3/2)/a^(1/2))/b^(5/2)
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 x^{3/2} \left (3 A b-3 a B+b B x^3\right )}{9 b^2}+\frac {2 \sqrt {a} (-A b+a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}} \] Input:
Integrate[(x^(7/2)*(A + B*x^3))/(a + b*x^3),x]
Output:
(2*x^(3/2)*(3*A*b - 3*a*B + b*B*x^3))/(9*b^2) + (2*Sqrt[a]*(-(A*b) + a*B)* ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(3*b^(5/2))
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {959, 843, 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.
\(\displaystyle \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(A b-a B) \int \frac {x^{7/2}}{b x^3+a}dx}{b}+\frac {2 B x^{9/2}}{9 b}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{b x^3+a}dx}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {2 a \int \frac {x}{b x^3+a}d\sqrt {x}}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {2 a \int \frac {1}{a+b x}dx^{3/2}}{3 b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{3/2}}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
Input:
Int[(x^(7/2)*(A + B*x^3))/(a + b*x^3),x]
Output:
(2*B*x^(9/2))/(9*b) + ((A*b - a*B)*((2*x^(3/2))/(3*b) - (2*Sqrt[a]*ArcTan[ (Sqrt[b]*x^(3/2))/Sqrt[a]])/(3*b^(3/2))))/b
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] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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 = 0.78 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {2 x^{\frac {3}{2}} \left (b B \,x^{3}+3 A b -3 B a \right )}{9 b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b^{2} \sqrt {a b}}\) | \(55\) |
derivativedivides | \(\frac {\frac {2 b B \,x^{\frac {9}{2}}}{9}+\frac {2 A b \,x^{\frac {3}{2}}}{3}-\frac {2 B a \,x^{\frac {3}{2}}}{3}}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b^{2} \sqrt {a b}}\) | \(58\) |
default | \(\frac {\frac {2 b B \,x^{\frac {9}{2}}}{9}+\frac {2 A b \,x^{\frac {3}{2}}}{3}-\frac {2 B a \,x^{\frac {3}{2}}}{3}}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 b^{2} \sqrt {a b}}\) | \(58\) |
Input:
int(x^(7/2)*(B*x^3+A)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/9*x^(3/2)*(B*b*x^3+3*A*b-3*B*a)/b^2-2/3*a*(A*b-B*a)/b^2/(a*b)^(1/2)*arct an(b*x^(3/2)/(a*b)^(1/2))
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.96 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{3} - 2 \, b x^{\frac {3}{2}} \sqrt {-\frac {a}{b}} - a}{b x^{3} + a}\right ) - 2 \, {\left (B b x^{4} - 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{9 \, b^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{\frac {3}{2}} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (B b x^{4} - 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}\right )}}{9 \, b^{2}}\right ] \] Input:
integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="fricas")
Output:
[-1/9*(3*(B*a - A*b)*sqrt(-a/b)*log((b*x^3 - 2*b*x^(3/2)*sqrt(-a/b) - a)/( b*x^3 + a)) - 2*(B*b*x^4 - 3*(B*a - A*b)*x)*sqrt(x))/b^2, 2/9*(3*(B*a - A* b)*sqrt(a/b)*arctan(b*x^(3/2)*sqrt(a/b)/a) + (B*b*x^4 - 3*(B*a - A*b)*x)*s qrt(x))/b^2]
Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (70) = 140\).
Time = 66.77 (sec) , antiderivative size = 428, normalized size of antiderivative = 5.86 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {9}{2}}}{9}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {9}{2}}}{9} + \frac {2 B x^{\frac {15}{2}}}{15}}{a} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: a = 0 \\- \frac {A a \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {A a \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {A a \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} - \frac {A a \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 A x^{\frac {3}{2}}}{3 b} + \frac {B a^{2} \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{3} \sqrt {- \frac {a}{b}}} - \frac {B a^{2} \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 b^{3} \sqrt {- \frac {a}{b}}} - \frac {B a^{2} \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{3} \sqrt {- \frac {a}{b}}} + \frac {B a^{2} \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 B a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 B x^{\frac {9}{2}}}{9 b} & \text {otherwise} \end {cases} \] Input:
integrate(x**(7/2)*(B*x**3+A)/(b*x**3+a),x)
Output:
Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(9/2)/9), Eq(a, 0) & Eq(b, 0)), (( 2*A*x**(9/2)/9 + 2*B*x**(15/2)/15)/a, Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x* *(9/2)/9)/b, Eq(a, 0)), (-A*a*log(sqrt(x) - (-a/b)**(1/6))/(3*b**2*sqrt(-a /b)) + A*a*log(sqrt(x) + (-a/b)**(1/6))/(3*b**2*sqrt(-a/b)) + A*a*log(-4*s qrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b**2*sqrt(-a/b)) - A*a*lo g(4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b**2*sqrt(-a/b)) + 2 *A*x**(3/2)/(3*b) + B*a**2*log(sqrt(x) - (-a/b)**(1/6))/(3*b**3*sqrt(-a/b) ) - B*a**2*log(sqrt(x) + (-a/b)**(1/6))/(3*b**3*sqrt(-a/b)) - B*a**2*log(- 4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b**3*sqrt(-a/b)) + B*a **2*log(4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b**3*sqrt(-a/b )) - 2*B*a*x**(3/2)/(3*b**2) + 2*B*x**(9/2)/(9*b), True))
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} + \frac {2 \, {\left (B b x^{\frac {9}{2}} - 3 \, {\left (B a - A b\right )} x^{\frac {3}{2}}\right )}}{9 \, b^{2}} \] Input:
integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="maxima")
Output:
2/3*(B*a^2 - A*a*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b^2) + 2/9*(B*b *x^(9/2) - 3*(B*a - A*b)*x^(3/2))/b^2
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} + \frac {2 \, {\left (B b^{2} x^{\frac {9}{2}} - 3 \, B a b x^{\frac {3}{2}} + 3 \, A b^{2} x^{\frac {3}{2}}\right )}}{9 \, b^{3}} \] Input:
integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a),x, algorithm="giac")
Output:
2/3*(B*a^2 - A*a*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b^2) + 2/9*(B*b ^2*x^(9/2) - 3*B*a*b*x^(3/2) + 3*A*b^2*x^(3/2))/b^3
Time = 0.78 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.52 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{9/2}}{9\,b}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {72\,b^{3/2}\,x^{3/2}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{\sqrt {a}\,\left (72\,A\,a^2\,b^2-72\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}\right )\,\left (A\,b-B\,a\right )}{3\,b^{5/2}} \] Input:
int((x^(7/2)*(A + B*x^3))/(a + b*x^3),x)
Output:
x^(3/2)*((2*A)/(3*b) - (2*B*a)/(3*b^2)) + (2*B*x^(9/2))/(9*b) - (2*a^(1/2) *atan((72*b^(3/2)*x^(3/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/(a^(1/2)* (72*A*a^2*b^2 - 72*B*a^3*b)*(A*b - B*a)))*(A*b - B*a))/(3*b^(5/2))
Time = 0.20 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.10 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {2 \sqrt {x}\, x^{4}}{9} \] Input:
int(x^(7/2)*(B*x^3+A)/(b*x^3+a),x)
Output:
(2*sqrt(x)*x**4)/9